$(W^{G_{L}})^{\hat{G}\text{$\mbox{-}\mathrm{la}$},\nabla_{\gamma}=0}$ is of
dimension $d$ over $L$, this is achieved via a monodromy descent. In Step 2,
via an (easy) étale descent, we further show the ($\gamma=1$)-invariant
$D_{{\mathrm{Sen}},{K_{\infty}}}(W)$ has dimension $d$.

Step 1 (monodromy descent). We claim the following:

  * •

Let $M$ be a $\hat{L}^{\hat{G}\text{$\mbox{-}\mathrm{la}$}}$-vector space of
dimension $d$ with a semi-linear and locally analytic $\hat{G}$-action. Then
the subspace $M^{\nabla_{\gamma}=0}$ is a $L$-vector space of dimension $d$
such that

(7.11)
$M^{\nabla_{\gamma}=0}\otimes_{L}\hat{L}^{\hat{G}\text{$\mbox{-}\mathrm{la}$}}=M$

This is a “$\theta$-specialization” of the argument [GP21, Rem. 6.1.7], hence
the proof is practically verbatim. To proceed, as in Prop. 7.11, we denote
$\beta=\theta(\mathfrak{t})$. There, we also made use of the operator
$\partial_{\gamma}=\frac{1}{\beta}\nabla_{\gamma}$, which is precisely
$\theta$-specialization of the operator (with same notation) in [GP21, 5.3.4].
Choose a basis of $M$, and let $D_{\gamma}=\mathrm{Mat}(\partial_{\gamma})$,
then it suffices to show that there exists
$H\in{\mathrm{GL}}_{d}(\hat{L}^{\hat{G}\text{$\mbox{-}\mathrm{la}$}})$ such
that

(7.12) $\partial_{\gamma}(H)+D_{\gamma}H=0$

For $k\in\mathbb N$, let $D_{k}=\mathrm{Mat}(\partial_{\gamma}^{k})$. For $n$
large enough, the series given by

$H=\sum_{k\geq 0}(-1)^{k}D_{k}\frac{(\beta-\beta_{n})^{k}}{k!}$

converges to the desired solution of (7.12). Here, $\beta-\beta_{n}$ is used
as a “variable” just as in the proof of Prop. 7.11.

Step 2 (etale descent). By [GP21, Prop. 3.1.6], we know

(7.13)
$(W^{G_{L}})^{\hat{G}\text{$\mbox{-}\mathrm{la}$}}=D_{{\mathrm{Sen}},{K_{p^{\infty}}}}(W)\otimes_{K_{p^{\infty}}}(\hat{L})^{\hat{G}\text{$\mbox{-}\mathrm{la}$}}.$

Apply Step 1 to the above vector space, and so

$X:=(W^{G_{L}})^{\hat{G}\text{$\mbox{-}\mathrm{la}$},\nabla_{\gamma}=0}$

is a vector space over $L$ of dimension $d$. In addition, $X$ is stable under
${\mathrm{Gal}}(L/{K_{\infty}})$-action as this action commutes with
$\nabla_{\gamma}$. (Note however $\tau$-action does not commute with
$\nabla_{\gamma}$, not even on the ring level: for example
$\tau\nabla_{\gamma}(u)=0\neq\nabla_{\gamma}\tau(u)$.) In summary, $X$ is a
$L$-vector space with a ${\mathrm{Gal}}(L/{K_{\infty}})$-action. Note
$L=\cup_{n}K(\pi_{n},\mu_{n})$ and ${\mathrm{Gal}}(L/{K_{\infty}})$ is
topologically finitely generated. Thus, for $n\gg 0$, $X$ descends to some
${\mathrm{Gal}}(L/{K_{\infty}})$-stable vector space $X_{n}$ over
$K(\pi_{n},\mu_{n})$. By Galois descent, $X_{n}^{\gamma=1}$ is a
$K(\pi_{n})$-vector space of dimension $d$, and hence
$X_{n}^{\gamma=1}\otimes_{K(\pi_{n})}{K_{\infty}}$ is precisely the desired
$D_{{\mathrm{Sen}},{K_{\infty}}}(W)$. Finally, apply [GP21, Prop. 3.1.6]
again, then we have

(7.14)
$(W^{G_{L}})^{\hat{G}\text{$\mbox{-}\mathrm{la}$}}=D_{{\mathrm{Sen}},{K_{\infty}}}(W)\otimes_{K_{\infty}}(\hat{L})^{\hat{G}\text{$\mbox{-}\mathrm{la}$}},$

which together with (7.13) proves (7.10). ∎

Let $W\in{\mathrm{Rep}}_{G_{K}}(C)$. Since
$D_{{\mathrm{Sen}},{K_{\infty}}}(W)$ are locally analytic vectors, we can
define (cf. Def. 7.5)

(7.15)
$N_{\nabla}:D_{{\mathrm{Sen}},{K_{\infty}}}(W)\rightarrow(W^{G_{L}})^{\hat{G}\text{$\mbox{-}\mathrm{la}$}}$

######  Theorem 7.13.

Let $W\in{\mathrm{Rep}}_{G_{K}}(C)$, then Eqn. (7.15), after linear scaling,
induces a ${K_{\infty}}$-linear operator, which we call the _Sen operator over
the Kummer tower_

(7.16) $\frac{1}{\theta(u\lambda^{\prime})}\cdot
N_{\nabla}:D_{{\mathrm{Sen}},{K_{\infty}}}(W)\rightarrow
D_{{\mathrm{Sen}},{K_{\infty}}}(W).$

(We also sometimes use the simplified terminology “${K_{\infty}}$-Sen
operator”). Extend it $C$-linearly to a $C$-linear operator on
$D_{{\mathrm{Sen}},{K_{\infty}}}(W)\otimes_{K_{\infty}}C=W$, and denote it by
the same notation:

(7.17) $\frac{1}{\theta(u\lambda^{\prime})}\cdot N_{\nabla}:W\rightarrow W$

Then this is precisely the (uniquely defined) _Sen operator_ in Eqn. (7.3).

###### Proof.

Note we have the relation $gN_{\nabla}=N_{\nabla}g,\forall
g\in{\mathrm{Gal}}(L/{K_{\infty}})$, hence $N_{\nabla}$ stabilizes
$D_{{\mathrm{Sen}},{K_{\infty}}}(W)$. Furthermore,
$\theta(u\lambda^{\prime})\in{K_{\infty}}$, hence
$\frac{1}{\theta(u\lambda^{\prime})}\cdot N_{\nabla}$ still stabilizes
$D_{{\mathrm{Sen}},{K_{\infty}}}(W)$. Thus (7.16) is well-defined. This is a
${K_{\infty}}$-linear operator because $\nabla_{\tau}$ hence
$\frac{1}{\theta(u\lambda^{\prime})}\cdot N_{\nabla}$ kills ${K_{\infty}}$.

By Cor. 7.7,
$\mathfrak{a}=\nabla_{\gamma}+\frac{1}{\theta(u\lambda^{\prime})}N_{\nabla}$
is an $(\hat{L})^{\hat{G}\text{$\mbox{-}\mathrm{la}$}}$-_linear_ operator on
both sides of the following:

$D_{{\mathrm{Sen}},{K_{\infty}}}(W)\otimes_{K_{\infty}}(\hat{L})^{\hat{G}\text{$\mbox{-}\mathrm{la}$}}=D_{{\mathrm{Sen}},{K_{p^{\infty}}}}(W)\otimes_{K_{p^{\infty}}}(\hat{L})^{\hat{G}\text{$\mbox{-}\mathrm{la}$}}.$

On the right hand side, $N_{\nabla}$ kills
$D_{{\mathrm{Sen}},{K_{p^{\infty}}}}(W)$ and hence
$\mathfrak{a}=\nabla_{\gamma}$ is precisely the
$(\hat{L})^{\hat{G}\text{$\mbox{-}\mathrm{la}$}}$-linear extension of the Sen
operator in Eqn. (7.2). Similarly, on the left hand side, $\nabla_{\gamma}$
kills $D_{{\mathrm{Sen}},{K_{\infty}}}(W)$, and hence $\mathfrak{a}$ is the
same as the $(\hat{L})^{\hat{G}\text{$\mbox{-}\mathrm{la}$}}$-linear extension
of (7.16). We can conclude by further extending $C$-linearly. ∎

######  Corollary 7.14.

The two operators $\frac{1}{\theta(u\lambda^{\prime})}\cdot N_{\nabla}$ in
Eqn. (7.16) and $\nabla_{\gamma}$ in Eqn. (7.2) have the same eigenvalues, and
have the same semi-simplicity property.

######  Definition 7.15.

Let $F$ be a field. Write $\mathrm{End}_{F}$ for the category consisting of
$(M,f)$ where $M$ is a finite dimensional $F$-vector space and $f:M\rightarrow
M$ is an $F$-linear endomorphism.

######  Proposition 7.16.

Let $W_{1},W_{2}\in{\mathrm{Rep}}_{G_{K}}(C)$. By abuse of notation, we use
$\phi_{\mathrm{Sen}}$ to denote the Sen operators on
$D_{{\mathrm{Sen}},{K_{p^{\infty}}}}(W_{i})$ as well as on $W_{i}$, cf.
Notation 7.3. Similarly, we use $\phi_{{K_{\infty}}-\mathrm{Sen}}$ to denote
the ${K_{\infty}}$-Sen operators on $D_{{\mathrm{Sen}},{K_{\infty}}}(W_{i})$,
cf. Thm. 7.13. The following are equivalent.

  1. (1)

$W_{1}$ and $W_{2}$ are isomorphic as objects in ${\mathrm{Rep}}_{G_{K}}(C)$.

  2. (2)

$(D_{{\mathrm{Sen}},{K_{p^{\infty}}}}(W_{1}),\phi_{\mathrm{Sen}})$ and
$(D_{{\mathrm{Sen}},{K_{p^{\infty}}}}(W_{2}),\phi_{\mathrm{Sen}})$ are
isomorphic as objects in $\mathrm{End}_{K_{p^{\infty}}}$.

  3. (3)

$(W_{1},\phi_{\mathrm{Sen}})$ and $(W_{2},\phi_{\mathrm{Sen}})$ are isomorphic
as objects in $\mathrm{End}_{C}$.

  4. (4)

$(D_{{\mathrm{Sen}},{K_{\infty}}}(W_{1}),\phi_{{K_{\infty}}-\mathrm{Sen}})$
and
$(D_{{\mathrm{Sen}},{K_{\infty}}}(W_{2}),\phi_{{K_{\infty}}-\mathrm{Sen}})$
are isomorphic as objects in $\mathrm{End}_{K_{\infty}}$.

###### Proof.

Clearly, (1) implies (2)-(4). (2) obviously implies (3); (4) implies (3) by
Thm. 7.13. The implication from (3) to (1) is proved in [Sen81, p. 101, Thm.
7]. ∎

We record a cohomology comparison theorem in Sen theory. Recall for
$W\in{\mathrm{Rep}}_{G_{K}}(C)$, we have the classical Sen module
$D_{{\mathrm{Sen}},{K_{p^{\infty}}}}(W)$ together with Sen operator
$\nabla_{\gamma}$ in Notation 7.3. We also have ${K_{\infty}}$-Sen module
$D_{{\mathrm{Sen}},{K_{\infty}}}(W)$ together with ${K_{\infty}}$-Sen operator
$\phi_{{\mathrm{Sen}},{K_{\infty}}}=\frac{1}{\theta(u\lambda^{\prime})}\cdot
N_{\nabla}$.

