Proof. The spectral sequence (3) is functorial in A, so it is compatible with multiplication by m map $[m]\colon A\to A$ , for any integer m. The induced map on $\mathrm {H}^r_{\mathrm {{\acute et}}}(A_{k_{\mathrm {s}}},{\mathbb Z}/\ell ^n)\cong \wedge ^r\mathrm {H}^1_{\mathrm {{\acute et}}}(A_{k_{\mathrm {s}}},{\mathbb Z}/\ell ^n)$ is multiplication by $m^r$ . Each $E_i^{p,q}$ for $i\geq 2$ is a subquotient of $E_{2}^{p,q}$ , hence $[m]^*$ acts on it via multiplication by $m^q$ . Since $[m]^*$ commutes with all differentials on the i-th page, we have $m^{q-i+1}\delta _{i}^{p,q}=\delta _{i}^{p,q}m^q$ for all $p,q\geq 0$ . Thus we have

$$ \begin{align*}(m^q-m^{q-i+1})\delta^{p,q}_i=m^{q-i+1}(m^{i-1}-1)\delta^{p,q}_i=0.\end{align*} $$

Taking $m=\ell $ we see that $\ell ^{q-i+1}$ annihilates $\delta ^{p,q}_i$ . From now on we assume that $(m,\ell )=1$ . Then we have $(m^{i-1}-1)\delta ^{p,q}_i=0$ .

Suppose $\ell \neq 2$ . If $\ell -1$ does not divide $i-1$ , then we can find an integer m such that $(m,\ell )=1$ and $m^{i-1}-1$ is non-zero modulo $\ell $ . Then $\delta ^{p,q}_i=0$ .

If $\ell -1$ divides $i-1$ , we claim that the lowest value of $\mathrm {val}_\ell (m^{i-1}-1)$ , where $(m,\ell )=1$ , is $n+1$ . If $(m,\ell )=1$ , then $m^{r\ell ^n(\ell -1)}-1$ is clearly divisible by $\ell ^{n+1}$ , so it is enough to check that $\mathrm {val}_\ell ((1+\ell )^{r\ell ^n(\ell -1)}-1)=n+1$ when $(r,\ell )=1$ . This is immediate for $n=0$ , and the general case follows by induction in n. This proves (a).

Suppose $\ell =2$ . If i is even, then taking $m=-1$ we prove the first statement of (b). If i is odd, so that $n\geq 1$ , we claim that the smallest value of $\mathrm {val}_2(m^{r 2^n}-1)$ , where r and m are odd, is $n+2$ . We have

$$ \begin{align*}m^{r2^n}-1=(m^{2^n}-1)(m^{(r-1)2^n}+\ldots+1).\end{align*} $$

Since r and m are odd, the second factor in the right hand side is odd. Thus we can assume that $r=1$ . For $n=1$ the statement is obvious, and the general case immediately follows by induction. This proves (b).

Proof. We begin by recasting étale cohomology of $A_{k_{\mathrm {s}}}$ in terms of group cohomology of the profinite abelian group $M^{\vee }$ , using that semiabelian varieties are étale $K(\pi ,1)$ -spaces:

Proof of Lemma 1.7.

For any connected scheme X with a geometric base point $\bar {x}$ and any integer $m\geq 1$ there is a natural map

(11) $$ \begin{align} \mathrm{R}\Gamma(\pi_1^{\mathrm{{\acute et}}}(X,\bar{x}),{\mathbb Z}/m)\to \mathrm{R}\Gamma_{\mathrm{{\acute et}}}(X,{\mathbb Z}/m) \end{align} $$

induced by the pullback along the map of sites $X_{\mathrm {{\acute et}}}\to X_{\mathrm {f\acute et}}$ , where the target site consists of schemes finite étale over X, cf. [Reference OlssonOls09, §5]. Moreover, if $X=Y_{k_{\mathrm {s}}}$ , where Y is a scheme over k, and $\bar {x}$ is above a k-point of Y, this map naturally lifts to a map in $D(\Gamma ,{\mathbb Z}/m)$ , where the action of $\Gamma $ on the continuous cochain complex $\mathrm {R}\Gamma (\pi _1^{\mathrm {{\acute et}}}(X,\bar {x}),{\mathbb Z}/m)$ is induced by the Galois action on the fundamental group $\pi _1^{\mathrm {{\acute et}}}(X,\bar {x})$ .

The map (11) induces a natural map

(12) $$ \begin{align} \mathrm{H}^1_{\mathrm{cont}}(\pi_1^{\mathrm{{\acute et}}}(X,\bar{x}),{\mathbb Z}/m)\to \mathrm{H}^1(X,{\mathbb Z}/m). \end{align} $$

Since étale X-torsors for ${\mathbb Z}/m$ are the same as Galois covers of X with Galois group ${\mathbb Z}/m$ , this map is an isomorphism [SGA1, Exposé XI, §5]. Consider the composition

(13) $$ \begin{align} \mathrm{R}\Gamma(\pi_1^{\mathrm{{\acute et}}}(X,\bar{x})^{(p')},{\mathbb Z}/m)\to \mathrm{R}\Gamma(\pi_1^{\mathrm{{\acute et}}}(X,\bar{x}),{\mathbb Z}/m)\to \mathrm{R}\Gamma_{\mathrm{{\acute et}}}(X,{\mathbb Z}/m), \end{align} $$

where the first map is induced by the surjection $\pi _1^{\mathrm {{\acute et}}}(X,\bar {x})\twoheadrightarrow \pi _1^{\mathrm {{\acute et}}}(X,\bar {x})^{(p')}$ onto the prime-to-p completion of the fundamental group where $p=\mathrm {char}(k)$ ; if $\mathrm {char}(k)=0$ we set $\pi _1^{\mathrm {{\acute et}}}(X,\bar {x})^{(p')}:=\pi _1^{\mathrm {{\acute et}}}(X,\bar {x})$ . The map (13) induces an isomorphism on $\mathrm {H}^1$ whenever m is not divisible by $\mathrm {char}(k)$ .

