Proof. Let

$$\begin{align*}0{\longrightarrow} H{\longrightarrow} M{\longrightarrow}{\mathscr{O}}_S{\longrightarrow}0 \end{align*}$$

be an exact sequence in ${{\mathrm {Fil}\text {-}F\text {-}\mathrm {MIC}}}(S,\sigma )$ . Since ${\mathrm {Fil}}^0H_{{\mathrm {d\hspace {-0.2pt}R}}}=0$ , there is the unique lifting $e\in {\mathrm {Fil}}^0M_{\mathrm {rig}}$ of $1\in \mathscr {O}_S$ . Then,

(2.11) $$ \begin{align} u(M)=(\nabla(e),(1-\Phi)e)\in H^1({\mathscr{S}}(H))\subset \Omega^1_{B^{\dagger}_K}{\otimes} {\mathrm{Fil}}^{-1}H_{\mathrm{rig}}{\oplus} H_{\mathrm{rig}} \end{align} $$

by definition of u. If $u(M)=0$ , then the datum $(B_Ke,B^{\dagger }_Ke, c,\Phi ,\nabla ,{\mathrm {Fil}}^{\bullet })$ forms a subobject of M which is isomorphic to the unit object ${\mathscr {O}}_S$ . This gives a splitting of the above exact sequence. The latter assertion is immediate as

$$\begin{align*}H^1_{\mathrm{syn}}(S,{\mathbb{Q}}_p(1))\subset {\mathbb{Q}}{\otimes}\varprojlim_n(\Omega^1_{B_n}{\oplus} B_n),\quad H^1_{\mathrm{syn}}(S,{\mathbb{Q}}_p(r))\subset {\mathbb{Q}}{\otimes}\varprojlim_n(\{0\}{\oplus} B_n),\,(r\geq2) \end{align*}$$

by (2.8) and

$$\begin{align*}H^1({\mathscr{S}}({\mathscr{O}}_S(1)))\subset \Omega^1_{B^{\dagger}_K}{\oplus} B^{\dagger}_K,\quad H^1({\mathscr{S}}({\mathscr{O}}_S(r)))\subset \{0\}{\oplus} B^{\dagger}_K,\,(r\geq2) \end{align*}$$

by (2.6).

Proof. By definition of u in (2.7), one has

$$\begin{align*}u({\mathop{\mathscr{L}\!\!\mathit{o}\mathit{g}}\nolimits}(f))=(\nabla(e_0),(1-\Phi)e_0)= \left(\frac{df}{f},\log^{(\sigma)}(f)\right) \end{align*}$$

as desired.

Proof. Because of the injectivity of (2.10) (Lemma 2.5), the assertion can be reduced to the compatibility of Kato’s symbol maps by Lemma 2.7.

Proof. Since $\ln ^{(p)}_r(z)$ has ${\mathbb {Z}}_p$ -coefficients, it is enough to show that $\ln ^{(p)}_r(z)\in (x-x^2){\mathbb {Q}}_p[x]^{\dagger }$ . We first note that

(2.15) $$ \begin{align} (x^2-x)\frac{d}{dx}\mathrm{ln}_{k+1}^{(p)}(z)=\mathrm{ln}_k^{(p)}(z). \end{align} $$

The limit in (2.14) can be rewritten as

$$\begin{align*}\lim_{s\to\infty}\frac{1}{x^{p^s}-(x-1)^{p^s}} \sum_{1\leq n<p^s,\,p\not{\hspace{2pt}|}\,n}\frac{x^{p^s}(1-x^{-1})^n}{n^r}. \end{align*}$$

This shows that $\ln ^{(p)}_r(z)$ vanishes at $x=0, 1$ . Therefore, it suffices to show that $\ln _r^{(p)}(z)\in {\mathbb {Q}}_p[x]^{\dagger }$ because then it is divisible by $x-x^2$ in ${\mathbb {Q}}_p[x]^{\dagger }$ by [Reference MacarroMa, Theorem 3.5].

Let $w(x)\in {\mathbb {Z}}_p[x]$ be defined by

$$\begin{align*}\frac{1-z^p}{(1-z)^p}=x^p-(x-1)^p=1-pw(x). \end{align*}$$

Then,

$$\begin{align*}\ln_1^{(p)}(z)=p^{-1}\log\left(\frac{1-z^p}{(1-z)^p}\right) =-p^{-1}\sum_{n=1}^{\infty}\frac{p^nw(x)^n}{n}\in {\mathbb{Z}}_p[x]^{\dagger}. \end{align*}$$

This shows that $\ln ^{(p)}_1(z)\in (x-x^2){\mathbb {Q}}_p[x]^{\dagger }$ , as required in case $r=1$ .

Now, let

$$\begin{align*}-(x-x^2)^{-1}\mathrm{ln}_1^{(p)}(z) =a_0+a_1x+a_2x^2+\cdots+a_nx^n+\cdots\in{\mathbb{Z}}_p[x]^{\dagger}. \end{align*}$$

By (2.15), one has

$$\begin{align*}\ln_2^{(p)}(z)=c+a_0x+\frac{a_1}{2}x^2+\cdots +\frac{a_n}{n+1}x^{n+1}+\cdots\in {\mathbb{Q}}_p[x]^{\dagger}, \end{align*}$$

and hence, $\ln _2^{(p)}(z)\in (x-x^2){\mathbb {Q}}_p[x]^{\dagger }$ , as required in case $r=2$ . Continuing the same argument, we obtain $\ln _r^{(p)}(z)\in (x-x^2){\mathbb {Q}}_p[x]^{\dagger }$ for every r.

Setting 2.14. $u\colon U\to S$ is a projective smooth morphism of smooth V-schemes with $S={\operatorname {Spec}}(B)$ .

Proof. The first statement follows from Berthelot’s result [Reference LazdaLaz, Theorem 4.1], [Reference BerthelotBer1, Théorème 5]. Now, since c is a morphism of coherent modules over the noetherian ring $B_K^{\dagger }$ , it suffices to prove that it is an isomorphism on the reduction by each maximal ideal of $B_K^{\dagger }$ , which is the extension of a maximal ideal of $B_K$ . Therefore, we may assume that $S=\overline {S}={\operatorname {Spec}}(k)$ (after a possible extension of k), and then the claim follows from comparison of the (absolute) algebraic de Rham cohomology and the rigid (or, in this case, crystalline) cohomology (e.g. [Reference AsakuraA-Bal, 4.2], [Reference GerkmannG, (7)]).

Setting 2.16. Let $S={\operatorname {Spec}}(B)$ be a smooth affine V-scheme and let $\overline {S}$ be a projective smooth V-scheme with an open immersion $S\hookrightarrow \overline {S}$ such that the complement T is a relative simple NCD on $\overline {S}$ over V. Let $\overline {X}$ be a projective smooth V-scheme and let $\overline {f}\colon \overline {X}\to \overline {S}$ be a projective morphism. Let $\overline {D}^{\mathrm {h}}$ be a relative NCD on $\overline {X}$ over V and put $\overline {D}^{\mathrm {v}}=\overline {f}^{-1}(T)$ and $\overline {D}=\overline {D}^{\mathrm {h}}\cup \overline {D}^{\mathrm {v}}$ . We put $X=\overline {X}\setminus \overline {D}^{\mathrm {v}}$ , $f=\overline {f}|_X$ and $U=\overline {X}\setminus \overline {D}$ . We then assume that the following conditions hold:

1. (1) $\overline {D}=\overline {D}^{\mathrm {h}}\cup \overline {D}^{\mathrm {v}}$ is also a relative NCD over V.