######  Theorem 7.17.

Let $W\in{\mathrm{Rep}}_{G_{K}}(C)$. We have a diagram of quasi-isomorphisms
between (continuous) group cohomologies and Lie algebra cohomologies:

(7.18)
${{\mathrm{R}\Gamma(G_{K},W)}}$${{\mathrm{R}\Gamma(\hat{G},W^{G_{L}})}}$${{\mathrm{R}\Gamma({\Gamma_{K}},W^{G_{K_{p^{\infty}}}})}}$${{\mathrm{R}\Gamma(\hat{G},(W^{G_{L}})^{\hat{G}\text{$\mbox{-}\mathrm{la}$}})}}$${{\mathrm{R}\Gamma({\Gamma_{K}},D_{{\mathrm{Sen}},{K_{p^{\infty}}}}(W))}}$${{(\mathrm{R}\Gamma(\mathrm{Lie}\hat{G},(W^{G_{L}})^{\hat{G}\text{$\mbox{-}\mathrm{la}$}}))^{\hat{G}}}}$${{(\mathrm{R}\Gamma(\mathrm{Lie}{\Gamma_{K}},D_{{\mathrm{Sen}},{K_{p^{\infty}}}}(W)))^{{\Gamma_{K}}}}}$$\scriptstyle{\simeq}$$\scriptstyle{\simeq}$$\scriptstyle{\simeq}$$\scriptstyle{\simeq}$$\scriptstyle{\simeq}$$\scriptstyle{\simeq}$$\scriptstyle{\simeq}$$\scriptstyle{\simeq}$

In addition, we have quasi-isomorphism

(7.19) $\displaystyle\mathrm{R}\Gamma(G_{K},W)\otimes_{K}{K_{p^{\infty}}}$
$\displaystyle\simeq[D_{{\mathrm{Sen}},{K_{p^{\infty}}}}(W)\xrightarrow{\nabla_{\gamma}}D_{{\mathrm{Sen}},{K_{p^{\infty}}}}(W)]$
(7.20) $\displaystyle\mathrm{R}\Gamma(G_{K},W)\otimes_{K}L$
$\displaystyle\simeq\mathrm{R}\Gamma(\mathrm{Lie}\hat{G},(W^{G_{L}})^{\hat{G}\text{$\mbox{-}\mathrm{la}$}})$
(7.21) $\displaystyle\mathrm{R}\Gamma(G_{K},W)\otimes_{K}{K_{\infty}}$
$\displaystyle\simeq[D_{{\mathrm{Sen}},{K_{\infty}}}(W)\xrightarrow{\phi_{{\mathrm{Sen}},{K_{\infty}}}}D_{{\mathrm{Sen}},{K_{\infty}}}(W)]$

###### Proof.

All the quasi-isomorphism in diagram (7.18) (as well as (7.19)) should be
well-known to experts; but we cannot locate an explicit reference.
Fortunately, these all follow from some recent general formalism developed by
[Por22] and [RJRC22].

Note objects in ${\mathrm{Rep}}_{G_{K}}(C)$ satisfy the axioms (TS1), (TS2),
(TS3), (TS4) in [Por22, §5.1], cf. the examples below [Por22, Cor. 5.4]; here
(TS1)-(TS3) are the usual Tate-Sen axioms introduced by Berger-Colmez, and the
extra (TS4) ensures that $C$-representations have no “ _higher locally
analytic vectors_ ”, cf. _loc. cit._ for detailed discussions. Thus all the
quasi-isomorphisms in diagram (7.18) follow from [RJRC22, Thm. 1.5, Thm. 1.7]
and [Por22, Thm. 5.1] (as already observed in [Por22, Cor. 5.4]). (We are also
informed by Rodríguez Camargo that results and arguments in this paragraph are
established in much broader generality in [RC22]).

Finally, (7.19) and (7.20) follow from standard Galois descent; and (7.21)
follows by taking ${\mathrm{Gal}}(L/{K_{\infty}})$-invariant of (7.20). ∎

##  8\. Hodge–Tate crystals and (log-) nearly Hodge–Tate representations

In this section, we classify rational Hodge–Tate crystals by (log-) nearly
Hodge–Tate representations, and compare their cohomologies. The main idea is
to show the “small endomorphism” corresponding to a Hodge–Tate crystal, after
some normalization, matches with the ${K_{\infty}}$-Sen operator constructed
in §7.

######  Notation 8.1.

Let $\ast\in\\{\emptyset,\mathrm{log}\\}$. Let
$\mathbb{M}\in{\mathrm{Vect}}(({\mathcal{O}}_{K})_{{{\mathbbl{\Delta}}},\ast},{\overline{{\mathcal{O}}}_{{\mathbbl{\Delta}}}}[1/p]).$
Consider the evaluations

$M={\mathbb M}(({\mathfrak{S}},E,\ast))$

In addition, one can associate a small endomorphism $(M,\phi_{M})$ by Thm.
4.2. Consider also

$W={\mathbb M}(({\mathbf{A}_{\mathrm{inf}}},(\xi),\ast)).$

We have $W\in{\mathrm{Rep}}_{G_{K}}(C)$. By Thm. 7.13, we can associate

$(D_{{\mathrm{Sen}},{K_{\infty}}}(W),\phi_{K_{\infty}})\in{\mathrm{End}}_{K_{\infty}}.$

We shall consider $M$ as a subspace of $W$ via the identification

$M\otimes_{K}C\xrightarrow{\simeq}W$

induced by the morphism

$({\mathfrak{S}},E,\ast)\rightarrow({\mathbf{A}_{\mathrm{inf}}},(\xi),\ast)$

Finally recall we use

$a=\begin{cases}-E^{\prime}(\pi),&\text{if }\ast=\emptyset\\\ -\pi
E^{\prime}(\pi),&\text{if }\ast=\mathrm{log}\end{cases}$

######  Theorem 8.2.

Use Notation 8.1.

  1. (1)

Via the identification $M\otimes_{K}C=W$, we have $M\subset
D_{{\mathrm{Sen}},{K_{\infty}}}(W)$; furthermore, this inclusion induces an
isomorphism

(8.1) $M\otimes_{K}{K_{\infty}}\simeq D_{{\mathrm{Sen}},{K_{\infty}}}(W)$

  2. (2)

Scaling $\phi_{M}:M\rightarrow M$ by $\frac{1}{a}$ and extending
${K_{\infty}}$-linearly gives rise to

$\frac{\phi_{M}}{a}:M\otimes_{K}{K_{\infty}}\rightarrow
M\otimes_{K}{K_{\infty}};$

then this is the _same_ (via (8.1)) as the ${K_{\infty}}$-Sen operator

$\phi_{K_{\infty}}=\frac{1}{\theta(u\lambda^{\prime})}\cdot
N_{\nabla}:D_{{\mathrm{Sen}},{K_{\infty}}}(W)\rightarrow
D_{{\mathrm{Sen}},{K_{\infty}}}(W)$

defined in Thm. 7.13.

###### Proof.

Item (1). Let $(\underline{e})$ be a basis of $M$. The formula (5.3) in Prop.
5.10 implies that ${\mathrm{Gal}}(L/{K_{\infty}})$ acts trivially on $M$. In
addition, that formula implies that

(8.2) $\tau^{i}(\underline{e})=(\underline{e})\sum_{n\geq
0}\left(\left(\prod_{i=0}^{n-1}(A_{1}-ia)\right)\cdot\left(c(\tau^{i})\frac{\pi
E^{\prime}(\pi)}{a}\theta(\frac{\xi}{E})(\mu_{1}-1)\right)^{[n]}\right).$

where $A_{1}$ is the matrix of $\phi_{M}$ with respect to $\underline{e}$, and

$c(\tau^{i})=i,\text{ resp. }2i,\text{ when
}{K_{\infty}}\cap{K_{p^{\infty}}}=K,\text{ resp. }K(\pi_{1}).$

This implies that elements of $M$ are $\tau$-locally analytic. In summary,

$M\subset W^{\gamma=1,\tau\text{$\mbox{-}\mathrm{la}$}}$

Thus [GP21, Prop. 3.1.6] implies that

$D_{{\mathrm{Sen}},{K_{\infty}}}=M\otimes_{K}{K_{\infty}}.$

Item (2). The formula (8.2) can be re-written as

(8.3) $\tau^{i}(\underline{e})=(\underline{e})\left(1+c(\tau^{i})\pi
E^{\prime}(\pi)\theta(\frac{\xi}{E})(\mu_{1}-1)\right)^{\frac{A_{1}}{a}}$

This implies that

$\nabla_{\tau}(\underline{e})=(\underline{e})\frac{\pi
E^{\prime}(\pi)}{a}\theta(\frac{\xi}{E})(\mu_{1}-1)A_{1},\quad\text{ resp.
}(\underline{e})2\frac{\pi
E^{\prime}(\pi)}{a}\theta(\frac{\xi}{E})(\mu_{1}-1)A_{1},$

when ${K_{\infty}}\cap{K_{p^{\infty}}}=K,\text{ resp. }K(\pi_{1})$. This
implies our ${K_{\infty}}$-Sen operator acts via

(8.4) $\frac{1}{\theta(u\lambda^{\prime})}\cdot
N_{\nabla}(\underline{e})=(\underline{e})\frac{\pi
E^{\prime}(\pi)\theta(\frac{\xi}{E})(\mu_{1}-1)A_{1}}{pa\theta(u\lambda^{\prime}{\mathfrak{t}})}=(\underline{e})\frac{\pi
E^{\prime}(\pi)\theta(\frac{\xi}{E})(\mu_{1}-1)}{p\theta(u\lambda^{\prime}{\mathfrak{t}})}\cdot\frac{A_{1}}{a}$

To finish the proof of this theorem, it suffices to check the “scalar term” is
$1$, namely:

$\frac{\pi
E^{\prime}(\pi)\theta(\frac{\xi}{E})(\mu_{1}-1)}{p\theta(u\lambda^{\prime}{\mathfrak{t}})}=1$

Using $\theta(\frac{\xi}{E})=\theta(\frac{\mu}{\varphi^{-1}(\mu)E})$ where
$\mu=[\epsilon]-1$, and $\theta(\varphi^{-1}(\mu))=\mu_{1}-1$, the above
identity simplifies as

$\theta(\frac{\mu E^{\prime}}{p\lambda^{\prime}{\mathfrak{t}}E})=1$

Note $\theta(\frac{t}{\mu})=1$ using $t=\mathrm{log}(\mu+1)$, it suffices to
check

(8.5) $\theta(\frac{tE^{\prime}}{p\lambda^{\prime}{\mathfrak{t}}E})=1$

Apply multiplication rule to $\lambda^{\prime}$, one sees that

$\theta(\lambda^{\prime})=\frac{E^{\prime}(\pi)}{E(0)}\theta(\varphi(\lambda))$

Hence (8.5) becomes

$\theta(\frac{t}{p\frac{E}{E(0)}\varphi(\lambda){\mathfrak{t}}})=1$

The denominator is
$p\frac{E}{E(0)}\varphi(\lambda){\mathfrak{t}}=p\lambda{\mathfrak{t}}=t$, thus
we can conclude. ∎

The following theorem (together with Thm. 5.7) completes the proof of all the
equivalences in Thm. 1.5(1). Recall
${\mathrm{Rep}}_{G_{K}}^{\ast-\mathrm{nHT}}(C)$ is defined in Def. 1.3.