Let us now specialise to the case $X=A_{k_{\mathrm {s}}}$ , where A is a semiabelian variety over k. We take $\bar x$ to be the origin of the group law on A and omit it from the notation.

We know that the cohomology ring of the target of (13) is the free exterior algebra on $\mathrm {H}^1(A_{k_{\mathrm {s}}},{\mathbb Z}/m)$ . We claim that the same is true for the source of (13). Indeed, [Reference Brion and SzamuelyBS13, Proposition 1.1 b)] implies that $\pi _1^{\mathrm {{\acute et}}}(A_{k_{\mathrm {s}}})^{(p')}$ is a finite product of copies of $\widehat {{\mathbb Z}}^{(p')}$ . The cohomological dimension of $\widehat {{\mathbb Z}}^{(p')}$ is 1 and the cohomology groups $\mathrm {H}^0(\widehat {{\mathbb Z}}^{(p')},{\mathbb Z}/m)\simeq \mathrm {H}^1(\widehat {{\mathbb Z}}^{(p')},{\mathbb Z}/m)\simeq {\mathbb Z}/m$ are free ${\mathbb Z}/m$ -modules, hence the claim follows from the Künneth formula. Thus for $X=A_{k_{\mathrm {s}}}$ the map (13) is a quasi-isomorphism in $D(\Gamma ,{\mathbb Z}/m)$ .

Next, $\pi _1^{\mathrm {{\acute et}}}(A_{k_{\mathrm {s}}})^{(p')}$ is the product of the $2$ -adic Tate module $T_2(A_{k_{\mathrm {s}}})\cong M^{\vee }$ and a 2-divisible profinite group. Passing to the limit over $m=2^n$ we obtain a quasi-isomorphism $\mathrm {R}\Gamma (\pi _1^{\mathrm {{\acute et}}}(A_{k_{\mathrm {s}}})^{(p')},{\mathbb Z}_2)\simeq \mathrm {R}\Gamma (M^{\vee },{\mathbb Z}_2)$ in $D(\Gamma ,{\mathbb Z}_2)$ , hence also a quasi-isomorphism $\mathrm {R}\Gamma _{\mathrm {{\acute et}}}(A_{k_{\mathrm {s}}},{\mathbb Z}_2)\simeq \mathrm {R}\Gamma (M^{\vee },{\mathbb Z}_2)$ .

Proof of Proposition 1.9.

We can calculate $\mathrm {R}\Gamma (M^{\vee },{\mathbb Z}_2)$ using the standard bar complex of ${\mathbb Z}_2$ -modules for computing continuous cohomology of the group $M^{\vee }$ :

$$ \begin{align*}0\longrightarrow{\mathbb Z}_2\stackrel{d_0=0}\longrightarrow \mathrm{Func}(M^\vee,{\mathbb Z}_2)\stackrel{d_1}\longrightarrow \mathrm{Func}(M^\vee\times M^\vee,{\mathbb Z}_2)\stackrel{d_2}\longrightarrow\ldots, \end{align*} $$

where $\mathrm {Func}$ denotes the ${\mathbb Z}_2$ -module of continuous functions. Considering this complex as a complex of $\Gamma $ -modules for the natural action of the Galois group $\Gamma $ induced by its action on M, we get a complex in $D(\Gamma ,{\mathbb Z}_2)$ that calculates the cohomology of the semi-direct product $M^\vee \rtimes \Gamma $ .

The differential $d_1$ sends a function $f\colon M^\vee \to {\mathbb Z}_2$ to the function

$$ \begin{align*}(x,y)\mapsto f(x+y)-f(x)-f(y).\end{align*} $$

The inclusion $Q(M)\subset \mathrm {Func}(M^\vee ,{\mathbb Z}_2)$ gives rise to a commutative diagram

(14)

The right-hand vertical map is the inclusion of bilinear functions on $M^\vee \times M^\vee $ into all continuous functions. Commutativity of the diagram is immediate from the definitions of the maps. The submodule $M\subset Q(M)$ of linear functions maps isomorphically to $\mathrm {H}^1_{\mathrm {cont}}(M^{\vee },{\mathbb Z}_2)$ . Moreover, the right vertical map lands in the kernel of the next differential $d_2$ , and the induced map $M^{\otimes 2}\to \mathrm {H}^2_{\mathrm {cont}}(M^{\vee },{\mathbb Z}_2)$ coincides with the cup-product map $\mathrm { H}^1_{\mathrm {cont}}(M^{\vee },{\mathbb Z}_2)^{\otimes 2}\to \mathrm {H}^2_{\mathrm {cont}}(M^{\vee },{\mathbb Z}_2)$ .