2. (2) $D:=\overline {D}^{\mathrm {h}}\cap X\hookrightarrow X$ is a relative NCD over S.

3. (3) The morphism $\overline {f}\colon (\overline {X},\overline {D})\to (\overline {S},T)$ is log smooth and integral, and $(\overline {S}, T)$ is of Zariski type.

Proof. Our setting assures us that we are in the situation of [Reference ShihoSh3, Section 2], i.e. the assumptions before [Reference ShihoSh3, Theorem 2.1] are satisfied. Moreover, since f is integral, the assumption ( $\star $ ) in [Reference ShihoSh3, Thereom 2.1] and ( $\star $ )’ in [Reference ShihoSh3, Theorem 2.3] are also satisfied [Reference ShihoSh2, Corollary 4.7]. Therefore, the first statement follows from [Reference ShihoSh3, Theorem 2.2] and [Reference ShihoSh3, Theorem 2.4].

Now that the coherence is proved for all i, the proof reduces to the absolute case as in the proof of Proposition 2.15. Then, since the algebraic de Rham cohomology (resp. the rigid cohomology) is isomorphic to the algebraic log de Rham cohomology (resp. log rigid cohomology by e.g. [Reference TsuzukiTz1, Theorem 3.5.1]), the claim follows from the comparison theorem between algebraic log de Rham cohomology and log rigid cohomology [Reference Baldassarri and ChiarellottoBal-Ch, Corollary 2.6].

Proof. First, $H^i(X/S)$ and $H^i(D/S)$ are objects of ${{\mathrm {Fil}\text {-}F\text {-}\mathrm {MIC}}}(S)$ by Proposition 2.15, and so is $H^1(U/S)$ by assumption. Next, it is a standard fact about de Rham cohomology that we have an exact sequence

$$\begin{align*}0\to H^1_{{{\mathrm{d\hspace{-0.2pt}R}}}}(X_K/S_K)\to H^1_{{{\mathrm{d\hspace{-0.2pt}R}}}}(U_K/S_K)\to H^0_{{{\mathrm{d\hspace{-0.2pt}R}}}}(D_K/S_K)(-1)\to H^2_{{{\mathrm{d\hspace{-0.2pt}R}}}}(X_K/S_K)\to 0 \end{align*}$$

whose morphisms are horizontal and compatible with respect to ${\mathrm {Fil}}$ . Therefore, by the comparison isomorphism on each term and by the flatness of $B_K^{\dagger }$ over $B_K$ , we get a corresponding exact sequence

$$\begin{align*}0\to H^1_{{\mathrm{rig}}}(X_k/S_k)\to H^1_{{\mathrm{rig}}}(U_k/S_k)\to H^0_{{\mathrm{rig}}}(D_k/S_k)(-1)\to H^2_{{\mathrm{rig}}}(X_k/S_k)\to 0 \end{align*}$$

for rigid cohomologies. The compatibility of this sequence with Frobenius structures on each term can be checked on each closed point of $\widehat {S}_K$ and therefore reduced to the absolute case [Reference BannaiBan2, Theorem 2.19].

Proof. This lemma is essentially given in the proof of [Reference StumLS, 6.2.12]. We recall the argument for the convenience for the reader. Choose a closed immersion $X\hookrightarrow {\mathbb {A}}_V^N$ to an affine space and let Y be the closure of X in $X\hookrightarrow {\mathbb {A}}_V^N\hookrightarrow {\mathbb {P}}_V^N$ . Then, $V_{\rho }:=X_K^{{\mathop {\mathrm {an}}\nolimits }}\cap \mathbb {B}^N(0,\rho ^+)$ for $\rho>1$ form a cofinal family of strict neighborhoods of $\widehat {X}_K$ .

By the coherence of ${\mathscr {M}}$ , we can take a coherent ${\mathscr {O}}_{V_{\rho _0}}$ -module M for some $\rho _0>1$ such that ${\mathscr {M}}|_{V_{\rho _0}}=j^{\dagger }_{X}M$ [Reference StumLS, 5.4.4]. Then, if $j_{\rho }\colon V_{\rho }\hookrightarrow V_{\rho _0}$ denotes the inclusion for $1<\rho <\rho _0$ , we have isomorphisms

$$\begin{align*}R\Gamma(X_K^{{\mathop{\mathrm{an}}\nolimits}}, {\mathscr{M}})=R\Gamma(V_{\rho_0}, {\mathscr{M}}|_{V_{\rho_0}})=\varinjlim_{\rho}R\Gamma(V_{\rho_0},(j_{\rho})_{\ast}M)=\varinjlim_{\rho}R\Gamma(V_{\rho},M|_{V_{\rho}}). \end{align*}$$

In fact, the first identification holds because ${\mathscr {M}}$ is a $j^{\dagger }_X{\mathscr {O}}_{X_K^{{\mathop {\mathrm {an}}\nolimits }}}$ -module by a standard argument (as in [Reference Baldassarri and BerthelotBal-Ber, 1.1]), the second one follows from quasi-compactness and separatedness of $V_{\rho _0}$ and the third one holds because $j_{\rho }$ is affinoid. Now, the claim follows because each $V_{\rho }$ is affinoid and $M|_{V_{\rho }}$ is coherent.

Assumption 2.20. For each $i\geq 0$ , the i-th relative rigid cohomology sheaf $R^iu_{{\mathrm {rig}}}j^{\dagger }_U{\mathscr {O}}_{U_K^{{\mathop {\mathrm {an}}\nolimits }}}$ is a coherent $j^{\dagger }_{S}{\mathscr {O}}_{S_K^{{\mathop {\mathrm {an}}\nolimits }}}$ -module with Frobenius structure and the natural morphism

(2.17) $$ \begin{align} H^i_{{{\mathrm{d\hspace{-0.2pt}R}}}}(U_K/S_K)\otimes_{B_K} B_K^{\dagger}{\longrightarrow} H^i_{{\mathrm{rig}}}(U_k/S_k) \end{align} $$

is bijective.

Proof. To prove this, it follows from Lemma 2.8 that we may assume $U={\operatorname {Spec}} V[T,T^{-1}]$ and $f=T$ for the vanishing $[f]_U\cup [f]_U=0$ and $U={\operatorname {Spec}} V[T,(T-T^2)^{-1}]$ and $f=T$ for the vanishing $[f]_U\cup [1-f]_U=0$ .

Here, we show the latter vanishing. Let $C=V[T,(T-T^2)^{-1}]$ and $U={\operatorname {Spec}}(C)$ . Recall the dilog object $D:={\mathop {\mathscr {D}\!\mathit {ilog}}\nolimits }(T)$ which has a unique increasing filtration $W_{\bullet }$ (as an object of ${\mathrm {Fil}}$ -F- ${\mathop {\mathrm {MIC}}\nolimits }(U)$ ) that satisfies

$$\begin{align*}W_jD_{{\mathrm{d\hspace{-0.2pt}R}}}=\begin{cases} 0 & \text{if } j\leq -5,\\ C_Ke_{-4} & \text{if } j=-4, -3,\\ C_Ke_{-4}\oplus C_Ke_{-2} & \text{if } j=-2, -1,\\ {\mathop{\mathscr{D}\!\mathit{ilog}}\nolimits}(T)_{{{\mathrm{d\hspace{-0.2pt}R}}}} & \text{if } j\geq 0 \end{cases} \end{align*}$$

and the filtration ${\mathrm {Fil}}^{\bullet }$ on $W_jD_{{\mathrm {d\hspace {-0.2pt}R}}}$ is given to be ${\mathrm {Fil}}^iW_jD_{{\mathrm {d\hspace {-0.2pt}R}}}=W_jD_{{\mathrm {d\hspace {-0.2pt}R}}}\cap {\mathrm {Fil}}^iD_{{{\mathrm {d\hspace {-0.2pt}R}}}}$ . Then, it is straightforward to check that $W_{-4}\cong {\mathscr {O}}_U(2), W_{-2}\cong {\mathop {\mathscr {L}\!\!\mathit {o}\mathit {g}}\nolimits }(T)(1), W_0/W_{-4}\cong {\mathop {\mathscr {L}\!\!\mathit {o}\mathit {g}}\nolimits }(1-T)$ and $W_0/W_{-2}\cong {\mathscr {O}}_U$ . Thus, $[T]_U\cup [1-T]_U$ is realized by the extension

$$\begin{align*}0{\longrightarrow} W_{-4}{\longrightarrow} W_{-2}{\longrightarrow} W_0/W_{-4}{\longrightarrow} W_0/W_{-2}{\longrightarrow} 0.\end{align*}$$