######  Theorem 8.3.

The evaluation functor ${\mathbb M}\mapsto W:={\mathbb
M}(({\mathbf{A}_{\mathrm{inf}}},(\xi),\ast))$ induces a bi-exact equivalence
of categories:

${\mathrm{Vect}}(({\mathcal{O}}_{K})_{{{\mathbbl{\Delta}}},\ast},\overline{\mathcal{O}}_{{{\mathbbl{\Delta}}}}[{1}/{p}])\simeq{\mathrm{Rep}}_{G_{K}}^{\ast-\mathrm{nHT}}(C).$

###### Proof.

Let ${\mathbb
M}\in{\mathrm{Vect}}(({\mathcal{O}}_{K})_{{{\mathbbl{\Delta}}},\ast},\overline{\mathcal{O}}_{{{\mathbbl{\Delta}}}}[{1}/{p}])$,
we first show $W({\mathbb M})$ is $\ast$-nearly Hodge–Tate. Let $(M,\phi_{M})$
be the associated Sen module, then Thm. 8.2 implies $\frac{\phi_{M}}{a}$ is
the Sen operator; hence the condition $\prod_{i=0}^{n-1}(\phi_{M}-ai)$
converges to zero translates to the fact that the eigenvalues of
$\frac{\phi_{M}}{a}$ —i.e., the Sen weights of $W(\mathbb{M})$— are in the
range as in Def. 1.3. This shows $W({\mathbb M})$ is $\ast$-nearly Hodge–Tate,
and gives our desired functor

(8.6)
${\mathrm{Vect}}(({\mathcal{O}}_{K})_{{{\mathbbl{\Delta}}},\ast},\overline{\mathcal{O}}_{{{\mathbbl{\Delta}}}}[{1}/{p}])\rightarrow{\mathrm{Rep}}_{G_{K}}^{\ast-\mathrm{nHT}}(C).$

We now prove the functor is essentially surjective. For _any_
$W\in{\mathrm{Rep}}_{G_{K}}(C)$, let $\phi_{W}$ be the (classical) Sen
operator on $D_{{\mathrm{Sen}},{K_{p^{\infty}}}}(W)$. By [Sen81, p.100, Thm.
5], there exists a sub-$K$-vector space of full dimension

$N\subset D_{{\mathrm{Sen}},{K_{p^{\infty}}}}(W)$

which is stable under $\phi_{W}$. 222Note this sub-$K$-vector space is in
general not unique. For example, consider the trivial $C$-representation whose
Sen operator is the zero map, then _any_ sub-$K$-vector space is stable under
the zero map. If $W$ is in addition $\ast$\- nearly Hodge–Tate, then the
eigenvalues of $\phi_{W}|_{N}$ has to satisfy the condition in Def. 1.3, and
hence $(N,f=a\phi_{W})$ is an object in
$\mathrm{End}_{K}^{\ast-{\mathrm{nHT}}}$. Via Thm. 4.2, this gives rise to a
rational Hodge–Tate crystal $\mathbb{M}$. Let
$W^{\prime}=\mathbb{M}(({\mathbf{A}_{\mathrm{inf}}},(\xi),\ast))\in{\mathrm{Rep}}_{G_{K}}(C)$
be the associated representation, one needs to check

$W^{\prime}\simeq W.$

By Thm. 8.2,
$D_{{\mathrm{Sen}},{K_{\infty}}}(W^{\prime})=N\otimes_{K}{K_{\infty}}$, and
its ${K_{\infty}}$-Sen operator is precisely $\frac{f}{a}=\phi_{W}|_{N}\otimes
1$; hence in particular, by Thm. 7.13, the $C$-linear Sen operator on
$W^{\prime}=N\otimes_{K}C$ is the _same_ as that on $W=N\otimes_{K}C$. Thus
$W^{\prime}$ is isomorphic to $W$ by Prop. 7.16. (We point to Rem. 8.4 for a
possible confusion in the construction above.)

We now prove the functor is fully faithful. Let
$\mathbb{M}_{1},\mathbb{M}_{2}\in{\mathrm{Vect}}(({\mathcal{O}}_{K})_{{{\mathbbl{\Delta}}},\ast},\overline{\mathcal{O}}_{{{\mathbbl{\Delta}}}}[{1}/{p}])$,
let $(M_{1},\phi_{1}),(M_{2},\phi_{2})$ be the corresponding Sen modules, and
let $W_{1},W_{2}\in{\mathrm{Rep}}_{G_{K}}^{\ast-\mathrm{nHT}}(C)$ be the
corresponding representations. It suffices to show

$\mathrm{Morph}((M_{1},\phi_{1}),(M_{2},\phi_{2}))\rightarrow\mathrm{Morph}(W_{1},W_{2})$

is a bijection. It is injective because $M_{i}$ contains a basis of $W_{i}$;
it hence suffices to show both sides have the same $K$-dimension. Note that
(8.6) is a tensor functor and note $f_{i}$’s are (scaled) Sen operators. By
considering the $C$-representation $\mathrm{Hom}_{C}(W_{1},W_{2})$, it
suffices to show the following statement: given
$\mathbb{M}\in{\mathrm{Vect}}(({\mathcal{O}}_{K})_{{{\mathbbl{\Delta}}},\ast},\overline{\mathcal{O}}_{{{\mathbbl{\Delta}}}}[1/p])$
with corresponding $(M,\phi)$ and $W$, then we have

$M^{\phi=0}=W^{G_{K}};$

but this follows from [Sen81, p.100, Thm. 6] (which implies the above two
spaces are isomorphic after tensoring with $C$). ∎

######  Remark 8.4.

Let us point out a possible confusion in the proof of essential surjectivity
above. In the proof, we obtained a $C$-linear isomorphim $W\rightarrow
W^{\prime}$ (of two vectors spaces _without_ considering Galois actions),
which is further compatible with the $C$-linear Sen operators. Indeed, it
comes from the following composite:

(8.7) $W\simeq
N\otimes_{K}C\simeq\mathbb{M}(({\mathfrak{S}},E))\otimes_{K}C\simeq
W^{\prime}$

where the first isomorphism follows from Sen’s original construction, the
second comes from construction of $\mathbb{M}$, and the third comes from
definition of $W^{\prime}$. Note however, the “identity” map
$N\otimes_{K}C\simeq\mathbb{M}(({\mathfrak{S}},E))\otimes_{K}C$ is _not_
“$G_{K}$-equivariant”! (Hence, in particular, the composite (8.7) is _not_
$G_{K}$-equivariant!) Indeed, the $\tau$-action on $N$ is trivial under
classical Sen theory; whereas the $\tau$-action on
$\mathbb{M}(({\mathfrak{S}},E))$ is in general not trivial by Prop. 5.10. The
powerful aspect of Sen’s theorem [Sen81, p. 101, Thm. 7], is that the Sen-
operator-equivariant map (8.7) already guarantees existence of a
$G_{K}$-equivariant isomorphism between $W$ and $W^{\prime}$, although in an
_implicit_ way!

We prove the cohomology comparison in Thm. 1.5(2).

######  Theorem 8.5.

Let $\ast\in\\{\emptyset,\mathrm{log}\\}$. Let
$\mathbb{M}\in{\mathrm{Vect}}({(\mathcal{O}_{K})_{{{\mathbbl{\Delta}}},\ast}},\overline{\mathcal{O}}_{{{\mathbbl{\Delta}}}}[1/p])$
be a Hodge–Tate crystal, and let
$W\in{\mathrm{Rep}}_{G_{K}}^{\ast-{\mathrm{nHT}}}(C)$ be the associated
representation. There exists a natural quasi-isomorphism

$\mathrm{R}\Gamma({(\mathcal{O}_{K})_{{{\mathbbl{\Delta}}},\ast}},\mathbb{M})\simeq\mathrm{R}\Gamma({G_{K}},W)$

where the right hand side is Galois cohomology.

###### Proof.

Restriction to the perfect site induces natural morphisms

$\mathrm{R}\Gamma({(\mathcal{O}_{K})_{{{\mathbbl{\Delta}}},\ast}},\mathbb{M})\rightarrow\mathrm{R}\Gamma({(\mathcal{O}_{K})^{\mathrm{perf}}_{{{\mathbbl{\Delta}}},\ast}},\mathbb{M})\simeq\mathrm{R}\Gamma({G_{K}},W)$

where the second quasi-isomorphism follows from Lem. 5.5. It suffices to show
the composite is also a quasi-isomorphism. Let $(M,\phi_{M})$ be the
corresponding Sen module. By Thm. 4.5, we have

${\mathrm{R}}\Gamma(({\mathcal{O}}_{K})_{{{\mathbbl{\Delta}}},*},{\mathbb
M})\simeq[M\xrightarrow{\phi_{M}}M].$

Apply Thm. 8.2 and Thm. 7.17 respectively, we have

$[M\xrightarrow{\phi_{M}}M]\otimes_{K}{K_{\infty}}\simeq[D_{{\mathrm{Sen}},{K_{\infty}}}(W)\xrightarrow{\phi_{{\mathrm{Sen}},{K_{\infty}}}}D_{{\mathrm{Sen}},{K_{\infty}}}(W)]\simeq\mathrm{R}\Gamma(G_{K},W)\otimes_{K}{K_{\infty}}.$

We can conclude. ∎

###  8.1. Comparison with a theorem of Sen

In this subsection, we recall a (rather peculiar) theorem of Sen which also
concerns about “$K$-rationality” of the Sen operator, and make some
comparisons with our result.

######  Theorem 8.6.

[Sen81, p.110, Thm. 10] Suppose the residue field $k$ is _algebraically
closed_. Then there exists a (_non-canonical_) equivalence between
${\mathrm{Rep}}_{G_{K}}(C)$ and the category of $K$-vector spaces equipped
with a linear operator. This non-canonical equivalence, which depend on
choices for each simple object in ${\mathrm{Rep}}_{G_{K}}(C)$, is _never_
compatible with tensor products.

######  Construction 8.7.

We give a very brief sketch of the rather technical proof of Thm. 8.6,
pointing out where _non-canonicity_ comes from.

  1. (1)

Recall, as we already used in the proof of our Thm. 8.3, Sen shows in [Sen81,
p.100, Thm. 5] that for any $C$-representation, the Sen operator is
“$K$-rational”: that is, it is stable on a sub-$K$-vector space.

  2. (2)

Conversely, when the residue field $k$ is algebraically closed, [Sen81, p.104,
Thm. 9] shows that any $K$-matrix can be recovered as a Sen operator for some
$C$-representation. Note however, as stated in [Sen81, p.105, Thm. 9’], this
only constructs a _set bijection_ between $H^{1}(G_{K},{\mathrm{GL}}_{n}(C))$
and the set of similarity classes of $n\times n$ matrices over $K$. One still
needs to build some _functoriality_ into this bijection.

  3. (3)

Sen then _chooses_ , for each _simple_ $C$-representation $U$, a $K$-vector
space $U_{0}$ stable under Sen operator; this builds a functor for _semi-
simple_ objects. To get the complete functor, one first map a general
$C$-representation $W$ to its semi-simple part $W_{S}$, then one notes
$(W_{S})_{0}$ is still stable under the Sen operator of $W$ (not just that of
$W_{S}$!), since Sen operator is Galois equivariant.

######  Remark 8.8.

In the remark following [Fon04, Thm. 2.14], Fontaine mentions that Thm. 8.6
can also be obtained using his _classification_ of $C$-representations: this
also points to the non-canonicity of Thm. 8.6.

######  Remark 8.9.

We give some yet further remarks about the non-canonicity of Thm. 8.6, and
compare it with our Thm. 8.3. In $p$-adic Hodge theory, many typical theorems
say that there is an equivalence of categories:

$\mathrm{MOD}\simeq\mathrm{REP}$

where $\mathrm{MOD}$ is certain “module category” and $\mathrm{REP}$ is
certain “representation category”. We invite the readers to consider the
examples such as results of Bhatt-Scholze [BS23], of Cherbonnier-Colmez
[CC98], or Thm. 8.3. In proofs of these results, it is usually easy to
construct a _functor_ from $\mathrm{MOD}$ to $\mathrm{REP}$ (sometimes perhaps
initially to a bigger representation category, such as in our Thm. 8.3). The
most difficult part almost always is to show the functor is essentially
surjective (or, to determine its essential image).

What happens with Sen’s result Thm. 8.6, is that there is not even an obvious
_map_ from $\mathrm{MOD}$ to $\mathrm{REP}$; Sen can still construct a such
map, by the _set bijection_ mentioned in 8.7(2): but then this is hopeless to
be a _functor_. This forces Sen to go the other way around, by constructing a
_functor_ from $\mathrm{REP}$ to $\mathrm{MOD}$, which then unfortunately
relies on many choices, and cannot be canonical. We regard this strong
contrast with our Thm. 8.3 as another hint (in addition to Rem. 1.6) that the
(log)- nearly Hodge–Tate representations deserve to be further studied.

##  9\. Cohomology vanishing: proof of Proposition 3.9(3)

In this section, we prove Proposition 3.9(3). The main strategy is explained
in Construction 9.3; the two steps there are then carried out in the two
subsections.

We start by introducing some notations.