Exactness of (5) then implies that (14) gives a quasi-isomorphism in $D(\Gamma ,{\mathbb Z}_2)$ between the two-term complex $Q(M)\to M^{\otimes 2}$ and the truncation $\tau ^{[1,2]}\mathrm {R}\Gamma (M^{\vee },{\mathbb Z}_2)$ , as desired. The description of the analogous truncation $\tau ^{[1,2]}\mathrm {R}\Gamma (M^{\vee },{\mathbb Z}/2^n)$ now follows formally because cohomology groups $\mathrm {H}^i_{\mathrm {cont}}(M^{\vee },{\mathbb Z}_2)$ as well as terms of the extension (5) are torsion-free.

Proof. (i) Let $u,v,w$ be the three non-zero elements of V. Then the unique non-zero element in $\wedge ^2 V$ lifts to $uv+vw+wu\in S^2V$ . This lifting is $\mathrm {{GL}}(2,{\mathbb F}_2)$ -equivariant.

(ii) Assume that there is a non-zero element $x\in (\wedge ^2 V)^\Gamma $ . We have a perfect bilinear pairing of ${\mathbb F}_2$ -vector spaces $V\times \wedge ^2V\to {\mathbb F}_2$ . The elements of V that pair trivially with x form an ${\mathbb F}_2$ -subspace $N\subset V$ of dimension 2 which is stable under the action of $\Gamma $ . Moreover, $x\in \wedge ^2N$ . We have a commutative diagram of $\Gamma $ -modules with exact rows

By part (i), the top sequence is split. Thus x lifts to an element of $(S^2V)^\Gamma $ .

Proof. (i) Let $e_1,\ldots ,e_n$ be a $\Gamma $ -stable basis of V. The union of sets $\{e_i^2\}$ and $\{e_ie_j|i<j\}$ is a $\Gamma $ -stable basis of $S^2V$ , which gives a $\Gamma $ -equivariant splitting of $(15)$ .

(ii) By (i) it is enough to show that any representation of ${\mathbb Z}/2$ in V is permutational, but this is well-known. Indeed, endow V with the structure of an ${\mathbb F}_2[x]$ -module by letting x act as the generator of ${\mathbb Z}/2$ . By the classification of finitely generated torsion modules over a PID, the ${\mathbb F}_2[x]$ -module V is a direct sum of cyclic submodules isomorphic to ${\mathbb F}_2={\mathbb F}_2[x]/(x-1)$ or ${\mathbb F}_2[x]/(x^2-1)$ , both of which are permutational.

Proof. The mod $p^m$ reduction of the map $Y\to X[2]$ is the Yoneda class of the complex $(X\xrightarrow {\psi } \widetilde {E})\otimes ^L_{{\mathbb Z}_p}{\mathbb Z}/p^m$ concentrated in degrees $-1$ and $0$ . It can be computed as the totalisation of the bicomplex

(18)

The compositions of (16) and (17) are represented, respectively, by the following Yoneda extensions

(19) $$ \begin{align} 0\to X/p^m\xrightarrow{a} R_m\xrightarrow{b} Y/p^{m+1}\xrightarrow{c} Y/p^m\to 0 \end{align} $$

(20) $$ \begin{align} 0\to X/p^m\xrightarrow{a'} X/p^{m+1}\xrightarrow{b'} T_m\xrightarrow{c'} Y/p^m\to 0, \end{align} $$

where $R_m$ (respectively, $T_m$ ) is the following pushout (respectively, pullback)

and the map $b:R_m\to Y/p^{m+1}$ is the composition $R_m\to Y/p\hookrightarrow Y/p^{m+1}$ , while $b':X/p^{m+1}\to T_m$ is the composition $X/p^{m+1}\twoheadrightarrow X/p\to T_m$ . The result of subtracting the map $Y/p^m\to X/p^m[2]$ corresponding to (19) from that corresponding to (20) is then represented by the extension

(21) $$ \begin{align} X/p^m\xrightarrow{(a,0)} \frac{R_m\oplus X/p^{m+1}}{(a(x),-a'(x))|x\in X/p^m}\xrightarrow{b\oplus b'} \ker(Y/p^{m+1}\oplus T_m\xrightarrow{c-c'} Y/p^m)\xrightarrow{c} Y/p^m \end{align} $$

We can now write down a chain-level map from the totalisation of (18) to the two-term complex from (21):

(22)

Let $f_1$ be the map induced by the direct sum of maps $\widetilde {E}\to E\to R_m$ and $X\twoheadrightarrow X/p^{m+1}$ , and $f_2$ is the sum of compositions $\widetilde {E}\to Y\to Y/p^{m+1}$ and $\widetilde {E}\twoheadrightarrow T_m$ . The map $f_2$ indeed lands in the kernel of $c-c'$ , the square in (22) commutes, and the composition $f_1\circ (\psi ,-p^m)$ is zero.

The complexes given by rows of (22) both have two non-zero cohomology groups given by $X/p^m$ and $Y/p^m$ . The map of complexes induced by $f_1,f_2$ induces the identity map on cohomology, giving the desired identification between the two classes in $\mathrm {Ext}^2_{\mathcal {A}(\Gamma ,{\mathbb Z}_p)}(Y/p^m,X/p^m)$ .