Consider a commutative diagram

(2.18)

where $\iota $ is the first inclusion, $\pi _1$ is the first projection, $\pi _2$ is the composition with the second projection and the inclusion $W_{-2}\hookrightarrow W_0$ , $\pi _3$ is the quotient $W_0\to W_0/W_{-2}\cong {\mathscr {O}}_U$ , and $\mathrm {add}\colon (x,y)\mapsto x+y$ . The above diagram shows the vanishing $[T]_U\cup [1-T]_U=0$ .

The proof of the vanishing $[T]_U\cup [T]_U=0$ goes in a similar way by replacing ${\mathop {\mathscr {D}\!\mathit {ilog}}\nolimits }(T)$ with $\mathrm {Sym}^2{\mathop {\mathscr {L}\!\!\mathit {o}\mathit {g}}\nolimits }(T)$ .

Proof. Write the homomorphism (2.21) by ${\widetilde {\rho }}$ . It is enough to show the following.

$$\begin{align*}{\widetilde{\rho}}(h_0{\otimes}\cdots{\otimes} f{\otimes} f{\otimes}\cdots{\otimes} h_n)=0,\quad {\widetilde{\rho}}(h_0{\otimes}\cdots{\otimes} f{\otimes} (1-f){\otimes}\cdots{\otimes} h_n)=0. \end{align*}$$

We show the latter. Let $f\in {\mathscr {O}}(U)^{\times }$ such that $1-f\in {\mathscr {O}}(U)^{\times }$ . Let $u:U\to {\operatorname {Spec}} V[T,(T-T^2)^{-1}]$ be the morphism given by $u^*T=f$ . Recall the diagram (2.18), and take the pullback by u. It follows that

$$\begin{align*}0\to{\mathscr{O}}_U(n+1)\to \overbrace{{\mathop{\mathscr{L}\!\!\mathit{o}\mathit{g}}\nolimits}(h_n)\to\cdots\to{\mathop{\mathscr{L}\!\!\mathit{o}\mathit{g}}\nolimits}(f)(i+1)\to{\mathop{\mathscr{L}\!\!\mathit{o}\mathit{g}}\nolimits}(1-f)(i)\to\cdots \to{\mathop{\mathscr{L}\!\!\mathit{o}\mathit{g}}\nolimits}(h_0)}^{{\mathscr{M}}}\to{\mathscr{O}}_U\to0 \end{align*}$$

is equivalent to

$$\begin{align*}0\to{\mathscr{O}}_U(n+1)\to \overbrace{{\mathop{\mathscr{L}\!\!\mathit{o}\mathit{g}}\nolimits}(h_n)\to\cdots\to{\mathscr{O}}_U(i+2){\overset{0}{\to}} \underbrace{{\mathscr{O}}_U(i)\to\cdots\to{\mathop{\mathscr{L}\!\!\mathit{o}\mathit{g}}\nolimits}(h_0)}_{\mathscr N}}^{{\mathscr{M}}'}\to{\mathscr{O}}_U\to0 \end{align*}$$

as an $(n+1)$ -extension in ${{\mathrm {Fil}\text {-}F\text {-}\mathrm {MIC}}}(U)$ . This implies that ${\mathscr {M}}$ is quasi-isomorphic to ${\mathscr {M}}'$ as mixed Hodge modules or F-isocrystals. Hence, $\mathscr N$ gives rise to a splitting of

$$\begin{align*}0{\longrightarrow} H^n(U/S)(n+1){\longrightarrow} M_{h_0,\ldots,f,1-f,\ldots,h_n}(U/S){\longrightarrow} {\mathscr{O}}_S{\longrightarrow}0. \end{align*}$$

This completes the proof of ${\widetilde {\rho }}(h_0{\otimes }\cdots {\otimes } f{\otimes } (1-f){\otimes }\cdots {\otimes } h_n)=0$ .

To see ${\widetilde {\rho }}(h_0{\otimes }\cdots {\otimes } f{\otimes } f{\otimes }\cdots {\otimes } h_n)=0$ , we consider the similar diagram to (2.18) obtained from $\mathrm {Sym}^2{\mathop {\mathscr {L}\!\!\mathit {o}\mathit {g}}\nolimits }(T)$ . Then, the rest is the same.

Proof. In this proof, we omit to write the symbol “ $\wedge $ ”.

First, we describe the extension $[\xi ]_{U/S}$ . Let

$$\begin{align*}0{\longrightarrow} H^n_{{\mathrm{d\hspace{-0.2pt}R}}}(U/S)(n+1){\overset{\iota}{{\longrightarrow}}} M_\xi(U/S){\longrightarrow} {\mathscr{O}}_S{\longrightarrow}0 \end{align*}$$

be the extension $[\xi ]_{U/S}$ in ${{\mathrm {Fil}\text {-}F\text {-}\mathrm {MIC}}}(S,\sigma )$ . Let $e_\xi \in {\mathrm {Fil}}^0M_\xi (U/S)_{{\mathrm {d\hspace {-0.2pt}R}}}$ be the unique lifting of $1\in {\mathscr {O}}(S_K)$ . Then,

(2.27) $$ \begin{align} \iota (D_{U/S}(\xi))=\nabla(e_\xi),\quad \iota({\mathrm{reg}}_{U/S}^{(\sigma)}(\xi))=(1-\Phi)e_\xi \end{align} $$

by definition (see also (2.11)), where $\nabla $ and $\Phi $ are the data in $M_\xi (U/S)\in {{\mathrm {Fil}\text {-}F\text {-}\mathrm {MIC}}}(S,\sigma )$ .