######  Notation 9.1.

First we set up some notation on indices. Let $s\geq 1$ (throughout this
section). For $I=(i_{1},\dots,i_{s})\in{\mathbb N}^{s}$, let
$|I|:=i_{1}+i_{2}+\cdots+i_{s}.$ Let $E_{q}=(0,\dots,0,1,0,\dots,0)\in{\mathbb
N}^{s}$ with $1$ at the $q$-th component. Thus,

$I=i_{1}E_{1}+\cdots+i_{s}E_{s}.$

For $I=(i_{1},\dots,i_{s})\in{\mathbb N}^{s}$ and
$J=(j_{1},\dots,j_{t})\in{\mathbb N}^{t}$, we define

$(I,J):=(i_{1},\dots,i_{s},j_{1},\dots,j_{t})\in{\mathbb N}^{s+t}.$

######  Notation 9.2.

From now on, we use notations in §3.2. Consider two elements

$f:=\sum_{I=(i_{1},\dots,i_{s})\in{\mathbb
N}^{s}}m_{I}\underline{X}^{[I]}\in{\mathrm{C}}^{s}$

and

$g:=\sum_{J=(j_{1},\dots,j_{s+1})\in{\mathbb
N}^{s+1}}n_{J}\underline{X}^{[J]}\in{\mathrm{C}}^{s+1}.$

If ${\mathrm{d}}^{s}(f)=g$, then by (3.8), we can write the equation as
follows

(9.1) $\begin{split}\sum_{J\in{\mathbb
N}^{s+1}}n_{J}\underline{X}^{[J]}=&\sum_{I\in{\mathbb
N}^{s}}(1+aX_{1})^{\frac{\phi_{M}}{a}-|I|}(m_{I})(X_{2}-X_{1})^{[i_{1}]}\cdots(X_{s+1}-X_{1})^{[i_{s}]}\\\
&+\sum_{l=1}^{s+1}(-1)^{l}\sum_{I\in{\mathbb N}^{s}}m_{I}X_{1}^{[i_{1}]}\dots
X_{l-1}^{[i_{l-1}]}X_{l+1}^{[i_{l}]}\cdots X_{s+1}^{[i_{s}]}.\end{split}$

######  Construction 9.3.

Now given $g\in{\mathrm{Ker}}({\mathrm{d}}^{s+1})$, to prove Proposition
3.9(3), we need to find an $f$ such that ${\mathrm{d}}^{s}(f)=g$. The proof
will be roughly divided into the following steps.

  1. Step 1: In §9.1 (precisely, Prop. 9.11), we prove that if $g=\sum_{K}n_{K}\underline{X}^{[J]}\in{\mathrm{Ker}}(d^{s+1})$, then the coefficients $n_{K}$ of $g$ are determined by those whose indices are in $\Lambda^{s+1}$, a specific subset of ${\mathbb N}^{s+1}$ (cf. Proposition 9.11). To construct the subset $\Lambda^{s+1}$, we need to investigate the coefficients in the Equation (9.1). The whole process will be divided into several parts with respect to the leading value $k_{1}$ of $K=(k_{1},k_{2},\cdots,k_{s+1})\in{\mathbb N}^{s+1}$.

(We warn that in §9.1, we will prove much more than claimed in Step 1 above,
cf. Construction 9.4. We further warn there will be a (necessary) switch of
superscripts between $s$ and $s+1$ there.)

  2. Step 2: cf. §9.2. Based on the subset $\Lambda^{s+1}$, we will give a concrete construction of a candidate $f$. It remains to verify ${\mathrm{d}}^{s}(f)=g$. Note both ${\mathrm{d}}^{s}(f),g\in\mathrm{Ker}({\mathrm{d}}^{s+1})$, thus by Step 1, it suffices to compare their coefficients over indices in $\Lambda^{s+1}$.

We remark that although the construction of $f$ and the verification of
$d^{s}(f)=g$ look quite complicated, it is indeed a direct generalisation of
the pattern we found in the case of $s=1$. (Namely, we encourage the
interested reader to try the computation for $s=1$.)

###  9.1. Step 1: construction of $\Lambda^{s}$ (and $\Lambda^{s+1}$)

######  Construction 9.4.

  1. (1)

We shall construct some subset $\Lambda^{s}\subset{\mathbb N}^{s}$ by solving
the equation ${\mathrm{d}}^{s}(f)=g$ (9.1), and show that the coefficients of
$f$—i.e., the $\\{m_{K}\\}_{K\in{\mathbb N}^{s}}$’s in (9.1)—are uniquely
determined by $\\{m_{I}\\}_{I\in\Lambda^{s}}$ and $\\{n_{J}\\}_{J\in{\mathbb
N}^{s+1}}$. We shall construct $\Lambda^{s}$ as a union of 3 parts, each part
enough to _determine some $m_{K}$_, depending on the leading value $k_{1}$ of
$K=(k_{1},k_{2},\cdots,k_{s})$; see Def. 9.10 for a summary of these parts.

  2. (2)

Note the above item achieves more than as stated in Step 1 in Construction 9.3
above; cf. next item.

  3. (3)

Indeed, by letting $g=0$, then ${\mathrm{d}}^{s}(f)=0$ implies the
coefficients of $f$ are uniquely determined by
$\\{m_{I}\\}_{I\in\Lambda^{s}}$. (Note there is shift from $s+1$ to $s$ in
this discussion; cf Prop. 9.11 and caution above Def. 9.10.)

######  Observation 9.5.

The basic idea to achieve Construction 9.4 is simple. Let

$F=\sum_{I\in{\mathbb N}^{s}}m_{I}\underline{X}^{[I]},\quad
G=\sum_{I\in{\mathbb N}^{s}}n_{I}\underline{X}^{[I]}.$

  1. (1)

If we want to compare the coefficients $m_{I}$ with $n_{I}$ for
$I=(i_{1},i_{2},\dots,i_{s})$ such that $i_{t_{1}}=\cdots=i_{t_{j}}=0$, then
it suffices to let $X_{i_{t_{1}}}=\cdots=X_{i_{t_{j}}}=0$, and consider the
coefficients of $Y:=\prod_{1\leq h\leq s,h\neq
t_{1},\dots,t_{j}}X_{h}^{[i_{h}]}$.

  2. (2)

Similarly, if we want to compare $m_{I}$ with $n_{I}$ for $I=(1,*)$ with the
leading term being $1$, it suffices to consider the coefficients (including
$X_{2},\dots,X_{s}$ as well) of $X_{1}$ in the expansions.

For example, in Lemma 9.6, we simply let $X_{1}=0$ in (9.1); for another
example, in Lemma 9.7, we compare coefficients of $X_{1}$.

We embark on the mission sketched in Construction 9.4(1). The next lemma deals
with the case where the index $K=(k_{1},\dots,k_{s})$ satisfies $k_{1}=0$. So
we let $X_{1}=0$ in (9.1) and then compare coefficients on both sides (cf.
Observation 9.5).

######  Lemma 9.6.

Consider the following subset $\Lambda_{1}^{s}\subset{\mathbb N}^{s}$ of
indices:

  * •

If $s\in\\{1,2\\}$, then $\Lambda_{1}^{s}=\\{(0,\dots,0)\\}$.

  * •

If $s\geq 3$, then
$\Lambda^{s}_{1}:=\\{(0,\dots,0,i_{2j+1},\dots,i_{s})\in{\mathbb N}^{s}\mid
j\geq 1,i_{2j+1}\geq 1\\}$.

Let $K=(0,k_{2},\dots,k_{s})\in{\mathbb N}^{s}$ be an index with the first
term being $0$. There are integers

$\\{z^{(s)}_{K,I}\\}_{I\in\Lambda^{s}_{1}}\text{ and
}\epsilon^{(s)}_{K,K},\text{ all in }\\{\pm 1,0\\}$

depending only on $s$ and $K$, such that if we have ${\mathrm{d}}^{s}(f)=g$,
then

$m_{K}=\sum_{I\in\Lambda^{s}_{1}}z^{(s)}_{K,I}m_{I}+\epsilon^{(s)}_{K,K}n_{(0,K)}$

(Thus in particular, these $m_{K}$’s are determined by $m_{I}$’s with
$I\in\Lambda_{1}^{s}$ and $n_{(0,K)}$). Furthermore, for any
$I\in\Lambda^{s}_{1}$, we have

  1. (1)

if $K\in\Lambda^{s}_{1}$, then $\epsilon^{(s)}_{K,K}=0$, and $z^{(s)}_{K,I}=1$
for $K=I$ while $z^{(s)}_{K,I}=0$ for $K\neq I$, and that

  2. (2)

if $K=(0,\dots,0,k_{2i},\dots,k_{s})$ for some $i\geq 1$ such that
$k_{2i}\cdots k_{s}\neq 0$, then $\epsilon^{(s)}_{K,K}=1$ and
$z^{(s)}_{K,I}=0$, and

  3. (3)

if $K=(0,\dots,0,k_{2i},\dots,k_{s})$ for some $i\geq 1$ such that $k_{2i}\neq
0$ and that $k_{2i}\cdots k_{s}=0$, then $\epsilon^{(s)}_{K,K}=1$, and that

  4. (4)

if $|K|\neq|I|$, then $z_{K,I}^{(s)}=0$.

###### Proof.

As results hold trivially for $s=1$, now assume $s\geq 2$.

If $K\in\Lambda_{1}^{s}$, then we define $z_{K,I}^{(s)}$ and
$\epsilon_{K,K}^{(s)}$ as stated in Item (1).

Now, we are going to construct $z_{K,I}^{(s)}$ and $\epsilon_{K,K}^{(s)}$ for
$K\notin\Lambda_{1}^{s}$. In this case, we must have
$K=(0,\dots,0,k_{2i},\dots,k_{s})$ for some $i\geq 1$ with $k_{2i}\neq 0$.

By letting $X_{1}=0$ in (9.1), we obtain that

(9.2) $\begin{split}\sum_{I=(i_{1},\dots,i_{s})\in{\mathbb
N}^{s}}n_{(0,I)}X_{2}^{[i_{1}]}\cdots
X_{s+1}^{[i_{s}]}=&\sum_{l=2}^{s+1}(-1)^{l}\sum_{I=(0,i_{2},\dots,i_{s})\in{\mathbb
N}^{s}}m_{I}X_{2}^{[i_{2}]}\dots X_{l-1}^{[i_{l-1}]}X_{l+1}^{[i_{l}]}\cdots
X_{s+1}^{[i_{s}]}\\\ =&\sum_{l=2}^{s+1}(-1)^{l}\sum_{(0,i_{1},\dots
i_{l-2},i_{l}\dots,i_{s})\in{\mathbb N}^{s}}m_{(0,i_{1},\dots
i_{l-2},i_{l}\dots,i_{s})}X_{2}^{[i_{1}]}\cdots
X_{l-1}^{[i_{l-2}]}X_{l+1}^{[i_{l}]}\cdots X_{s+1}^{[i_{s}]}.\end{split}$

For any $2\leq t_{1}<t_{2}<\dots<t_{r}\leq s+1$ (with $1\leq r\leq s$),
letting $X_{t_{1}}=\dots=X_{t_{r}}=0$ in (9.2) and considering the
coefficients of $\prod_{2\leq t\leq s+1,t\neq
t_{1},\dots,t_{r}}X_{t}^{i_{t-1}}$ with all $i_{t-1}\geq 1$ (so we only need
to consider $l\in\\{t_{1},\dots,t_{r}\\}$ in the above summation for $l$), we
have

(9.3) $\sum_{k=1}^{r}(-1)^{t_{k}}m_{(\sum_{2\leq q\leq t_{k}-1,q\neq
t_{1},\dots,t_{k-1}}i_{q-1}E_{q}+\sum_{t_{k}+1\leq q^{\prime}\leq
s+1,q^{\prime}\neq
t_{k+1},\dots,t_{r}}i_{q^{\prime}-1}E_{q^{\prime}-1})}=n_{(\sum_{2\leq h\leq
s+1,h\neq t_{1},\dots,t_{r}}i_{h-1}E_{h})}.$

Now, assume $k_{2i}\cdots k_{s}\neq 0$. In (9.3), let $r=2i-1$ with $i\leq 1$
and let $t_{1}=2,t_{2}=3,\dots,t_{2i-1}=2i\leq s+1$, and $i_{h-1}=k_{h-1}$ for
all $2i+1\leq h\leq s+1$. Then we get

$n_{(0,0,\dots,0,k_{2i},\dots,k_{s})}=\sum_{k=2}^{2i}(-1)^{k}m_{(\sum_{2i+1\leq
q^{\prime}\leq
s+1}k_{q^{\prime}-1}E_{q^{\prime}-1})}=m_{(0,\dots,0,k_{2i},\dots,k_{s})},$

as claimed in Item (2).