Proof. By Theorem 1.6 the degree $2$ extension class corresponding to $\tau ^{[1,2]}\mathrm {R}\Gamma _{\mathrm {{\acute et}}}(A_{k_{\mathrm {s}}},{\mathbb Z}_2)$ is equal to the composition

$$ \begin{align*}\wedge^2M\to\wedge^2(M/2)\xrightarrow{\alpha(M/2)}M/2[1]\xrightarrow{\mathrm{Bock}_M}M[2] \end{align*} $$

To calculate the induced map on mod $2^m$ reductions, we apply Lemma 1.12 to the case $p=2$ with $X=M$ , $Y=\wedge ^2 M$ , and the extension E given by

$$ \begin{align*}0\to M/2\to S^2(M/2)\to \wedge^2(M/2)\to 0.\end{align*} $$

We obtain that the extension class of $\tau ^{[1,2]}\mathrm {R}\Gamma _{\mathrm {{\acute et}}}(A_{k_{\mathrm {s}}},{\mathbb Z}/2^m)$ is equal to the difference of the following compositions:

(23) $$ \begin{align} \wedge^{2}(M/2^{m})\longrightarrow \wedge^{2}(M/2)\xrightarrow{\alpha(M/2)} (M/2)[1] \overset{\mathrm{Bock}}{\longrightarrow}(M/2^{m})[2], \end{align} $$

(24) $$ \begin{align} \wedge^{2}(M/2^{m})\overset{\mathrm{Bock}}{\longrightarrow} \wedge^{2}(M/2)[1]\xrightarrow{\alpha(M/2)[1]} (M/2)[2] \longrightarrow(M/2^{m})[2]. \end{align} $$

The differential $\delta _2^{p,2}$ is equal to the map induced by this difference on the degree p derived functor of $\Gamma $ -invariants, which gives the desired result.

Proof. We will deduce this decomposition from the fact that J admits integral Künneth projectors defined over the field k, as proved in [Reference SuhSuh17, Theorem 1.4].

Proof. This is proved in [Reference SuhSuh17, Theorem 4.2.3] in the case $k_{\mathrm {s}}=\mathbb {C}$ . The proof goes through over an arbitrary separably closed field: its only geometric input is Poincaré’s formula $r!\cdot w^{[r]}=(w^{[1]})^{r}$ in cohomology of J, see [Reference KleimanKle68, Remark 2A13] for a proof of this over arbitrary fields.

Proof. The two projections $p_1,p_2:Y\times Y\to Y$ define maps in $D(\Gamma ,{\mathbb Z}/n)$ :

$$ \begin{align*}p_1^*:\mathrm{R}\Gamma_{\mathrm{{\acute et}}}(Y_{k_{\mathrm{s}}},{\mathbb Z}/n)\to \mathrm{R}\Gamma_{\mathrm{{\acute et}}}((Y\times Y)_{k_{\mathrm{s}}},{\mathbb Z}/n)\end{align*} $$

and

$$ \begin{align*}p_{2*}:\mathrm{R}\Gamma_{\mathrm{{\acute et}}}((Y\times Y)_{k_{\mathrm{s}}},{\mathbb Z}/n)\to \mathrm{R}\Gamma_{\mathrm{{\acute et}}}(Y_{k_{\mathrm{ s}}},{\mathbb Z}/n)(-d)[-2d],\end{align*} $$

where $(-d)$ refers to the twist by the $(-d)$ -th power of the cyclotomic character $\Gamma \to \mathrm {Aut}_{{\mathbb Z}/n}(\mu _n)$ .

There are cycle classes $\mathrm {cl}(c)\in \mathrm {H}^{2i}_{\mathrm {{\acute et}}}(Y\times Y,({\mathbb Z}/n)(i))$ in absolute étale cohomology of $Y\times Y$ , as defined in [Reference Cohomologie étale, Boutot, Grothendieck and Verdier.SGA4½, Cycle, 2.2.10]. The absolute étale cohomology complex $\mathrm {R}\Gamma _{\mathrm {{\acute et}}}(Y\times Y,({\mathbb Z}/n)(i))$ can be identified with the continuous group cohomology complex $\mathrm {R}\Gamma _{\mathrm {cont}}(\Gamma , \mathrm {R}\Gamma _{\mathrm {{\acute et}}}((Y\times Y)_{k_{\mathrm {s}}},({\mathbb Z}/n)(i)))$ with coefficients in geometric étale cohomology, hence $\mathrm {cl}(c)$ corresponds to a map $\mathrm {cl}(c):({\mathbb Z}/n)[-2i]\to \mathrm {R}\Gamma _{\mathrm {{\acute et}}}((Y\times Y)_{k_{\mathrm { s}}},({\mathbb Z}/n)(i))$ in the derived category $D(\Gamma ,{\mathbb Z}/n)$ .

We can now define the endomorphism of the complex $\mathrm {R}\Gamma _{\mathrm {{\acute et}}}(Y_{k_{\mathrm {s}}},{\mathbb Z}/n)$ induced by c as the composition

$$ \begin{align*} \mathrm{R}\Gamma_{\mathrm{{\acute et}}}(Y_{k_{\mathrm{s}}},{\mathbb Z}/n)\xrightarrow{p_1^*}& \mathrm{R}\Gamma_{\mathrm{{\acute et}}}((Y\times Y)_{k_{\mathrm{s}}},{\mathbb Z}/n)\xrightarrow{\cup \mathrm{cl}(c)}\\ &\mathrm{R}\Gamma_{\mathrm{{\acute et}}}((Y\times Y)_{k_{\mathrm{s}}},{\mathbb Z}/n)(i)[2i]\xrightarrow{p_{2*}}\mathrm{R}\Gamma_{\mathrm{{\acute et}}}(Y_{k_{\mathrm{s}}},{\mathbb Z}/n)(i-d)[2i-2d] \end{align*} $$

Here the middle map $\cup \mathrm {cl}(c)$ denotes the composition

$$ \begin{align*} \mathrm{R}\Gamma_{\mathrm{{\acute et}}}((Y\times Y)_{k_{\mathrm{s}}},{\mathbb Z}/n)\xrightarrow{\mathrm{id}\otimes \mathrm{cl}(c)}\mathrm{R}\Gamma_{\mathrm{{\acute et}}}((Y\times Y)_{k_{\mathrm{s}}},{\mathbb Z}/n)^{\otimes 2}(i)[2i]\xrightarrow{}\\ \mathrm{R}\Gamma_{\mathrm{{\acute et}}}((Y\times Y)_{k_{\mathrm{s}}},{\mathbb Z}/n)(i)[2i] \end{align*} $$

where the second map is the cup-product on the level of cohomology complexes.