We first write down the term $M_\xi (U/S)$ explicitly. Write $l_i:=p^{-1}\log (h_i^p/h_i^{\varphi })$ . Let $\{e_{i,0},e_{i,-2}\}$ be the basis of ${{\mathscr {L}{og}}}(h_i)(i)$ such that ${\mathrm {Fil}}^{-i}{\mathop {\mathscr {L}\!\!\mathit {o}\mathit {g}}\nolimits }(h_i)(i)_{{\mathrm {d\hspace {-0.2pt}R}}}=A_Ke_{i,0}$ and

$$\begin{align*}\nabla(e_{i,0})=\frac{dh_i}{h_i} e_{i,-2},\quad \Phi(e_{i,0})=p^{-i}e_{i,0}-p^{-i}l_ie_{i,-2},\quad \Phi(e_{i,-2})=p^{-i-1}e_{i,-2}.\end{align*}$$

Recall the $(n+1)$ -extension

$$\begin{align*}0\to {\mathscr{O}}_{U}(n+1)\to {{\mathscr{L}{og}}}(h_n)(n)\to\cdots\to{{\mathscr{L}{og}}}(h_1)(1)\to{{\mathscr{L}{og}}}(h_0)\to{\mathscr{O}}_{U}\to0. \end{align*}$$

Let $(T^{\bullet }_{A_K/B_K},D)$ be the total complex of the double complex $\Omega ^{\bullet }_{A_K/B_K}{\otimes } {{\mathscr {L}{og}}}(h_\star )_{{\mathrm {d\hspace {-0.2pt}R}}}$ . In a more down-to-earth manner, we have

$$\begin{align*}T^q_{A_K/B_K}:=\bigoplus_{i=0}^d\Omega^i_{A_K/B_K}{\otimes}{{\mathscr{L}{og}}}(h_{i-q})(i-q)_{{\mathrm{d\hspace{-0.2pt}R}}},\quad q\in{\mathbb{Z}}, \end{align*}$$

where we denote ${{\mathscr {L}{og}}}(h_j)(j):=0$ if $j<0$ or $j>n$ , and the differential $D:T^q\to T^{q+1}$ is defined by

$$\begin{align*}D(\omega^i{\otimes} x_j)=d\omega^i{\otimes} x_j+(-1)^i\omega^i\wedge \nabla(x_j)+(-1)^i\omega^i{\otimes} \pi(x_j) \end{align*}$$

for $\omega ^i{\otimes } x_j\in \Omega ^i_{A_K/B_K}{\otimes }{{\mathscr {L}{og}}}(h_j)(j)_{{\mathrm {d\hspace {-0.2pt}R}}}$ , where $\pi \colon {{\mathscr {L}{og}}}(h_i)(i)\to {{\mathscr {L}{og}}}(h_{i-1})(i-1)$ is the composite of the projection ${{\mathscr {L}{og}}}(h_i)(i)\to A_Ke_{i,0}$ and the injection $A_Ke_{i,0}\cong A_Ke_{i-1,-2}\hookrightarrow {{\mathscr {L}{og}}}(h_{i-1})(i-1)$ defined by $e_{i,0}\mapsto e_{i-1,-2}$ . We have the exact sequence

(2.28)

where the first arrow is induced from ${\mathscr {O}}_{U_K}\cong {\mathscr {O}}_{U_K}e_{d,-2}{\hookrightarrow } {{\mathscr {L}{og}}}(h_d)_{{\mathrm {d\hspace {-0.2pt}R}}}$ , the second arrow is induced from the projection ${{\mathscr {L}{og}}}(h_0)_{{\mathrm {d\hspace {-0.2pt}R}}}\to {\mathscr {O}}_{U_K}e_{0,0}\cong {\mathscr {O}}_{U_K}$ and the differential on $\Omega ^{\bullet +n}_{A_K/B_K}$ is the usual differential operator d (not $(-1)^nd$ ). For the rigid part, let $(T_{A^{\dagger }_K/B^{\dagger }_K}^{\bullet },D)$ be defined in the same way by replacing $\Omega ^{\bullet }_{A_K/B_K}$ with $\Omega ^{\bullet }_{A^{\dagger }_K/B^{\dagger }_K}$ (see Remark 2.21 (3)). Then, we also have an exact sequence corresponding to (2.28). Now, we have a description

$$\begin{align*}M_\xi(U/S)_{{\mathrm{d\hspace{-0.2pt}R}}}=H^0(T^{\bullet}_{A_K/B_K}),\quad M_\xi(U/S)_{\mathrm{rig}}=H^0(T^{\bullet}_{A^{\dagger}_K/B^{\dagger}_K}) \end{align*}$$

(the description of the rigid part follows from the exact sequence (2.28) on two sides with Remark 2.21 (3)), and we have an exact sequence

in ${{\mathrm {Fil}\text {-}F\text {-}\mathrm {MIC}}}(S,\sigma )$ .

Before going to the proof of Theorem 2.25, we give an explicit description of $e_\xi $ in (2.27). Put

$$\begin{align*}\omega^0:=1,\qquad \omega^i:=\frac{dh_0}{h_0}\frac{dh_1}{h_1}\cdots \frac{dh_{i-1}}{h_{i-1}}\in \Omega^i_{A_K/B_K}\quad (1\leq i\leq n+1). \end{align*}$$

However, we note that $\omega ^{n+1}=\omega ^n\wedge \frac {dh_n}{h_n}=0$ as $\omega ^n \in {\mathrm {Fil}}^{n+1}H^{n+1}_{{\mathrm {d\hspace {-0.2pt}R}}}(U_K/S_K)=0$ . Put

$$\begin{align*}e^{\prime}_\xi=\sum_{i=0}^{n}\omega^ie_{i,0}\in T^0_{A_K/B_K}. \end{align*}$$

It is a direct computation to show $D(e^{\prime }_\xi )=0$ , so that one has $e^{\prime }_\xi \in M_\xi (U/S)_{{\mathrm {d\hspace {-0.2pt}R}}}$ . This is obviously a lifting of $1\in {\mathscr {O}}(S_K)$ . We claim

(2.29) $$ \begin{align} e_\xi=e^{\prime}_\xi. \end{align} $$

To do this, it is enough to show that $e^{\prime }_\xi \in {\mathrm {Fil}}^0M_\xi (U/S)_{{\mathrm {d\hspace {-0.2pt}R}}}$ . Let $j:U_K{\hookrightarrow } X_K$ . We think $T^{\bullet }_{A_K/B_K}$ of being a complex of ${\mathscr {O}}_{U_K}$ -modules. Let ${\mathrm {Fil}}^0{\mathop {\mathscr {L}\!\!\mathit {o}\mathit {g}}\nolimits }(h_i)_{X_K}={\mathscr {O}}_{X_K} e_{i,0}$ , ${\mathrm {Fil}}^k{\mathop {\mathscr {L}\!\!\mathit {o}\mathit {g}}\nolimits }(h_i)_{X_K}={\mathscr {O}}_{X_K} e_{i,0}+{\mathscr {O}}_{X_K} e_{i,-2}$ for $k<0$ , and ${\mathrm {Fil}}^k{\mathop {\mathscr {L}\!\!\mathit {o}\mathit {g}}\nolimits }(h_i)_{X_K}=0$ for $k>0$ . Let ${\mathscr T}^{\bullet }_{U/S}\subset j_*T^q_{A_K/B_K}$ be the subcomplex of locally free ${\mathscr {O}}_{X_K}$ -modules such that

$$\begin{align*}{\mathscr T}^q_{U/S}=\bigoplus_{i=0}^d\Omega^i_{X_K/S_K}(\log D_K){\otimes} {{\mathscr{L}{og}}}(h_{i-q})_{X_K}\subset j_*T^q_{A_K/B_K}, \end{align*}$$

and put

$$\begin{align*}{\mathrm{Fil}}^k{\mathscr T}^q_{U/S}=\bigoplus_{i=0}^d\Omega^i_{X_K/S_K}(\log D_K){\otimes} {\mathrm{Fil}}^{k-q}{{\mathscr{L}{og}}}(h_{i-q})_{X_K}\subset {\mathscr T}^q_{U/S}. \end{align*}$$

Then,

$$\begin{align*}{\mathrm{Fil}}^kM_\xi(U/S)_{{\mathrm{d\hspace{-0.2pt}R}}}={{\mathrm{Im}}}[H^0(X_K,{\mathrm{Fil}}^k{\mathscr T}_{U/S}^{\bullet}) \to H^0(X_K,{\mathscr T}_{U/S}^{\bullet})\cong M_\xi(U/S)]. \end{align*}$$

Notice that ${\mathrm {Fil}}^0{\mathscr T}^q_{U/S}=0$ for $q<0$ as ${\mathrm {Fil}}^k{\mathop {\mathscr {L}\!\!\mathit {o}\mathit {g}}\nolimits }(h_i)_{X_K}=0$ for $k>0$ . Hence, there is a natural map

$$\begin{align*}{\varGamma}(X_K,\ker[{\mathrm{Fil}}^0{\mathscr T}^0_{U/S}\overset{D}{\to} {\mathrm{Fil}}^0{\mathscr T}^1_{U/S}]) {\longrightarrow} H^0(X_K,{\mathrm{Fil}}^k{\mathscr T}_{U/S}^{\bullet}), \end{align*}$$

and the element $e^{\prime }_\xi $ lies in the image of the left-hand side by definition. This shows $e^{\prime }_\xi \in {\mathrm {Fil}}^0M_\xi (U/S)_{{\mathrm {d\hspace {-0.2pt}R}}}$ and hence completes the proof of (2.29).