Assume $k_{2i}\cdots k_{s}=0$. In the spirit of Observation 9.5, in the
equation (9.3) above, we need to choose $r\geq 2i$ with $i\geq 1$ and
$t_{1}=2,\dots,t_{2i-1}=2i,2i+2\leq t_{2i}<\dots<t_{r}\leq s+1$ such that for
any $2i+2\leq h\leq s$, $k_{h}=0$ if and only if
$h+1\in\\{t_{2i},\dots,t_{r}\\}$. Then we have

$K=k_{2i}E_{2i}+\sum_{2i+2\leq q^{\prime}\leq s+1,q^{\prime}\neq
t_{2i},\dots,t_{r}}k_{q^{\prime}-1}E_{q^{\prime}-1}.$

Let $i_{h-1}=k_{h-1}$ for all
$h\in\\{2i+1,\dots,s+1\\}\setminus\\{t_{2i},\dots,t_{r}\\}$ in (9.3), and then
we get

$\begin{split}&n_{(k_{2i}E_{2i+1}+\sum_{2i+2\leq h\leq s+1,h\neq
t_{2i},\dots,t_{r}}k_{h-1}E_{h})}\\\
=&\sum_{l=2}^{2i}(-1)^{l}m_{(k_{2i}E_{2i}+\sum_{2i+2\leq q^{\prime}\leq
s+1,q^{\prime}\neq t_{2i},\dots,t_{r}}k_{q^{\prime}-1}E_{q^{\prime}-1})}\\\
&+\sum_{h=2i}^{r}(-1)^{t_{h}}m_{(\sum_{2i+1\leq q\leq t_{h}-1,q\neq
t_{2i},\dots,t_{h-1}}k_{q-1}E_{q}+\sum_{t_{h}+1\leq q^{\prime}\leq
s+1,q^{\prime}\neq t_{h+1},\dots,t_{r}}k_{q^{\prime}-1}E_{q^{\prime}-1})}\\\
=&m_{(k_{2i}E_{2i}+\sum_{2i+2\leq q^{\prime}\leq s+1,q^{\prime}\neq
t_{2i},\dots,t_{r}}k_{q^{\prime}-1}E_{q^{\prime}-1})}\\\
&+\sum_{h=2i}^{r}(-1)^{t_{h}}m_{(k_{2i}E_{2i+1}+\sum_{2i+2\leq q\leq
t_{h}-1,q\neq t_{2i},\dots,t_{h-1}}k_{q-1}E_{q}+\sum_{t_{h}+1\leq
q^{\prime}\leq s+1,q^{\prime}\neq
t_{h+1},\dots,t_{r}}k_{q^{\prime}-1}E_{q^{\prime}-1})},\end{split}$

which implies that

(9.4) $\begin{split}&m_{(k_{2i}E_{2i}+\sum_{2i+2\leq q^{\prime}\leq
s+1,q^{\prime}\neq t_{2i},\dots,t_{r}}k_{q^{\prime}-1}E_{q^{\prime}-1})}\\\
=&\sum_{h=2i}^{r}(-1)^{t_{h}+1}m_{(k_{2i}E_{2i+1}+\sum_{2i+2\leq q\leq
t_{h}-1,q\neq t_{2i},\dots,t_{h-1}}k_{q-1}E_{q}+\sum_{t_{h}+1\leq
q^{\prime}\leq s+1,q^{\prime}\neq
t_{h+1},\dots,t_{r}}k_{q^{\prime}-1}E_{q^{\prime}-1})}\\\
&+n_{(k_{2i}E_{2i+1}+\sum_{2i+2\leq q^{\prime}\leq s+1,q^{\prime}\neq
t_{2i},\dots,t_{r}}k_{q^{\prime}-1}E_{q^{\prime}})}.\end{split}$

For any $2i\leq h\leq r$, define

$J_{h}:=k_{2i}E_{2i+1}+\sum_{2i+2\leq q\leq t_{h}-1,q\neq
t_{2i},\dots,t_{h-1}}k_{q-1}E_{q}+\sum_{t_{h}+1\leq q^{\prime}\leq
s+1,q^{\prime}\neq t_{h+1},\dots,t_{r}}k_{q^{\prime}-1}E_{q^{\prime}-1},$

and then $J_{h}\in\Lambda_{1}^{s}$ (as $k_{2i}\neq 0$) satisfying

$|K|=k_{2i}+\sum_{2i+1\leq q^{\prime}\leq s,q^{\prime}+1\neq
t_{2i},\dots,t_{r}}k_{q^{\prime}}=|J_{h}|.$

Using this, we see that (9.4) amounts to

(9.5) $m_{K}=\sum_{h=2i}^{r}(-1)^{t_{h}+1}m_{J_{h}}+n_{(0,K)}.$

So one can construct $z_{K,I}^{(s)}$’s in the obvious way by applying (9.5)
and obtain the Item (3).

Finally, Item (4) follows from the above constructions directly. ∎

Now we consider the case where the index $K=(k_{1},\dots,k_{s})$ of $m_{K}$
satisfies $k_{1}\geq 2$. It turns out that the $m_{k_{1},\ast}$’s with
$k_{1}\geq 2$ are uniquely determined by $m_{1,*}$’s and $n_{1,k_{1}-1,*}$. We
will show this by comparing the coefficient of $X_{1}$ on both sides of (9.1).

######  Lemma 9.7.

For any $s\geq 1$, consider the subset of indices with leading term being $1$,
i.e.,

$\Lambda_{2}^{s}=\\{(1,i_{2},\dots,i_{s})\mid i_{2},\dots,i_{s}\geq
0\\}\subset{\mathbb N}^{s}.$

Let $K=(k_{1},\dots,k_{s})\in{\mathbb N}^{s}$ with $k_{1}\geq 2$. For each
$I\in\Lambda_{2}^{(s)}$, there exists

$z_{K,I}^{(s)}\in\\{\pm 1,0\\}$

depending only on $s$ and $K$, such that if $d^{s}(f)=g$, then

$m_{K}=\sum_{I\in\Lambda_{2}^{s}}z_{K,I}^{(s)}m_{I}+(\phi_{M}-a(|K|-1))m_{K-E_{1}}-\sum_{l=2}^{s}m_{K-E_{1}+E_{l}}-n_{(1,K-E_{1})}.$

Furthermore, we have $z_{K,I}^{(s)}=0$ in the case either $k_{2}\cdots
k_{s}\neq 0$ or $|K|\neq|I|$.

###### Proof.

In (9.1), comparing the coefficient of $X_{1}$, we see that

(9.6) $\begin{split}&\sum_{I=(i_{1},\dots,i_{s})\in{\mathbb
N}^{s}}n_{(1,I)}X_{2}^{[i_{1}]}\cdots X_{s+1}^{[i_{s}]}\\\
=&\sum_{I=(i_{1},\dots,i_{s})\in{\mathbb
N}^{s}}((\phi_{M}-a|I|)(m_{I})-\sum_{l=1}^{s}m_{I+E_{l}})X_{2}^{[i_{1}]}\cdots
X_{s+1}^{[i_{s}]}+\sum_{l=2}^{s+1}(-1)^{l}\sum_{I=(1,i_{2},\dots,i_{s})}m_{I}X_{2}^{[i_{2}]}\cdots
X_{l-1}^{[i_{l-1}]}X_{l+1}^{[i_{l}]}\cdots X_{s+1}^{[i_{s}]}.\end{split}$

Let $K=(k_{1},k_{2},\dots,k_{s})$ satisfy $k_{1}\geq 2$.

Assume $k_{2}\cdots k_{s}\neq 0$. In (9.6), comparing the coefficients of
$X_{2}^{[i_{1}]}\cdots X_{s+1}^{[i_{s}]}$ with
$(i_{1},\dots,i_{s})=(k_{1}-1,\dots,k_{s})$ (namely, $I=K-E_{1}$), we have

$n_{(1,K-E_{1})}=(\phi_{M}-a|K-E_{1}|)(m_{K-E_{1}})-\sum_{l=1}^{s}m_{K-E_{1}+E_{l}}.$

In other words, we conclude that

(9.7)
$m_{K}=(\phi_{M}-a(|K|-1))(m_{K-E_{1}})-\sum_{l=2}^{s}m_{K-E_{1}+E_{l}}-n_{(1,K-E_{1})}.$

Now, we can define $z_{K,I}^{(s)}$’s in the obvious way by applying (9.7), and
conclude the “furthermore” part in this case.

Now, assume $k_{2}\cdots k_{s}=0$. Then there is an $r\geq 1$ and $3\leq
t_{1}<\cdots<t_{r}\leq s+1$ such that $k_{h}=0$ if and only if
$h+1\in\\{t_{1},\dots,t_{r}\\}$. In particular, we have

$K=E_{1}+\sum_{2\leq t\leq s+1,t\neq t_{1},\dots,t_{r}}j_{t-1}E_{t-1}$

where $j_{1}=k_{1}-1$ and $j_{t-1}=k_{t-1}$ for any
$t\notin\\{2,t_{1},\dots,t_{r}\\}$. Thus, $j_{t-1}\geq 1$ for all $t$.

Letting $X_{t_{1}}=\cdots=X_{t_{r}}=0$ in (9.6) and considering the
coefficient of $\prod_{2\leq t\leq s+1,t\neq
t_{1},\dots,t_{r}}X_{t}^{[j_{t-1}]}$ (cf. Observation 9.5), we have

(9.8) $\begin{split}&n_{(E_{1}+\sum_{2\leq h\leq s+1,h\neq
t_{1},\dots,t_{r}}j_{h-1}E_{h})}\\\ =&(\phi_{M}-a(\sum_{2\leq t\leq s+1,t\neq
t_{1},\dots,t_{r}}j_{t-1}))(m_{(\sum_{2\leq t\leq s+1,t\neq
t_{1},\dots,t_{r}}j_{t-1}E_{t-1})})-\sum_{l=1}^{s}m_{(E_{l}+\sum_{2\leq t\leq
s+1,t\neq t_{1},\dots,t_{r}}j_{t-1}E_{t-1})}\\\
&+\sum_{k=1}^{r}(-1)^{t_{k}}m_{(E_{1}+\sum_{2\leq q\leq t_{k}-1,q\neq
t_{1},\dots,t_{k-1}}j_{q-1}E_{q}+\sum_{t_{k}+1\leq q^{\prime}\leq
s+1,q^{\prime}\neq
t_{k+1},\dots,t_{r}}j_{q^{\prime}-1}E_{q^{\prime}-1})},\end{split}$

which implies that

(9.9) $\begin{split}&m_{(E_{1}+\sum_{2\leq t\leq s+1,t\neq
t_{1},\dots,t_{r}}j_{t-1}E_{t-1})}\\\ =&(\phi_{M}-a(\sum_{2\leq t\leq
s+1,t\neq t_{1},\dots,t_{r}}j_{t-1}))(m_{(\sum_{2\leq t\leq s+1,t\neq
t_{1},\dots,t_{r}}j_{t-1}E_{t-1})})-\sum_{l=2}^{s}m_{(E_{l}+\sum_{2\leq t\leq
s+1,t\neq t_{1},\dots,t_{r}}j_{t-1}E_{t-1})}\\\
&+\sum_{k=1}^{r}(-1)^{t_{k}}m_{(E_{1}+\sum_{2\leq q\leq t_{k}-1,q\neq
t_{1},\dots,t_{k-1}}j_{q-1}E_{q}+\sum_{t_{k}+1\leq q^{\prime}\leq
s+1,q^{\prime}\neq
t_{k+1},\dots,t_{r}}j_{q^{\prime}-1}E_{q^{\prime}-1})}-n_{(E_{1}+\sum_{2\leq
h\leq s+1,h\neq t_{1},\dots,t_{r}}j_{h-1}E_{h})}.\end{split}$