Proof. Without loss of generality we can assume $a=0$ . Proceed by induction on n. If $n=0$ , then C is isomorphic in $D^b(A)$ to an object represented by the one-term complex $\mathrm {H}^0(C)$ in degree $0$ . Under this isomorphism, f corresponds to an endomorphism induced by $\mathrm {H}^0(f)\in \mathrm {{Hom}}_A(\mathrm {H}^0(C),\mathrm {H}^0(C))$ , but $\mathrm {H}^0(f)=0$ .

Now suppose that $f^n$ is zero on $\tau ^{<n}C$ . The composition $C\xrightarrow {f} C \to \mathrm {H}^n(C)[-n]$ is zero, because it is equal to the composition $C\to \mathrm {H}^n(C)[-n]\to \mathrm {H}^n(C)[-n]$ , where the last map is $\mathrm {H}^n(f)=0$ . Given the distinguished triangle

$$ \begin{align*}\tau^{<n}C\to C\to\mathrm{H}^n(C)[-n],\end{align*} $$

this implies that $f\colon C\to C$ factors as $C\xrightarrow {g}\tau ^{<n}C\to C$ for some map g. Then $f^{n+1}$ is the composition $C\xrightarrow {g}\tau ^{<n}C\to C\xrightarrow {f^{n}}C$ . The truncation $\tau ^{<n}C$ is functorial in C, so the composition of the last two maps is equal to $\tau ^{<n}C\xrightarrow {f^{n}}\tau ^{<n}C\to C$ . By the induction assumption $f^n$ is zero on $\tau ^{<n}C$ , so the iterate $f^{n+1}$ is zero on C.

Proof. The antipodal involution $[-1]\colon A\to A$ induces an action of ${\mathbb Z}/2$ on $\mathrm {{Pic}}(A_{k_{\mathrm {s}}})$ which turns (29) into an exact sequence of abelian groups with an action of ${\mathbb Z}/2$ . The induced action on $\mathrm {NS\,}(A_{k_{\mathrm {s}}})$ is trivial, cf. [Reference MilneMil84, Remark 12.13]. The involution $[-1]_A$ induces the involution $[-1]_{A^\vee }$ on $A^\vee $ . Recall that for an abelian group M with an action of ${\mathbb Z}/2=\langle \sigma \rangle $ we have $\mathrm {H}^1({\mathbb Z}/2,M)\cong M^{\sigma =-1}/(1-\sigma )M$ . Since $A^\vee (k_{\mathrm {s}})$ is 2-divisible, the endomorphism $1-[-1]_{A^{\vee }}=[2]_{A^{\vee }}$ is surjective, and we obtain $\mathrm {H}^1({\mathbb Z}/2,A^\vee (k_{\mathrm {s}}))=0$ . Thus the long exact sequence of cohomology for the group ${\mathbb Z}/2$ gives an exact sequence of $\Gamma $ -modules

(30) $$ \begin{align} 0 \longrightarrow A^\vee[2] \longrightarrow \mathrm{{Pic}}(A_{k_{\mathrm{s}}})^{[-1]^*} \longrightarrow \mathrm{NS\,}(A_{k_{\mathrm{s}}}) \longrightarrow 0. \end{align} $$

Let $c_\lambda \in \mathrm {H}^1(k,A^\vee [2])$ be the image of $\lambda $ under the connecting map of $(30)$ . It is clear that $c^{\prime }_\lambda $ is the image of $c_\lambda $ in $\mathrm {H}^1(k,A^\vee )$ .

Let $T_2(A^\vee )$ be the 2-adic Tate module, and let $e_2\colon A[2]\times A^\vee [2]\to {\mathbb Z}/2$ be the Weil pairing. The short exact sequences (30) and (15) are compatible, so that there is a commutative diagram (see diagram (16) in [Reference Poonen and RainsPR11]):

Furthermore, the right-hand vertical map sends $\lambda $ to the element of $\wedge ^2(A^\vee [2])$ corresponding to

$$ \begin{align*}e_2(x,\lambda(y))\in \mathrm{{Hom}}(\wedge^2A[2],{\mathbb Z}/2).\end{align*} $$