We prove (2.26). Apply $1-\Phi $ on (2.29). We have

(2.30) $$ \begin{align} (1-\Phi)e_\xi=\sum_{i=0}^n \omega^ie_{i,0}-\varphi_i(\omega^i)(e_{i,0}-l_ie_{i,-2}) =\sum_{i=0}^n (1-\varphi_i)(\omega^i)e_{i,0}+l_i\varphi_i(\omega^i)e_{i,-2}. \end{align} $$

Put

$$\begin{align*}Q_0:=l_0,\qquad Q_k:=\sum_{i=0}^k(-1)^il_i \left(\frac{dh_0}{h_0}\right)^{\varphi_1}\cdots\left(\frac{dh_{i-1}}{h_{i-1}}\right)^{\varphi_1} \frac{dh_{i+1}}{h_{i+1}}\cdots\frac{dh_k}{h_k}\quad (1\leq k\leq n). \end{align*}$$

A direct calculation yields

$$\begin{align*}dQ_k=(1-\varphi_{k+1})\left(\frac{dh_0}{h_0}\cdots\frac{dh_k}{h_k}\right) =(1-\varphi_{k+1})(\omega^{k+1}). \end{align*}$$

It follows

$$ \begin{align*} &D(Q_ke_{k+1,0})\\ &=(1-\varphi_{k+1})(\omega^{k+1}) e_{k+1,0}+(-1)^kQ_k\frac{dh_{k+1}}{h_{k+1}}e_{k+1,-2} +(-1)^kQ_ke_{k,-2}\\ &=(1-\varphi_{k+1})(\omega^{k+1})e_{k+1,0} +(-1)^k(Q_{k+1}-(-1)^{k+1}l_{k+1}\varphi_{k+1}(\omega^{k+1})) e_{k+1,-2} +(-1)^kQ_ke_{k,-2}\\ &= [(1-\varphi_{k+1})(\omega^{k+1})e_{k+1,0}+l_{k+1}\varphi_{k+1}(\omega^{k+1})e_{k+1,-2}] +(-1)^kQ_ke_{k,-2}- (-1)^{k+1}Q_{k+1}e_{k+1,-2} \end{align*} $$

for $0\leq k\leq n-1$ . Hence,

$$ \begin{align*} D\left(\sum_{k=0}^{n-1}Q_ke_{k+1,0}\right)&= (1-\Phi)(e_\xi) -l_0e_{0,-2}+(-1)^n(1-\varphi_{n+1})(\eta)e_{n,-2} +Q_0e_{0,-2}-(-1)^nQ_ne_{n,-2}\\ &=(1-\Phi)(e_\xi)-(-1)^nQ_ne_{n,-2} \end{align*} $$

by (2.30). This shows (cf. (2.27))

$$\begin{align*}{\mathrm{reg}}^{(\sigma)}_{U/S}(\xi)=(-1)^nQ_n =(-1)^n\sum_{i=0}^n(-1)^il_i \left(\frac{dh_0}{h_0}\right)^{\varphi_1}\cdots\left(\frac{dh_{i-1}}{h_{i-1}}\right)^{\varphi_1} \frac{dh_{i+1}}{h_{i+1}}\cdots\frac{dh_n}{h_n}, \end{align*}$$

as required.

We prove (2.25). Note that $D_{U/S}(\xi )$ is characterized by $D_{U/S}(\xi )e_{n,-2}=\nabla (e_\xi )$ , where $\nabla $ is the Gauss-Manin connection on $M_\xi (U/S)_{{\mathrm {d\hspace {-0.2pt}R}}}$ (cf. (2.27)). Let $T_{A_K}^{\bullet }$ be the complex defined in the same way as $T_{A_K/B_K}^{\bullet }$ by replacing $\Omega ^{\bullet }_{A_K/B_K}$ with $\Omega ^{\bullet }_{A_K}$ . Let

$$\begin{align*}{\widetilde{\omega}}^0:=1,\qquad {\widetilde{\omega}}^i:=\frac{dh_0}{h_0}\frac{dh_1}{h_1}\cdots \frac{dh_{i-1}}{h_{i-1}}\in \Omega^i_{A_K}\quad (1\leq i\leq n+1), \end{align*}$$

and $\widetilde {e}_\xi :=\sum _{i=0}^{n}{\widetilde {\omega }}^ie_{i,0}\in T^0_{A_K}$ . Then,

$$\begin{align*}D(\widetilde{e}_\xi)=\sum_{i=0}^n(-1)^i\left( \frac{dh_0}{h_0}\cdots\frac{dh_i}{h_i}e_{i,-2}+ \frac{dh_0}{h_0}\cdots\frac{dh_{i-1}}{h_{i-1}}e_{i-1,-2}\right) =(-1)^n\frac{dh_0}{h_0}\cdots\frac{dh_n}{h_n}e_{n,-2} \end{align*}$$

and this shows (cf. (2.25)),

(2.31) $$ \begin{align} \nabla(e_\xi)=(-1)^n\frac{dh_0}{h_0}\cdots\frac{dh_n}{h_n}e_{n,-2} \in \Omega_{B_K}^1{\otimes}{\varGamma}(X_K,\Omega^n_{X_K/S_K}(\log D_K))e_{n,-2} \end{align} $$

as required.

Condition 3.3.

1. (1) For each $n\geq 1$ , $(Z_n,N_n)$ has p-bases over $(S_n,L_n)$ locally,

2. (2) Let $(D_n,M_{D_n})\to (Z_n,N_n)$ be the DP-envelope of $(Y_n,M_n)$ compatible with the DP-structure on $(S_n,L_n)$ , and let $J^{[r]}_{D_n}\subset {\mathscr {O}}_{D_n}$ be the r-th DP-ideal. Then, $J^{[r]}_{D_n}$ is flat over ${\mathbb {Z}}/p^n{\mathbb {Z}}$ and $J^{[r]}_{D_{n+1}}{\otimes }{\mathbb {Z}}/p^n{\mathbb {Z}}\cong J^{[r]}_{D_n}$ .

Condition 3.4. There is a hypercovering $\{Y_n^{\star }\}_n\to \{Y_n\}_n$ in the etale topology which admits an embedding $\{(Y^{\star }_n,M_n^{\star })\}_n{\hookrightarrow } \{(Z^{\star }_n,N^{\star }_n)\}_n$ into an adic inductive system of simplicial log schemes over $\{(S_n,L_n)\}_n$ , such that each $\{(Y^{\nu }_n,M_n^{\nu })\}_n{\hookrightarrow } \{(Z^{\nu }_n,N^{\nu }_n)\}_n$ satisfies Condition 3.3, and there is a p-th Frobenius $\{\varphi ^{\nu }_{Z_n}:(Z^{\nu }_n,N^{\nu }_n)\to (Z^{\nu }_n,N^{\nu }_n)\}_n$ compatible with $\sigma $ .