For any $1\leq k\leq r$, put

$J_{k}:=E_{1}+\sum_{2\leq q\leq t_{k}-1,q\neq
t_{1},\dots,t_{k-1}}j_{q-1}E_{q}+\sum_{t_{k}+1\leq q^{\prime}\leq
s+1,q^{\prime}\neq t_{k+1},\dots,t_{r}}j_{q^{\prime}-1}E_{q^{\prime}-1}.$

and then for any $1\leq k\leq r$, we have $J_{k}\in\Lambda_{2}^{s}$ and

$|J_{k}|=1+j_{1}+\sum_{3\leq q\leq s+1,q\neq
t_{1},\dots,t_{r}}j_{q-1}=k_{1}+\sum_{3\leq q\leq s+1}k_{q-1}=|K|.$

Using this, we see that (9.9) amounts to

(9.10)
$\begin{split}m_{K}=&(\phi_{M}-a|K-E_{1}|)(m_{K-E_{1}})-\sum_{l=2}^{s}m_{E_{l}+K-E_{1}}+\sum_{k=1}^{r}(-1)^{t_{k}}m_{J_{k}}-n_{(1,K-E_{1})}\\\
=&\sum_{k=1}^{r}(-1)^{t_{k}}m_{J_{k}}+(\phi_{M}-a(|K|-1))(m_{K-E_{1}})-\sum_{l=2}^{s}m_{E_{l}+K-E_{1}}-n_{(1,K-E_{1})}.\end{split}$

Now, we can construct $z_{K,I}^{(s)}$’s in the obvious way by using (9.10),
and conclude the “furthermore” part in this case. ∎

As a corollary of Lemma 9.7, we have

######  Corollary 9.8.

Let $\Lambda_{2}^{s}$ be as above. Then for any
$K=(k_{1},\dots,k_{s})\in{\mathbb N}^{s}$ with $k_{1}\geq 1$, there are
integers

  * •

$z_{K,I}^{(s)}$ for all $I\in\Lambda_{2}^{s}$ with $|I|\leq|K|$ and

  * •

$\epsilon_{K,J}^{(s)}$ for all $J\in\Lambda_{2}^{s}$ with $|J|<|K|$

such that if ${\mathrm{d}}^{s}(f)=g$, we have

$m_{K}=\sum_{I\in\Lambda_{2}^{s},|I|\leq|K|}z_{K,I}^{(s)}\prod_{i=|I|}^{|K|-1}(\phi_{M}-ia)m_{I}+\sum_{J\in\Lambda_{2}^{s},|J|<|K|}\epsilon_{K,J}^{(s)}\prod_{i=|J|+1}^{|K|-1}(\phi_{M}-ia)n_{(1,J)},$

where by convention, $\prod_{i=|I|}^{|I|-1}(\phi_{M}-ia):={\mathrm{id}}_{M}$.

###### Proof.

We have to construct integers
$\\{z^{(s)}_{K,I}\\}_{I\in\Lambda_{2}^{s},|I|\leq|K|}$ for any
$K=(k_{1},k_{2},\dots,k_{s})\in{\mathbb N}^{s}$ with $k_{1}\geq 1$, which
depends only on $K$. We do induction on $k_{1}$.

When $k_{1}=1$, we put $\epsilon_{K,J}^{(s)}=0$ for any $J\in\Lambda_{s}^{2}$
with $|J|<|K|$, and $z^{(s)}_{K,I}=1$ if $I=K$ and otherwise
$z^{(s)}_{K,I}=0$. Then the result is obviously true in this case.

Now, assume $k_{1}\geq 2$, and then $K=E_{1}+L$ with
$L=(l_{1},k_{2},\dots,k_{s})$ such that $l_{1}\geq 1$. When
$I\in\Lambda_{2}^{s}$ with $|I|=|K|$, let $x_{K,I}^{(s)}$ denote the
corresponding $z_{K,I}^{(s)}$ in Lemma 9.7, and in particular,
$x_{K,I}^{(s)}=0$ as long as $|K|\neq|I|$. Then by Lemma 9.7 again, we have

$\begin{split}m_{K}=&\sum_{J\in\Lambda_{2}^{s},|J|=|K|}x_{K,J}^{(s)}m_{J}+(\phi_{M}-a|L|)m_{L}-\sum_{l=2}^{s}m_{L+E_{l}}-n_{(1,L)}\\\
=&\sum_{J\in\Lambda_{2}^{s},|J|=|K|}x_{K,J}^{(s)}m_{J}+(\phi_{M}-a|L|)(\sum_{I\in\Lambda_{2}^{s},|I|\leq|L|}z_{L,I}^{(s)}\prod_{h=|I|}^{|L|-1}(\phi_{M}-ah)m_{I}+\sum_{J\in\Lambda_{2}^{s},|J|<|L|}\epsilon_{L,J}^{(s)}\prod_{i=|J|+1}^{|L|-1}(\phi_{M}-ia)n_{(1,J)})\\\
&-\sum_{l=2}^{s}(\sum_{I\in\Lambda_{2}^{s},|I|\leq|L+E_{l}|=|K|}z_{L+E_{l},I}^{(s)}\prod_{h=|I|}^{|K|-1}(\phi_{M}-ah)m_{I}+\sum_{J\in\Lambda_{2}^{s},|J|<|K|}\epsilon_{L+E_{l},J}^{(s)}\prod_{i=|J|+1}^{|K|-1}(\phi_{M}-ia)n_{(1,J)})\\\
&-n_{(1,L)}\\\
=&\sum_{J\in\Lambda_{2}^{s},|J|=|K|}x_{K,J}^{(s)}m_{J}+\sum_{I\in\Lambda_{2}^{s},|I|\leq|L|}z_{L,I}^{(s)}\prod_{h=|I|}^{|K|-1}(\phi_{M}-ah)m_{I}-\sum_{l=2}^{s}\sum_{I\in\Lambda_{2}^{s},|I|\leq|K|}z_{L+E_{l},I}^{(s)}\prod_{h=|I|}^{|K|-1}(\phi_{M}-ah)m_{I}\\\
&+\sum_{J\in\Lambda_{2}^{s},|J|<|L|}\epsilon_{L,J}^{(s)}\prod_{h=|J|+1}^{|K|-1}(\phi_{M}-ah)m_{(1,J)}-\sum_{l=2}^{s}\sum_{J\in\Lambda_{2}^{s},|J|<|K|}\epsilon_{L+E_{l},J}^{(s)}\prod_{h=|J|+1}^{|K|-1}(\phi_{M}-ah)n_{(1,J)}-n_{(1,L)}\\\
=&\sum_{I\in\Lambda_{2}^{s},|I|\leq|K|}z_{K,I}^{(s)}\prod_{h=|I|}^{|K|-1}(\phi_{M}-ah)m_{I}+\sum_{J\in\Lambda_{2}^{s},|J|<|K|}\epsilon_{K,J}^{(s)}\prod_{h=|J|+1}^{|K|-1}(\phi_{M}-ah)n_{(1,J)},\end{split}$

where $z_{K,I}^{(s)}$’s and $\epsilon_{K,J}^{(s)}$’s are defined such that the
last equality above holds true. The proof is complete. ∎

Thanks to Lemma 9.7, we are reduced to studying the $m_{K}$’s with
$K=(1,k_{2},*)$. The next lemma tells us if $k_{2}=0$, then $m_{K}$ is
determined by $m_{I}$’s with $I\in\Lambda_{3}^{s}$ (which will be defined
below) and $n_{1,*}$’s as long as ${\mathrm{d}}^{s}(f)=g$.

######  Lemma 9.9.

Consider the following subset $\Lambda_{3}^{s}$ of indices:

  * •

If $1\leq s\leq 3$, let $\Lambda_{3}^{s}=\\{(1,0,\dots,0)\\}$.

  * •

If $s\geq 4$, let
$\Lambda_{3}^{s}:=\\{(1,0,\dots,0,i_{2j+2},\dots,i_{s})\in{\mathbb N}^{s}\mid
j\geq 1,i_{2j+2}\geq 1\\}$.

Let $K=(1,0,\dots,0,k_{2j+1},\dots,k_{s})$ with $j\geq 1$ and $k_{2j+1}\geq
1$. For each $I\in\Lambda_{3}^{s}$, there exists an integer

$z_{K,I}^{(s)}\in\\{\pm 1,0\\}$

depending only on $s$ and $K$ satisfying the following: If
${\mathrm{d}}^{s}(f)=g$, then we have

$m_{K}=\sum_{I\in\Lambda_{3}^{s}}z_{K,I}^{(s)}m_{I}+(\phi_{M}-(|K|-1)a)m_{K-E_{1}}-\sum_{l=2}^{s}m_{K-E_{1}+E_{l}}-n_{(1,K-E_{1})}.$

Furthermore, we have $z_{K,I}^{(s)}=0$ in the case either $k_{2j+2}\cdots
k_{s}\neq 0$ or $|K|\neq|I|$.

###### Proof.

All results hold trivially for $s=1,2$. From now on, assume $s\geq 3$.

Note that for any $2\leq t_{1}<t_{2}<\dots<t_{r}\leq s+1$ (with $1\leq r\leq
s$), we still have (9.8); that is,

$\begin{split}&(\phi_{M}-a(\sum_{2\leq t\leq s+1,t\neq
t_{1},\dots,t_{r}}j_{t-1}))(m_{(\sum_{2\leq t\leq s+1,t\neq
t_{1},\dots,t_{r}}j_{t-1}E_{t-1})})-\sum_{l=1}^{s}m_{(E_{l}+\sum_{2\leq t\leq
s+1,t\neq t_{1},\dots,t_{r}}j_{t-1}E_{t-1})}\\\
+&\sum_{k=1}^{r}(-1)^{t_{k}}m_{(E_{1}+\sum_{2\leq q\leq t_{k}-1,q\neq
t_{1},\dots,t_{k-1}}j_{q-1}E_{q}+\sum_{t_{k}+1\leq q^{\prime}\leq
s+1,q^{\prime}\neq
t_{k+1},\dots,t_{r}}j_{q^{\prime}-1}E_{q^{\prime}-1})}=n_{(E_{1}+\sum_{2\leq
h\leq s+1,h\neq t_{1},\dots,t_{r}}i_{h-1}E_{h})},\end{split}$

Assume $i\geq 1$. By letting $r\geq 2i$ and $t_{1}=2,\dots,t_{2i-1}=2i$, we
get

(9.11) $\begin{split}&n_{(E_{1}+\sum_{2i+1\leq h\leq s+1,h\neq
t_{2i},\dots,t_{r}}j_{h-1}E_{h})}\\\ =&(\phi_{M}-a(\sum_{2i+1\leq t\leq
s+1,t\neq t_{2i},\dots,t_{r}}j_{t-1}))(m_{(\sum_{2i+1\leq t\leq s+1,t\neq
t_{2i},\dots,t_{r}}j_{t-1}E_{t-1})})\\\
&-\sum_{l=1}^{s}m_{(E_{l}+\sum_{2i+1\leq t\leq s+1,t\neq
t_{2i},\dots,t_{r}}j_{t-1}E_{t-1})}+\sum_{l=2}^{2i}(-1)^{l}m_{(E_{1}+\sum_{2i+1\leq
q^{\prime}\leq s+1,q^{\prime}\neq
t_{2i},\dots,t_{r}}j_{q^{\prime}-1}E_{q^{\prime}-1})}\\\
&+\sum_{h=2i}^{r}(-1)^{t_{h}}m_{(E_{1}+\sum_{2i+1\leq q\leq t_{h}-1,q\neq
t_{2i},\dots,t_{h-1}}j_{q-1}E_{q}+\sum_{t_{h}+1\leq q^{\prime}\leq
s+1,q^{\prime}\neq t_{h+1},\dots,t_{r}}j_{q^{\prime}-1}E_{q^{\prime}-1})}\\\
=&(\phi_{M}-a(\sum_{2i+1\leq t\leq s+1,t\neq
t_{2i},\dots,t_{r}}j_{t-1}))(m_{(\sum_{2i+1\leq t\leq s+1,t\neq
t_{2i},\dots,t_{r}}j_{t-1}E_{t-1})})\\\
&-\sum_{l=1}^{s}m_{(E_{l}+\sum_{2i+1\leq t\leq s+1,t\neq
t_{2i},\dots,t_{r}}j_{t-1}E_{t-1})}+m_{(E_{1}+\sum_{2i+1\leq q^{\prime}\leq
s+1,q^{\prime}\neq t_{2i},\dots,t_{r}}j_{q^{\prime}-1}E_{q^{\prime}-1})}\\\
&+\sum_{h=2i}^{r}(-1)^{t_{h}}m_{(E_{1}+\sum_{2i+1\leq q\leq t_{h}-1,q\neq
t_{2i},\dots,t_{h-1}}j_{q-1}E_{q}+\sum_{t_{h}+1\leq q^{\prime}\leq
s+1,q^{\prime}\neq
t_{h+1},\dots,t_{r}}j_{q^{\prime}-1}E_{q^{\prime}-1})}.\end{split}$

Let $K=(1,0,\dots,0,k_{2i+1},\dots,k_{s})$ with $k_{2i+1}\neq 0$.