This map factors as the first Chern class map $\mathrm {c}_1$

$$ \begin{align*}\mathrm{NS\,}(A_{k_{\mathrm{s}}})\cong\mathrm{{Hom}}(A_{k_{\mathrm{s}}},A^\vee_{k_{\mathrm{s}}})^{\mathrm{sym}}\stackrel{\mathrm{c}_1}\longrightarrow\mathrm{ {Hom}}(T_2(A),T_2(A^\vee))^{\mathrm{sym}}\cong\wedge^2 T_2(A^\vee)(-1) \cong\mathrm{H}^2_{\mathrm{{\acute et}}}(A_{k_{\mathrm{s}}},{\mathbb Z}_2(1))\end{align*} $$

followed by the reduction modulo 2 map $\mathrm {H}^2_{\mathrm {{\acute et}}}(A_{k_{\mathrm {s}}},{\mathbb Z}_2(1))\to \mathrm {H}^2_{\mathrm {{\acute et}}}(A_{k_{\mathrm {s}}},{\mathbb Z}/2)$ , see [Reference Orr, Skorobogatov and ZarhinOSZ21, Lemma 2.6]. Theorem 1.6 gives

(31) $$ \begin{align} \delta_2^{0,2}(\mathrm{c}_1(\lambda))=\mathrm{Bock}_{T_2(A^\vee)}(c_\lambda), \end{align} $$

where $\mathrm {Bock}_{T_2(A^\vee )}$ is the connecting map attached to the exact sequence

$$ \begin{align*}0\to T_2(A^\vee)\xrightarrow{2} T_2(A^\vee)\to A^\vee[2]\to 0.\end{align*} $$

This sequence fits into the following commutative diagram with exact rows:

This implies a relation between connecting homomorphisms induced by the first and last rows: the image of $\mathrm {Bock}_{T_2(A^{\vee })}(c_{\lambda })\in \mathrm { H}^2(k, T_2A^{\vee })$ in $\mathrm {H}^2(k, A^{\vee }[2])$ equals $\beta _2(c^{\prime }_{\lambda })$ , where $\beta _2:\mathrm {H}^1(k, A^{\vee })\to \mathrm {H}^2(k, A^{\vee }[2])$ is the connecting homomorphism induced by the top exact row, as defined above. Combining this with (31) gives the desired equality.

To prove the final assertion of the proposition, note that the short exact sequence of Galois modules

$$ \begin{align*}0\to \mathrm{H}^1_{\mathrm{{\acute et}}}(A_{k_{\mathrm{s}}},{\mathbb Z}_2(1))\xrightarrow{2}\mathrm{H}^1_{\mathrm{{\acute et}}}(A_{k_{\mathrm{s}}},{\mathbb Z}_2(1))\to \mathrm{H}^1_{\mathrm{{\acute et}}}(A_{k_{\mathrm{s}}},{\mathbb Z}/2)\to 0\end{align*} $$

shows that $\delta ^{0,2}_2(c_1(\lambda ))\in \mathrm {H}^2(k,\mathrm {H}^1_{\mathrm {{\acute et}}}(A_{k_{\mathrm {s}}},{\mathbb Z}_2(1)))$ is divisible by $2$ if and only if its image in $\mathrm {H}^2(k, \mathrm {H}^1_{\mathrm {{\acute et}}}(A_{k_{\mathrm {s}}},{\mathbb Z}/2))\cong \mathrm {H}^2(k, A^{\vee }[2])$ is zero. Similarly, the short exact sequence of Galois modules

$$ \begin{align*}0\to A^{\vee}[2]\to A^{\vee}(k_{\mathrm{s}})\xrightarrow{2}A^{\vee}(k_{\mathrm{s}})\to 0\end{align*} $$

gives that $c_{\lambda }'\in \mathrm {H}^1(k,A^{\vee })$ is divisible by $2$ if and only if its image $\beta _2(c^{\prime }_{\lambda })\in \mathrm {H}^2(k,A^{\vee }[2])$ under the connecting homomorphism $\beta _2$ is zero. This implies the equivalence of $2$ -divisibility of $\delta _{2}^{0,2}(c_1(\lambda ))$ and $c^{\prime }_{\lambda }$ .

Proof. This is a particular case of [Reference CreutzCre13, Theorem 4], which is based on the non-degeneracy of the Poitou–Tate pairing [Reference MilneMil86, Ch. I, §4]

$$ \begin{align*}{\mathbf{\text{Ш}}}^1(k,J[2])\times {\mathbf{\text{Ш}}}^2(k,J^\vee[2])\to{\mathbb Z}/2\end{align*} $$

and its compatibility with the Cassels–Tate pairing $ {\mathbf{\text{Ш}}} (J)\times {\mathbf{\text{Ш}}} (J^\vee )\to {\mathbb Q}/{\mathbb Z}$ . For the second claim we consider the pairing of $c^{\prime }_\lambda =i(c_\lambda )$ with itself.

Proof. Since C is hyperelliptic of even genus, we have $\mathbf {Pic}_{C/k}^{g-1}\cong \mathbf {Pic}_{C/k}^{1}$ . For each place of k choose a local point $P_v\in C(k_v)\subset \mathbf {Pic}_{C/k}^{1}(k_v)$ . By a theorem of Manin [Reference SkorobogatovSko01, Theorem 6.2.3], we have

$$ \begin{align*}\sum_v\mathrm{inv}_v({\mathcal A}(P_v))=\langle x,c^{\prime}_\lambda\rangle\in\frac{1}{2}{\mathbb Z}/{\mathbb Z},\end{align*} $$

for any $x\in {\mathbf{\text{Ш}}} (J)$ and ${\mathcal A}\in \mathrm {{Br}}(C)$ associated to x, using that $c^{\prime }_\lambda =[\mathbf {Pic}_{C/k}^{1}]\in {\mathbf{\text{Ш}}} (J)$ . Taking $x\in i({\mathbf{\text{Ш}}} ^1(k,J[2]))$ we conclude by Lemma 3.2.