Proof. Since $(Y,M)\to (S,L)$ is smooth, it has p-bases locally by Lemma 3.2. Therefore, the natural morphism (3.1) is a quasi-isomorphism, and the exact sequence follows.

Proof. For an abelian group M and an integer n, we denote by $M[n]$ the kernel of the multiplication by n. An exact sequence

$$\begin{align*}0{\longrightarrow} \omega^{\bullet\geq k}_{Y/S}{\overset{p^n}{{\longrightarrow}}} \omega^{\bullet\geq k}_{Y/S} {\longrightarrow} \omega^{\bullet\geq k}_{Y_n/S_n}{\longrightarrow}0 \end{align*}$$

gives rise to an exact sequence

$$\begin{align*}0{\longrightarrow} H^i_{\mathrm{zar}}(Y,\omega^{\bullet\geq k}_{Y/S})/p^n {\longrightarrow} H^i_{\mathrm{zar}}(Y_n,\omega^{\bullet\geq k}_{Y_n/S_n}) {\longrightarrow} H^{i+1}_{\mathrm{zar}}(Y,\omega^{\bullet\geq k}_{Y/S})[p^n]{\longrightarrow} 0 \end{align*}$$

of finitely generated ${\mathscr {O}}(S)$ -modules as $Y\to S$ is proper. Therefore, the assertion is reduced to show that, for any p-torsion free noetherian ring A and any finitely generated A-module M,

(3.3) $$ \begin{align} A^{\wedge}{\otimes}_AM{\overset{\cong}{{\longrightarrow}}}\varprojlim_n M/p^nM, \end{align} $$

where the transition map $M/p^{n+1}\to M/p^n$ is the natural surjection, and

(3.4) $$ \begin{align} \varprojlim_n M[p^n]=0, \end{align} $$

where the transition map $M[p^{n+1}]\to M[p^n]$ is multiplication by p. The isomorphism (3.3) is well-known ([Reference AsakuraA-M, Prop. 10.13]). We show (3.4). Let

$$\begin{align*}0{\longrightarrow} M_1{\longrightarrow} M_2{\longrightarrow} M_3{\longrightarrow}0 \end{align*}$$

be any exact sequence of finitely generated A-modules. Then,

$$\begin{align*}0\to M_1[p^n]\to M_2[p^n]\to M_3[p^n]\to M_1/p^n\to M_2/p^n\to M_3/p^n\to0 \end{align*}$$

is exact. Suppose that $M_2[p^n]=0$ for all n. Then, by taking the projective limit, we have

Since $A\to A^{\wedge }$ is flat ([Reference AsakuraA-M, Prop. 10.14]), the bottom arrow is injective, and hence, the vanishing $\varprojlim _n M_3[p^n]=0$ follows. For the proof of (3.4), apply this to $M_3=M$ and $M_2$ a free B-module of finite rank.

Proof. For $a_n\in V(W_n)$ , we define $\phi (a_n)$ to be the morphism which makes the following diagram commutative:

One easily verifies that $\phi (a_n)$ is a W-morphism (i.e. $\phi (a_n)\in V(W_n)$ ) and commutes with the reduction map $\rho _n:V(W_n)\to V(W_{n-1})$ . Put

$$\begin{align*}V(W_n)_{b_{n-1}}:=\{a_n\in V(W_n)\mid \rho_n(a_n)=b_{n-1}\} \end{align*}$$

for $b_{n-1}\in V(W_{n-1})$ . This is a nonempty set as the reduction map $\rho _n$ is surjective (formal smoothness property). We claim that a map

$$\begin{align*}\phi:V(W_n)_{b_{n-1}}\to V(W_n)_{\phi(b_{n-1})} \end{align*}$$

is a constant map. Indeed, let $C_n=W_n[t_1,\ldots ,t_m]/I$ . One can write $\sigma _n(t_i)=t^p_i+pf_i(t)$ . Let $a_n$ be given by $t_i\mapsto \alpha _i$ . Then, $\phi (a_n)$ is given by the ring homomorphism

$$\begin{align*}t_i\longmapsto F_W^{-1}(\alpha_i^p+pf(\alpha_1,\ldots,\alpha_m))\quad \mod p^nW. \end{align*}$$

which depends only on $\{\alpha _i$ mod $p^{n-1}W\}_i$ , namely on $\rho _n(a_n)=b_{n-1}$ . This means that $\phi (a_n)$ is constant on the set $V(W_n)_{b_{n-1}}$ .

We prove Lemma 3.8. We have shown that if $b_{n-1}\in V(W_{n-1})$ satisfies $\phi (b_{n-1})=b_{n-1}$ , then $\phi ^m(a_n)=\phi (a_n)$ for all $a_n\in V(W_n)_{b_{n-1}}$ and $m\geq 1$ . Let $a_1\in V(W_1)$ be an arbitrary element and take an arbitrary sequence

$$\begin{align*}(a_1,c_2,\ldots,c_n,\ldots)\in \varprojlim_n V(W_n)=V(W). \end{align*}$$

Since $\phi $ is the identity on $V(W_1)$ , one has $\phi ^2(c_2)=\phi (c_2)$ as $c_2\in V(W_2)_{a_1}$ . Then, $\phi ^3(c_3)=\phi ^2(c_3)$ as $\phi (c_3)\in V(W_3)_{\phi (c_2)}$ . Continuing this, $a_n:=\phi ^{n-1}(c_n)$ satisfies that $\phi (a_n)=a_n$ in $V(W_n)$ and $\rho _n(a_n)=\phi ^{n-1}(c_{n-1})=\phi ^{n-2}(c_{n-1})=a_{n-1}\in V(W_{n-1})$ . Hence, the sequence $\{a_n\}_n$ defines a W-valued point $a=(a_n)\in V(W)$ which makes a diagram

commutative for all $n\geq 1$ . We show that such a is unique. Suppose that another W-valued point $a'=(a_1,a^{\prime }_2,\ldots ,a^{\prime }_n,\ldots )$ makes the above diagram commutative. This means $\phi (a^{\prime }_n) =a^{\prime }_n$ for all $n\geq 1$ . Since $\phi $ is a constant map on $V(W_n)_{a^{\prime }_{n-1}}$ , the point $a^{\prime }_n=\phi (a^{\prime }_n)$ is uniquely determined by $a^{\prime }_{n-1}$ , and hence by $a_1$ . Therefore, $a'$ is uniquely determined by $a_1$ , which means $a=a'$ .

Proof. Since there is an inclusion $H_K{\hookrightarrow } E_K$ into a free $\widehat {C}_K$ -module of finite rank which has a splitting, we may replace $H_K$ with $E_K$ , and hence, we may assume that $H_K$ is a free $\widehat {C}_K$ -module. Then, one can reduce the proof to the case of rank one, i.e. $H_K=\widehat {C}_K$ . Let $x\in \widehat {C}_K$ satisfy $x|_{P_a}=0$ for all $P_a$ . Suppose that $x\ne 0$ . There is an integer n such that $x\in p^n\widehat {C}\setminus p^{n+1}\widehat {C}$ . Write $x=p^n y$ with $y\in \widehat {C}\setminus p\widehat {C}$ that satisfies $y|_{P_a}=0$ for all $a\in V(k)$ . Then, y is zero in $\widehat {C}/p\widehat {C}\cong C/pC$ by Hilbert nullstellensatz, namely $y\in p\widehat {C}$ . This is a contradiction.