Assume $k_{2i+1}\cdots k_{s}\neq 0$. Then let $r=2i$ and $t_{2i}=2i+1$ in
(9.11). For any $2i+2\leq h\leq s+1$, by letting $j_{h-1}=k_{h-1}$ for all
$h$, we get

$\begin{split}(\phi_{M}-a(\sum_{t=2i+2}^{s}j_{t-1}))(m_{(\sum_{t=2i+2}^{s+1}j_{t-1}E_{t-1})})-\sum_{l=1}^{s}m_{(E_{l}+\sum_{t=2i+2}^{s+1}j_{t-1}E_{t-1})}=n_{(E_{1}+\sum_{h=2i+2}^{s+1}j_{h-1}E_{h})}.\end{split}$

In other words, we have

$m_{K}=(\phi_{M}-(|K|-1)a)m_{K-E_{1}}-\sum_{l=2}^{s}m_{K-E_{1}+E_{l}}-n_{(1,K-E_{1})}.$

Now, one can construct $z_{K,I}^{(s)}$’s in the obvious way by applying the
above equation and conclude the “furthermore” part in this case.

Now, assume $k_{2i+2}\cdots k_{s}=0$. Then in (9.11), we put $r\geq 2i+1$ and
furthermore let $t_{2i}=2i+1$ and $2i+3\leq t_{2i+1}<\cdots<t_{r}\leq s+1$
such that $k_{h}=0$ if and only if $h+1\in\\{t_{2i+1},\dots,t_{r}\\}$. Now,
let $j_{h}=k_{h}$ for any
$h\in\\{2i+1,\dots,s\\}\setminus\\{t_{2i+1}-1,\dots,t_{r}-1\\}$ in (9.11), and
then we get

$\begin{split}&n_{(E_{1}+j_{2i+1}E_{2i+2}+\sum_{2i+3\leq t\leq s+1,t\neq
t_{2i+1},\dots,t_{r}}j_{t-1}E_{t})}\\\ =&(\phi_{M}-a(\sum_{2i+2\leq t\leq
s+1,t\neq t_{2i+1},\dots,t_{r}}j_{t-1}))(m_{(j_{2i+1}E_{2i+1}+\sum_{2i+3\leq
t\leq s+1,t\neq t_{2i+1},\dots,t_{r}}j_{t-1}E_{t-1})})\\\
&-\sum_{l=1}^{s}m_{(E_{l}+j_{2i+1}E_{2i+1}+\sum_{2i+3\leq t\leq s+1,t\neq
t_{2i+1},\dots,t_{r}}j_{t-1}E_{t-1})}\\\
&+\sum_{h=2i+1}^{r}(-1)^{t_{h}}m_{(E_{1}+j_{2i+1}E_{2i+2}+\sum_{2i+3\leq q\leq
t_{h}-1,q\neq t_{2i+1},\dots,t_{h-1}}j_{q-1}E_{q}+\sum_{t_{h}+1\leq
q^{\prime}\leq s+1,q^{\prime}\neq
t_{h+1},\dots,t_{r}}j_{q^{\prime}-1}E_{q^{\prime}-1})}.\end{split}$

In other words, we have

(9.12) $\begin{split}&m_{(E_{1}+j_{2i+1}E_{2i+1}+\sum_{2i+3\leq t\leq
s+1,t\neq t_{2i+1},\dots,t_{r}}j_{t-1}E_{t-1})}\\\
=&(\phi_{M}-a(\sum_{2i+2\leq t\leq s+1,t\neq
t_{2i+1},\dots,t_{r}}j_{t-1}))(m_{(j_{2i+1}E_{2i+1}+\sum_{2i+3\leq t\leq
s+1,t\neq t_{2i+1},\dots,t_{r}}j_{t-1}E_{t-1})})\\\
&-\sum_{l=2}^{s}m_{(E_{l}+j_{2i+1}E_{2i+1}+\sum_{2i+3\leq t\leq s+1,t\neq
t_{2i+1},\dots,t_{r}}j_{t-1}E_{t-1})}\\\
&+\sum_{h=2i+1}^{r}(-1)^{t_{h}}m_{(E_{1}+j_{2i+1}E_{2i+2}+\sum_{2i+3\leq q\leq
t_{h}-1,q\neq t_{2i+1},\dots,t_{h-1}}j_{q-1}E_{q}+\sum_{t_{h}+1\leq
q^{\prime}\leq s+1,q^{\prime}\neq
t_{h+1},\dots,t_{r}}j_{q^{\prime}-1}E_{q^{\prime}-1})}\\\
&-n_{(E_{1}+j_{2i+1}E_{2i+2}+\sum_{2i+3\leq t\leq s+1,t\neq
t_{2i+1},\dots,t_{r}}j_{t-1}E_{t})}.\end{split}$

For any $h$, we define

$J_{h}:=E_{1}+j_{2i+1}E_{2i+2}+\sum_{2i+3\leq q\leq t_{h}-1,q\neq
t_{2i+1},\dots,t_{h-1}}j_{q-1}E_{q}+\sum_{t_{h}+1\leq q^{\prime}\leq
s+1,q^{\prime}\neq t_{h+1},\dots,t_{r}}j_{q^{\prime}-1}E_{q^{\prime}-1}.$

Then we have $J_{h}\in\Lambda_{3}^{s}$ (as $j_{2i+1}\neq 0$) satisfying

$|J_{h}|=1+j_{2i+1}+\sum_{2i+3\leq t\leq s+1,t\neq
t_{2i+1},\dots,t_{r}}j_{t-1}j_{t-1}=1+\sum_{2i+1\leq h\leq s+1,h+1\neq
t_{2i+1},\dots,t_{r}\\}}k_{h}=|K|$

by noting that

$K=E_{1}+j_{2i+1}E_{2i+1}+\sum_{2i+3\leq t\leq s+1,t\neq
t_{2i+1},\dots,t_{r}}j_{t-1}E_{t-1}.$

Using this, (9.12) amounts to that

(9.13)
$m_{K}=\sum_{h=2i+1}^{r}(-1)^{t_{h}}m_{J_{h}}+(\phi_{M}-a(|K|-1))m_{K-E_{1}}-\sum_{l=2}^{s}m_{K-E_{1}+E_{l}}-n_{(1,K-E_{1})}.$

Now one can construct the desired $z_{K,I}^{(s)}$’s in the obvious way by
applying (9.13) and conclude the “furthermore” part in this case immediately.
∎

Now, we are able to construct the desired set $\Lambda^{s+1}$ as promised in
Step 1 of Construction 9.3. To do so, we need to _change the superscript from
$s$ to $s+1$_ (and let $n_{J}=0$ for all $J$). To avoid confusion, we define
these sets for any $t\geq 1$.

######  Definition 9.10.

Let $t\geq 1$. Let $\Lambda^{t}\subset{\mathbb N}^{t}$ be the following union

$\Lambda^{t}:=\Lambda_{1}^{t}\cup\Lambda_{3}^{t}\cup\Lambda_{4}^{t},$

where $\Lambda_{1}^{t}$ and $\Lambda_{3}^{t}$ are defined in Lemmas 9.6 and
9.9 (for $s=t$) respectively, and $\Lambda_{4}^{t}$ is defined as follows:

  * •

If $t=1$, let $\Lambda_{4}^{t}=\\{1\\}$.

  * •

It $t\geq 2$, let $\Lambda_{4}^{t}=\\{(1,i_{2},\dots,i_{t})\mid i_{2}\geq
1\\}$.

We caution: the subset $\Lambda_{2}^{t}:=\\{(1,i_{2},\dots,i_{t})\mid
i_{2},\dots,i_{t}\geq 0\\}\subset{\mathbb N}^{t}$ (cf. Lemma 9.7) is not
related to the subset $\Lambda^{t}$; but it will be important for the
construction of $f$ in Construction 9.12.

The following proposition finishes Step 1 in Construction 9.3, showing that an
element $f$ in ${\mathrm{Ker}}({\mathrm{d}}^{s+1})$ is determined by its
coefficients over indices in $\Lambda^{s+1}$.

######  Proposition 9.11.

Recall $s\geq 1$. Consider $\Lambda^{s+1}$ as in Def. 9.10. Then for any
$I\in{\mathbb N}^{s+1}$, there are unique integers

$\\{z_{I,K}^{(s+1)}\\}_{K\in\Lambda^{s+1},|K|\leq|I|}$

depending only on $s$ and $I$ such that if $f=\sum_{I\in{\mathbb
N}^{s+1}}m_{I}\underline{X}^{[I]}\in{\mathrm{Ker}}({\mathrm{d}}^{s+1})$, i.e.,

${\mathrm{d}}^{s+1}(f)=g=0,$

then

$m_{I}=\sum_{K\in\Lambda^{s+1},|K|\leq|I|}z_{I,K}^{(s+1)}\prod_{i=|K|}^{|I|-1}(\phi_{M}-ia)m_{K}.$

Furthermore, when $I\in\Lambda^{s+1}$, we have

  * •

$z_{I,K}^{(s+1)}=1$ if $K=I$, and

  * •

$z_{I,K}^{(s+1)}=0$ if $K\neq I$.

###### Proof.

Fix an $I=(i_{1},\dots,i_{s+1})\in{\mathbb N}^{s+1}$. If $I\in\Lambda^{s+1}$,
then we define $z_{I,K}^{(s+1)}$ as stated. Now, we assume
$I\notin\Lambda^{s+1}$.

If $i_{1}=0$, we define $z_{I,K}^{(s+1)}$ by applying Lemma 9.6 (after
replacing $s$ by $s+1$ there) in the $g=0$ case for $K\notin\Lambda_{1}^{s+1}$
and $z_{I,K}^{(s+1)}=0$ for $K\in\Lambda^{s+1}\setminus\Lambda_{1}^{s+1}$.