Proof. (a) By Creutz [Reference CreutzCre13, Reference Petrovp. 941] and Creutz and Viray [Reference Creutz and VirayCV15, Theorem 6.7], this curve satisfies the assumption of Lemma 3.4. Here is a sketch of this calculation.

Let C be the hyperelliptic curve over a field k of characteristic different from 2 given by $y^2=cf(x)$ where $c\in k^\times $ and $f(x)\in k[x]$ is a separable monic polynomial of even degree. Let J be the Jacobian of C. Let $L=k[x]/(f(x))$ and let $\theta \in L$ be the image of x. The well-known identification of the group k-scheme $J[2]$ with the quotient of the kernel of the norm map $R_{L/k}(\mu _2)\to \mu _2$ by the image of $\mu _2$ gives a natural inclusion

$$ \begin{align*}(L^\times/k^\times L^{\times 2})_1\subset\mathrm{H}^1(k,J[2])/\langle c_\lambda\rangle,\end{align*} $$

where the subscript $1$ denotes the subgroup of elements with norm $1\in k^\times /k^{\times 2}$ . To $l\in L^\times $ we associate the class of the quaternion algebra $(l,x-\theta )\in \mathrm {{Br}}(L(x))$ , which only depends on the image of l in $L^\times /L^{\times 2}$ . The corestriction

$$ \begin{align*}\mathrm{cores}_{L(x)/k(x)}\big((l,x-\theta)\big)\in \mathrm{{Br}}(k({\mathbb P}^1_k))\end{align*} $$

is unramified away from the ramification locus of $C\to {\mathbb P}^1_k$ and the point at infinity. If the norm of l is a square in $k^\times $ , then it is unramified at infinity. It follows that the pullback of this element to $\mathrm {{Br}}(k(C))$ is unramified, and so belongs to the subgroup $\mathrm {{Br}}(C)\subset \mathrm {{Br}}(k(C))$ . We denote the resulting element by ${\mathcal A}_l\in \mathrm {{Br}}(C)$ . Finally, multiplying l by $s\in k^\times $ gives ${\mathcal A}_{ls}={\mathcal A}_l+(s,c)$ , so this does not change the image of ${\mathcal A}_l$ in $\mathrm {{Br}}(C)/\mathrm {{Br}}_0(C)$ . An important property of this construction is that the map sending l to ${\mathcal A}_l$ coincides with the composition

$$ \begin{align*}\mathrm{H}^1(k,J[2])/\langle c_\lambda\rangle\stackrel{i}\longrightarrow \mathrm{H}^1(k,J)/\langle c^{\prime}_\lambda\rangle\longrightarrow \mathrm{{Br}}(C)/\mathrm{{Br}}_0(C),\end{align*} $$

where the second arrow comes from (32).

Now let $k={\mathbb Q}$ and $L={\mathbb Q}(\sqrt {-1})\oplus {\mathbb Q}(\sqrt {-17})\oplus {\mathbb Q}(\sqrt {17})$ . The 0-dimensional scheme $\mathrm {{Spec}}(L)$ over ${\mathbb Q}$ is everywhere locally soluble, hence the curve C given by

$$ \begin{align*}y^2=c(x^2+1)(x^2+17)(x^2-17)\end{align*} $$

is everywhere locally soluble for any $c\in {\mathbb Q}^\times $ and we have $c_\lambda \in {\mathbf{\text{Ш}}} ^1({\mathbb Q},J[2])$ . Take $l=(1,1,-1)\in L$ . One checks that l gives rise to an everywhere locally trivial element of $(L^\times /{\mathbb Q}^\times L^{\times 2})_1$ , and thus to an element of ${\mathbf{\text{Ш}}} ^1({\mathbb Q},J[2])/\langle c_\lambda \rangle $ . We have ${\mathcal A}_l=(-1,x^2-17)$ . A local computation [Reference Creutz and VirayCV15, Lemma 6.8] shows that if the number of odd prime factors p of $c>0$ such that neither $17$ nor $-1$ is a square modulo p, is odd, then $\sum _v\mathrm {inv}_v({\mathcal A}_l(P_v))=1/2$ . For example, one can take $c=3$ . Thus Lemma 3.4 can be applied, so that $c^{\prime }_\lambda $ is not divisible by $2$ in $\mathrm {H}^1({\mathbb Q},J)$ , hence $\delta ^{0,2}_2$ is non-zero by Proposition 3.1.

(b) This curve was also communicated to us by Creutz (personal communication). It is the quadratic twist by $-1$ of the curve in [Reference Poonen and RainsPR11, Example 3.12]. The arguments in loc. cit. show that for the primes p where C has bad reduction, and also for $p=2$ , the polynomial $f(x)$ has a root in ${\mathbb Q}_p$ . In view of Lemma 3.2 and Remark 3.3 it remains to show that C is odd. By [Reference Poonen and StollPS99, Corollary 12] this happens if and only if the number of places v such that C does not have a ${\mathbb Q}_v$ -rational divisor of degree 1, is odd. Such places are necessarily real or places of bad reduction, so we only need to consider the real place. We check that $C({\mathbb {R}})=\emptyset $ , so C has no real divisor of degree 1, and hence is odd.