Proof. Write $U_K=U\times _WK$ and $(Y_K,M_K)=(Y,M)\times _W K$ . There is a canonical isomorphism

$$\begin{align*}R{\varGamma}_{\text{log-syn}}(U,{\mathbb{Q}}_p(j))\cong \text{Cone}[{\mathrm{Fil}}^jR{\varGamma}_{\text{log-dR}}((Y_K,M_K))\xrightarrow{1-p^{-j}\phi_{\mathrm{crys}}} R{\varGamma}_{\text{log-crys}}((Y_1,M_1)/K)][-1] \end{align*}$$

arising from (3.1) (see also Proposition 3.5), where $\phi _{\mathrm {crys}}$ is the p-th Frobenius which is defined thanks to the comparison of log de Rham and log crystalline cohomology. There is also a canonical isomorphism ([Reference BesserBes1, Remark 8.7, 3])

(3.12) $$ \begin{align} R{\varGamma}_{{{\mathrm{rig}\text{-}\mathrm{syn}}}}(U,{\mathbb{Q}}_p(j))\cong \text{Cone}[{\mathrm{Fil}}^jR{\varGamma}_{{\mathrm{d\hspace{-0.2pt}R}}}(U_K/K)\xrightarrow{1-p^{-j}\phi_{\mathrm{rig}}}R{\varGamma}_{\mathrm{rig}}(U_1/k)][-1], \end{align} $$

where $\phi _{\mathrm {rig}}$ is the p-th Frobenius which is defined thanks to the comparison of the algebraic de Rham cohomology and the rigid cohomology. Now, the isomorphism c is induced by the comparison morphism between the rigid and the log crystalline cohomology [Reference ShihoSh1] and the one between the de Rham cohomology and the log de Rham cohomology.

By the construction, c is compatible with respect to the cup-product and the period map to the etale cohomology,

where $\rho _{\mathrm {syn}}$ is as in [Reference TsujiTs1, §3.1], and $\rho _{{{\mathrm {rig}\text {-}\mathrm {syn}}}}$ is as in [Reference BesserBes1, Corollary 9.10]. Moreover, both $\rho _{\mathrm {syn}}\circ [-]_{\mathrm {syn}}$ and $\rho _{{{\mathrm {rig}\text {-}\mathrm {syn}}}}\circ {\mathrm {reg}}_{{{\mathrm {rig}\text {-}\mathrm {syn}}}}$ agree with the etale symbol map ([Reference TsujiTs1, Proposition 3.2.4], [Reference BesserBes1, Corollary 9.10]). We show the commutativity of the diagram (3.11). Thanks to the compatibility with respect to the cup-product, one can reduce the assertion to the case $i=1$ ; namely, it is enough to show that for $f\in {\mathscr {O}}(U)^{\times }$ an element $u:=c[f]_{\mathrm {syn}}-{\mathrm {reg}}_{{\mathrm {rig}\text {-}\mathrm {syn}}}(f)$ is zero. There is an exact sequence

from (3.12). By the construction of c and the definition of $[-]_{\mathrm {syn}}$ in (3.8) and [Reference BesserBes1, Def.6.5 and Prop.10.3], one has $h(u)=df/f-df/f=0$ . Therefore, u lies in the image of $H^0_{{\mathrm {d\hspace {-0.2pt}R}}}(U_K/K)$ . We claim that the composition of the maps

is injective. We may assume that U is connected. Moreover, we may replace W with $W(\overline {k})$ , so that we may further assume that U has a W-valued point. Then, $H^0_{{\mathrm {d\hspace {-0.2pt}R}}}(U_K/K)\cong K$ and the injectivity can be reduced to the case $U={\operatorname {Spec}} W$ , which can be easily verified. We turn to the proof of $u=0$ . It is enough to show $\rho _{{\mathrm {rig}\text {-}\mathrm {syn}}}(u)=0$ . However,

$$ \begin{align*} \rho_{{\mathrm{rig}\text{-}\mathrm{syn}}}(u)&=\rho_{{\mathrm{rig}\text{-}\mathrm{syn}}}({\mathrm{reg}}_{{\mathrm{rig}\text{-}\mathrm{syn}}}(f))-\rho_{{\mathrm{rig}\text{-}\mathrm{syn}}}(c[f]_{\mathrm{syn}})\\ &=\rho_{{\mathrm{rig}\text{-}\mathrm{syn}}}({\mathrm{reg}}_{{\mathrm{rig}\text{-}\mathrm{syn}}}(f))-\rho_{\mathrm{syn}}([f]_{\mathrm{syn}})=[f]_{{\mathrm{\acute{e}t}}}-[f]_{{\mathrm{\acute{e}t}}}=0, \end{align*} $$

where $[-]_{{\mathrm {\acute {e}t}}}$ is the etale symbol map, so we are done.

Proof of Theorem 3.7

By replacing W with $W(\overline {k})$ , we may assume that k is an algebraically closed field. We fix a p-th Frobenius $\varphi $ on ${\mathscr {O}}(U)^{\dagger }$ compatible with $\sigma $ . Recall the diagram (3.9). Let $\xi \in K_{n+1}^M({\mathscr {O}}(U))$ and let

$$\begin{align*}\langle\xi\rangle_{U/S},\quad \langle\xi\rangle_{\mathrm{syn}} \in {\mathscr{O}}(S)_K^{\wedge}{\otimes}_{{\mathscr{O}}(S_K)}H^n_{{\mathrm{d\hspace{-0.2pt}R}}}(U_K/S_K) \end{align*}$$

be the elements sent along the diagram in a counterclockwise direction and a clockwise direction, respectively. We want to show $\langle \xi \rangle _{U/S}=(-1)^n\langle \xi \rangle _{\mathrm {syn}}$ . For a closed point $a\in S_1(k)$ , we take the unique lifting $P_a\in S(W)$ as in Lemma 3.8. By Lemma 3.9, it is enough to show

$$\begin{align*}\langle\xi\rangle_{U/S}|_{P_a}= \langle\xi\rangle_{\mathrm{syn}}|_{P_a}\in \kappa(P_a){\otimes}_{B_K}H^n_{{\mathrm{d\hspace{-0.2pt}R}}}(U_K/S_K^*) \end{align*}$$

for every a, where $\kappa (P_a)\cong K$ denotes the residue field of $P_a$ . Therefore, to show the $(-1)^n$ -commutativity of the diagram (3.9), we may specialize the diagram at $P_a$ so that the proof is reduced to the case $(Q,L,\sigma )=({\operatorname {Spec}} W,W^{\times },F_W)$ . Summing up the above, the proof of Theorem 3.7 is reduced to showing $(-1)^n$ -commutativity of the following diagram:

(3.13)

Here, Y is a projective smooth scheme over W, D a relative NCD over W, $U=Y\setminus D$ and M is the log structure on Y defined by D. The extension group is taken in the category of ${{\mathrm {Fil}\text {-}F\text {-}\mathrm {MIC}}}({\operatorname {Spec}} W,F_W)$ .