If $i_{1}=1$, then there exists some $j\geq 1$ such that
$i_{2}=\cdots=i_{2j}=0$ and $i_{2j+1}\neq 0$ (as $I\notin\Lambda^{s+1}$). Then
we define $z_{I,K}^{(s+1)}$ by applying Lemma 9.9 (after replacing $s$ by
$s+1$ there) in the $g=0$ case for $K\in\Lambda_{3}^{s+1}$ and
$z_{I,K}^{(s+1)}=0$ for $K\in\Lambda^{s+1}_{4}$. Then we have

$\begin{split}m_{I}=&\sum_{K\in\Lambda^{s+1}\setminus\Lambda^{s+1}_{1},|K|\leq|I|}z_{I,K}^{(s+1)}\prod_{L=|K|}^{|I|-1}(\phi_{M}-la)(m_{K})+(\phi_{M}-(|I|-1)a)m_{I-E_{1}}-\sum_{l=2}^{s+1}m_{I-E_{1}+E_{l}}\\\
=&\sum_{K\in\Lambda^{s+1}\setminus\Lambda^{s+1}_{1},|K|\leq|I|}z_{I,K}^{(s+1)}\prod_{L=|K|}^{|I|-1}(\phi_{M}-la)(m_{K})+(\phi_{M}-(|I|-1)a)\sum_{J\in\Lambda_{1}^{s+1},|J|\leq|I|-1}z_{I-E_{1},J}^{(s+1)}\prod_{l=|J|}^{|I|-2}(\phi_{M}-la)m_{J}\\\
&-\sum_{l=2}^{s+1}\sum_{J_{l}\in\Lambda_{1}^{s+1},|J_{l}|\leq|I|}z_{I-E_{1}+E_{l},J_{l}}^{(s+1)}\prod_{l=|J_{l}|}^{|I|-1}(\phi_{M}-la)m_{J_{l}}\\\
=&\sum_{K\in\Lambda^{s+1}\setminus\Lambda^{s+1}_{1},|K|\leq|I|}z_{I,K}^{(s+1)}\prod_{L=|K|}^{|I|-1}(\phi_{M}-la)(m_{K})+\sum_{J\in\Lambda_{1}^{s+1},|J|\leq|I|-1}z_{I-E_{1},J}^{(s+1)}\prod_{l=|J|}^{|I|-1}(\phi_{M}-la)m_{J}\\\
&-\sum_{l=2}^{s+1}\sum_{J_{l}\in\Lambda_{1}^{s+1},|J_{l}|\leq|I|}z_{I-E_{1}+E_{l},J_{l}}^{(s+1)}\prod_{l=|J_{l}|}^{|I|-1}(\phi_{M}-la)m_{J_{l}}\\\
=&\sum_{K\in\Lambda^{s+1}\setminus\Lambda^{s+1}_{1},|K|\leq|I|}z_{I,K}^{(s+1)}\prod_{L=|K|}^{|I|-1}(\phi_{M}-la)(m_{K})+\sum_{J\in\Lambda_{1}^{s+1},|J|\leq|I|}(z_{I-E_{1},J}^{(s+1)}-\sum_{l=2}^{s+1}z_{I-E_{1}+E_{l},J}^{(s+1)})\prod_{l=|J|}^{|I|-1}(\phi_{M}-la)m_{J},\end{split}$

where we put $z_{I-E_{1},J}^{(s+1)}=0$ when $|J|=|I|$. Put
$z_{I,K}^{(s+1)}:=z_{I-E_{1},K}^{(s+1)}-\sum_{l=2}^{s+1}z_{I-E_{1}+E_{l},K}^{(s+1)}$
for $K\in\Lambda_{s+1}^{1}$ with $|K|\leq|I|$. Then the result holds true in
this case.

If $i_{1}\geq 2$, by Corollary 9.8 in the $g=0$ case, there are integers
$\\{w^{(s)}_{I,J}\\}_{J=(j_{1},j_{2},\dots,j_{s+1})\in{\mathbb
N}^{s+1},|J|\leq|I|,j_{1}=1}$ depending only on $s$ and $I$ such that if
${\mathrm{d}}^{s+1}(f)=0$, then

$m_{I}=\sum_{J=(1,j_{2},\dots,j_{s+1})\in{\mathbb
N}^{s+1},|J|\leq|I|}w^{(s)}_{I,J}\prod_{l=|J|}^{|I|-1}(\phi_{M}-la)(m_{J}).$

For any $J$ with $j_{1}=1$ and $|J|\leq|I|$, by what we have proved, we have

$m_{J}=\sum_{K\in\Lambda^{s+1},|K|\leq|J|}z_{J,K}^{(s+1)}\prod_{j=|K|}^{|J|-1}(\phi_{M}-ja)m_{K}.$

Now, one can construct the desired $z_{I,K}^{(s+1)}$’s in this case by
proceeding as in the “$i_{1}=1$” case above. ∎

###  9.2. Step 2: vanishing of cocycles

We proceed to carry out Step 2 in Construction 9.3. Here we will make use of
both $\Lambda^{s}$ and $\Lambda^{s+1}$. Given $g\in{\mathrm{C}}^{s+1}$
satisfying ${\mathrm{d}}^{s+1}(g)=0$,

  * •

we make use of $\Lambda^{s}$ to construct the desired $f\in{\mathrm{C}}^{s}$,
cf. Construction 9.12;

  * •

we then make use of $\Lambda^{s+1}$ to check ${\mathrm{d}}^{s}(f)=g$, cf.
Construction 9.13.

######  Construction 9.12.

Let $g=\sum_{I\in{\mathbb
N}^{s+1}}n_{I}\underline{X}^{[I]}\in{\mathrm{Ker}}({\mathrm{d}}^{s+1})$. We
will define an element $f=\sum_{I\in{\mathbb N}^{s}}m_{I}\underline{X}^{[I]}$
in the following; equivalently, we define elements $m_{I}$ for all
$I=(i_{1},\dots,i_{s})$.

  1. (a)

Suppose $i_{1}=0$.

     * •

If there exists some $j\geq 1$ such that $i_{1}=\cdots=i_{2j-1}=0$ and
$i_{2j}\neq 0$ (equivalently, $(0,I)\in\Lambda^{s+1}_{1}$), then define
$m_{I}=\epsilon_{I,I}^{(s)}n_{(0,I)}$ with $\epsilon_{I,I}^{s}$ defined in
Lemma 9.6 (which turns out to be $1$).

     * •

If there exists some $j\geq 1$ such that $i_{1}=\cdots=i_{2j}=0$ and
$i_{2j+1}\neq 0$ (equivalently, $I\in\Lambda_{1}^{s}$), then define $m_{I}=0$.

  2. (b)

Suppose $i_{1}=1$.

     * •

If there exists some $j\geq 1$ such that $i_{2}=\cdots=i_{2j-1}=0$ and
$i_{2j}\neq 0$, then define $m_{I}=0$.

     * •

If there exists some $j\geq 1$ such that $i_{2}=\cdots=i_{2j}=0$ and
$i_{2j+1}\neq 0$ (equivalently, $(1,I-E_{1})\in\Lambda^{s+1}_{3}$), then
define

$m_{I}=(\phi_{M}-(|I|-1)a)m_{I-E_{1}}-\left(\sum_{l=2}^{s}m_{I-E_{1}+E_{l}}\right)-n_{(1,I-E_{1})}.$

Here, $m_{I-E_{1}}$ and $m_{I-E_{1}+E_{l}}$ are already defined in (a).

  3. (c)

Suppose $i_{1}\geq 2$ (equivalently, $(1,I-E_{1})\in\Lambda_{4}^{s+1}$).

Suppose that we have constructed all $m_{I}$’s with $i_{1}\leq l$ for some
$l\geq 1$. We are going to construct $m_{I}$’s with $i_{1}=l+1$. Note that we
must have $I=E_{1}+J$ with $J=(j_{1}=l,j_{2},\dots,j_{s})$. Let
$\\{z_{I,K}^{(s)}\\}_{K\in\Lambda_{2}^{s}}$ be the integers defined in Lemma
9.7. Then define

$m_{I}=\left(\sum_{K\in\Lambda_{3}^{s}}z_{I,K}^{(s)}m_{K}\right)+(\phi_{M}-(|I|-1)a)m_{J}-\sum_{l=2}^{s}m_{J+E_{l}}-n_{(1,J)}.$

Here, $m_{J}$ and $m_{J+E_{l}}$ are defined by inductive hypothesis.

Now, we have constructed $m_{I}$ for all $I$, and hence obtain an element
$f=\sum_{I\in{\mathbb N}^{s}}m_{I}\underline{X}^{[I]}$. We have to show $f$ is
a well-defined element in ${\mathrm{C}}^{s}$; that is,
$\lim_{|I|\rightarrow+\infty}m_{I}=0$.

Let $I=(i_{1},i_{2},\dots,i_{s})$. When $i_{1}=0,1$, by Items (a) and (b)
above, we can conclude by using $\lim_{|J|\rightarrow+\infty}n_{J}=0$. Now,
assume $i_{1}\geq 2$ and let $z_{I,J}^{(s)}$ and $\epsilon_{I,K}^{(s)}$ for
$J,K\in\Lambda_{2}^{s}$ with $|J|,|K|+1\leq|I|$ be the integers defined in
Corollary 9.8. Then the proof of Corollary 9.8 together with Items (c) above
implies that in this case, we have

$m_{I}=\sum_{J\in\Lambda_{2}^{s},|J|\leq|I|}z_{I,J}^{(s)}\prod_{i=|J|}^{|I|-1}(\phi_{M}-ia)m_{J}+\sum_{K\in\Lambda_{2}^{s},|K|<|I|}\epsilon_{I,K}^{(s)}\prod_{i=|K|+1}^{|I|-1}(\phi_{M}-ia)n_{(1,K)}.$

As $\lim_{h\rightarrow+\infty}\prod_{i=m}^{h-1}(\phi_{M}-ia)=0$ and
$\lim_{|J|\rightarrow+\infty}n_{J}=0$, we see that
$\lim_{|I|\rightarrow+\infty}m_{I}=0$ as desired. So we see that $f$ is a
well-defined element in ${\mathrm{C}}^{s}$.

######  Construction 9.13 (Proof of Prop. 3.9(3)).

We now conclude the proof. Let $f$ be as in Construction 9.12. Write

$g^{\prime}:={\mathrm{d}}^{s}(f)=\sum_{I\in{\mathbb
N}^{s+1}}n_{I}^{\prime}\underline{X}^{[I]}.$

It remains to prove $g^{\prime}=g$. By Prop. 9.11, it suffices to check

$n_{L}=n_{L}^{\prime},\forall L\in\Lambda^{s+1}.$

We will freely use the Items (a), (b), (c) in Construction 9.12 above.

  * •

If $L=(0,I)\in\Lambda^{s+1}_{1}$ with $I=(i_{1},\dots,i_{s})$. Then there
exists some $j\geq 1$ such that $i_{1}=\cdots=i_{2j-1}=0$ and $i_{2j}\neq 0$.
By Item (a) above, we have $n_{L}=\epsilon_{I,I}^{(s)}m_{I}$. On the other
hand, by Lemma 9.6, we have $n^{\prime}_{L}=\epsilon_{I,I}^{(s)}m_{I}$.
Therefore, $n_{L}=n_{L}^{\prime}$.

  * •

If $L=(1,I-E_{1})\in\Lambda^{s+1}_{3}$ with $I=(1,i_{2},\dots,i_{s})$ such
that $i_{2}=\cdots=i_{2j}=0$ and $i_{2j+1}\neq 0$ for some $j\geq 1$. By Lemma
9.9 and Item (b) above, we have

$n_{L}^{\prime}=\sum_{J\in\Lambda_{3}^{s}}z_{I,J}^{(s)}m_{J}+(\phi_{M}-(|I|-1)a)m_{I-E_{1}}-\sum_{l=2}^{s}m_{I-E_{1}+E_{l}}-m_{I}=\sum_{J\in\Lambda_{3}^{s}}z_{I,J}^{(s)}m_{J}+n_{L}.$

By Item (a) above, we have $m_{J}=0$, for any $J\in\Lambda_{3}^{s}$. Thus, we
conclude that $n_{L}=n_{L}^{\prime}$.

  * •

If $L=(1,I)\in\Lambda_{4}^{s+1}$ with $I=(i_{1},\dots,i_{s})$ such that
$i_{1}\geq 1$, put $J=I+E_{1}$. Let
$\\{z_{J,K}^{(s)}\\}_{K\in\Lambda_{2}^{s},|K|=|J|}$ be the integers defined in
Lemma 9.7. By Lemma 9.7, we have

$n_{L}=\sum_{K\in\Lambda_{2}^{s}}z_{J,K}^{(s)}m_{K}+(\phi_{M}-(|J|-1)a)m_{I}-\sum_{l=1}^{s}m_{I+E_{l}}.$

By Item (c), we see that for any $L\in\Lambda_{4}^{(s+1)}$, we have
$n_{L}=n_{L}^{\prime}$.

In conclusion, we have proved ${\mathrm{d}}^{s}(f)=g$. This implies
${\mathrm{H}}^{s+1}({\mathrm{C}}(M,\varepsilon_{M}))=0$ for all $s\geq 1$.

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