To compare $[-]_{U/W}$ and $[-]_{\mathrm {syn}}$ , we use the explicit formula for $[-]_{U/W}$ (Theorem 2.25) and the theory of rigid syntomic regulators by Besser [Reference BesserBes1]. Consider a diagram

(3.14)

where ${\mathrm {reg}}_{U/W}:=\phi _{F_W}\circ [-]_{U/W}$ . Here, the isomorphism i follows from [Reference BesserBes1, (8.5)] and c is the canonical isomorphism (3.10) in Lemma 3.10. The commutativity of the right upper triangle is proven in Lemma 3.10, and that of the right lower square is immediate from the construction of c. We show the commutativity of the left. Let $\xi =\{h_0,\ldots ,h_n\}\in K^M_{n+1}({\mathscr {O}}(U))$ . Then, using [Reference BesserBes1, Def.6.5 and Prop.10.3], one can show that

(3.15) $$ \begin{align} {\mathrm{reg}}_{\text{rig-syn}}(\xi)= \sum_{i=0}^n(-1)^ip^{-1}\log\left(\frac{h_i^p}{h_i^{\varphi}}\right) \left(\frac{dh_0}{h_0}\right)^{\varphi_1}\wedge\cdots\wedge\left(\frac{dh_{i-1}}{h_{i-1}}\right)^{\varphi_1}\wedge \frac{dh_{i+1}}{h_{i+1}}\wedge\cdots\wedge\frac{dh_n}{h_n} \end{align} $$

under the inclusion $H^n_{{\mathrm {d\hspace {-0.2pt}R}}}(U_K/K){\hookrightarrow } \Omega ^n_{A_K/K}/d\Omega ^{n-1}_{A_K/K}\cong \Omega ^n_{A^{\dagger }_K/K}/d\Omega ^{n-1}_{A^{\dagger }_K/K}$ in a similar way to the proof of [Reference KatoKa1, Cor. 2.9]. This agrees with $(-1)^n{\mathrm {reg}}_{U/W}(\xi )$ by Theorem 2.25, and hence, the commutativity of the left square follows. This completes the proof of $(-1)^n$ -commutativity of the diagram (3.13), and hence Theorem 3.7.

Proof. The conditions imply that the morphism

$$\begin{align*}g:(\overline{Z},\overline{Z}\setminus V){\longrightarrow} (\overline{C}, T')\end{align*}$$

of log schemes is smooth so that $V/C$ is adapted to Setting 2.16. Therefore, $R^iv_{{\mathrm {rig}}}j^{\dagger }_{V}{\mathscr {O}}_{V_K^{{\mathop {\mathrm {an}}\nolimits }}}$ is a coherent $j^{\dagger }_C{\mathscr {O}}_{C_K^{{\mathop {\mathrm {an}}\nolimits }}}$ -module for each $i\geq 0$ by Proposition 2.17.

Proof. We first note that $K_2^M({\mathscr {O}}(U))$ is generated by symbols $\{f,g\}$ with ${\mathrm {ord}}_{D_X}(f)=0$ . Suppose ${\mathrm {ord}}_{D_X}(f)=0$ , namely $f\in {\mathscr {O}}(U\cup D_X)^{\times }$ . The natural morphism

$$\begin{align*}p:{\mathop{\mathrm{Cone}}\nolimits}\left[{\mathop{\mathscr{L}\!\!\mathit{o}\mathit{g}}\nolimits} f(1)\to{\mathop{\mathscr{L}\!\!\mathit{o}\mathit{g}}\nolimits} (g)\right][-1]\to {\mathop{\mathscr{L}\!\!\mathit{o}\mathit{g}}\nolimits} f(1)\end{align*}$$

and the residue map induce the following commutative diagram:

(4.3)

with exact rows where $\delta _{f,g}(1)=\frac {df}{f}\frac {dg}{g}$ , $\delta _f(1)=\frac {df}{f}$ and $\delta _{f|D_X}(1)=\frac {df}{f}|_{D_X}$ . The maps “Res”s are residue maps appearing in the Gysin exact sequence; as for the middle one, we are considering the Gysin exact sequence with coefficient in ${\mathop {\mathscr {L}\!\!\mathit {o}\mathit {g}}\nolimits }(f)(1)$ which is constructed in the same manner as Proposition 2.18. We show that $p'$ agrees with the map $\delta _g:1\mapsto \frac {dg}{g}$ . Let $[A\to B]$ denote the complex

$$\begin{align*}\cdots{\longrightarrow} 0{\longrightarrow} A{\longrightarrow} B{\longrightarrow} 0{\longrightarrow}\cdots\end{align*}$$

with A placed in degree zero. Consider a commutative diagram

of complexes in ${{\mathrm {Fil}\text {-}F\text {-}\mathrm {MIC}}}(U)$ where the two horizontal morphisms on the left-hand side are the canonical surjections, and the vertical ones are the projections. Under the identification ${\mathop {\mathscr {L}\!\!\mathit {o}\mathit {g}}\nolimits }(f)_{{{\mathrm {d\hspace {-0.2pt}R}}}}/{\mathscr {O}}_{U_K}e_{-2,f}\cong {\mathscr {O}}_{U_K}$ (which sends $e_{0,f}$ to $1$ ), the homomorphism $p'$ is nothing but the connecting morphism arising from the extension $0\to {\mathscr {O}}_{U_K}\to {\mathop {\mathscr {L}\!\!\mathit {o}\mathit {g}}\nolimits }(g)_{{{\mathrm {d\hspace {-0.2pt}R}}}}\to {\mathscr {O}}_{U_K}\to 0$ . This means $p'=\delta _g$ .

Since the composition ${\mathrm {Res}}\circ p'$ is multiplication by ${\mathrm {ord}}_D(g)$ , the diagram (4.3) induces

(4.4)

Now, we show the lemma. Let $\xi \in K^M_2({\mathscr {O}}(U))\cap {\mathrm {Ker}}({d{\mathrm {log}}})$ be arbitrary. Fix $t\in {\mathscr {O}}(U)^{\times }$ such that ${\mathrm {ord}}_{D_X}(t)>0$ . Replacing $\xi $ with $m\xi $ for some $m>0$ , one can express

$$\begin{align*}\xi=\{f,t\}+\sum_j\{u_j,v_j\} \end{align*}$$

with ${\mathrm {ord}}_{D_X}(f)={\mathrm {ord}}_{D_X}(u_j)={\mathrm {ord}}_{D_X}(v_j)=0$ . Then, (4.4) induces a commutative diagram

(4.5)

Since the tame symbol of $\xi $ is $f|_{D_X}$ , the assertion follows.

Proof. Let N be the kernel of the connecting homomorphism $H^0(D_X/S)(-1)\to H^2(X/S)$ in the Gysin exact sequence. Since $H^0_{{\mathrm {rig}}}(D_X/S)$ is an overconvergent F-isocrystal on S of weight $0$ , the weight of $N_{{\mathrm {rig}}}$ is $-2$ , and therefore, we have ${{\mathrm {Hom}}}_{{\mathrm {Fil}}\text {-} F\text {-} {\mathop {\mathrm {MIC}}\nolimits }(S)}\big ({\mathscr {O}}_{S}, N(2)\big )=0$ . This shows that the Gysin exact sequence induces an exact sequence

$$\begin{align*}0\longrightarrow {\mathrm{Ext}}^1\left({\mathscr{O}}_{S}, H^1(X/S)(2)\right)\longrightarrow {\mathrm{Ext}}^1\left({\mathscr{O}}_{S}, H^1(U/S)(2)\right)\longrightarrow{\mathrm{Ext}}^1\big({\mathscr{O}}_{S}, N(2)\big), \end{align*}$$

where the extension group is taken in the category of ${\mathrm {Fil}}\text {-} F\text {-}{\mathop {\mathrm {MIC}}\nolimits }(S)$ . Now, the construction of $[-]_{X/S}$ is immediate from Lemma 4.2.

Proof. To show the former, it is enough to show that a diagram