Proof. We will show that the restriction of $d_3^{0,2}$ to the $2^n$ -torsion subgroup can be read off from the Hochschild–Serre spectral sequence for the cohomology of T with coefficients in $\mu _{2^n}$ . The natural map $\tau ^{\leq 2}\mathrm {R}\Gamma _{\mathrm {{\acute et}}}(T_{k_{\mathrm {s}}},\mu _{2^n})\to \tau ^{\leq 2}\mathrm {R}\Gamma _{\mathrm {{\acute et}}}(T_{k_{\mathrm {s}}},{\mathbb G}_m)$ of truncations in degrees $\leq 2$ induces a commutative diagram relating the extensions between $\mathrm {H}^2$ and $\tau ^{\leq 1}$ in these complexes:

(35)

where we used that $\mathrm {H}^1_{\mathrm {{\acute et}}}(T_{k_{\mathrm {s}}},{\mathbb G}_m)=\mathrm {{Pic}}(T_{k_{\mathrm {s}}})$ vanishes. The differential $d_{3}^{0,2}$ is obtained by applying the functor $\mathrm {H}^0(k,-)$ to the bottom horizontal map in the diagram.

Applying the functor of Galois invariants to the left vertical map we get the inclusion of the $2^n$ -torsion $\wedge ^2(\widehat {T}/2^n)(-1)^{\Gamma }\cong \mathrm {H}^2_{\mathrm {{\acute et}}}(T_{k_{\mathrm {s}}},\mu _{2^n})$ into the Galois invariants in the Brauer group of $T_{k_{\mathrm {s}}}$ , so we are looking to compute the result of applying $\mathrm {H}^0(k,-)$ to the counter-clockwise composition in (35), composed with the projection $\mathrm {H}^3(k,\mathrm {H}^0_{\mathrm {{\acute et}}}(T_{k_{\mathrm {s}}},{\mathbb G}_m))\to \mathrm {H}^3(k,\widehat {T})$ .

From the Kummer sequence, we see that $\tau ^{\leq 1}\mathrm {R}\Gamma _{\mathrm {{\acute et}}}(T_{k_{\mathrm {s}}},\mu _{2^n})$ can be represented by the two-term complex $\mathrm {H}^0_{\mathrm {{\acute et}}}(T_{k_{\mathrm {s}}},{\mathbb G}_m)\xrightarrow {2^n}\mathrm {H}^0_{\mathrm {{\acute et}}}(T_{k_{\mathrm {s}}},{\mathbb G}_m)$ with the right vertical arrow in (35) given by shift by $[3]$ of the projection onto the $0$ th term of this complex. This map followed by the projection to $\widehat {T}$ sends $\mathrm {H}^0_{\mathrm {{\acute et}}}(T_{k_{\mathrm {s}}},\mu _{2^n})$ to zero, so the composition factors through $\tau ^{[1,1]}\mathrm {R}\Gamma _{\mathrm {{\acute et}}}(T_{k_{\mathrm {s}}},\mu _{2^n})$ . Applying $\mathrm {H}^0(k,-)$ to this composition, we obtain a map that factors as follows:

$$ \begin{align*}\mathrm{H}^0(k, \tau^{\leq 1}\mathrm{R}\Gamma_{\mathrm{{\acute et}}}(T_{k_{\mathrm{s}}},\mu_{2^n})[3])\to \mathrm{H}^2(k, \mathrm{H}^1_{\mathrm{{\acute et}}}(T_{k_{\mathrm{s}}},\mu_{2^n}))\cong \mathrm{H}^2(k,\widehat{T}/2^n)\xrightarrow{-\mathrm{Bock}_{\widehat{T}}}\mathrm{H}^3(k, \widehat{T}).\end{align*} $$

Therefore the clockwise composition in (35) evaluated on $\mathrm {H}^0(k,-)$ , composed with the projection onto $\mathrm { H}^3(k,\widehat {T})$ , is given by

$$ \begin{align*}\mathrm{H}^2_{\mathrm{{\acute et}}}(T_{k_{\mathrm{s}}},\mu_{2^n})^{\Gamma}\xrightarrow{\delta_{2}^{0,2}} \mathrm{H}^2(k, \mathrm{H}^1_{\mathrm{{\acute et}}}(T_{k_{\mathrm{s}}},\mu_{2^n}))\xrightarrow{-\mathrm{Bock}_{\widehat{T}}}\mathrm{H}^3(k,\widehat{T}),\end{align*} $$

where $\delta ^{0,2}_2$ is a differential of the spectral sequence

$$ \begin{align*}E^{p,q}_2=\mathrm{H}^p(k,\mathrm{H}^q_{\mathrm{{\acute et}}}(T_{k_{\mathrm{s}}},\mu_{2^n}))\Rightarrow\mathrm{H}^{p+q}_{\mathrm{{\acute et}}}(T,\mu_{2^n}).\end{align*} $$

By Theorem 1.13, $\delta ^{0,2}_2$ is obtained by applying $-\otimes _{{\mathbb Z}_2}^L\mu _{2^n}$ to the terms of the 2-extension given by the difference of the maps (23) and (24) with $M=\mathrm {H}^1_{\mathrm {{\acute et}}}(T_{k_{\mathrm {s}}},{\mathbb Z}_2)$ . The composition of two Bockstein maps is zero, so only (24) contributes to $\bar d_3^{\,0,2}$ , thus proving that $\bar d_3^{\,0,2}$ is the composition of the four maps in the theorem.

Proof. Let us show that $\bar d^{\,0,2}_3=0$ . The case of a quasi-trivial torus is immediate from Theorem 4.1 and Lemma 1.11 (i). If k is a p-adic field, then $\mathrm {H}^3(k,\widehat T)=0$ since k has strict cohomological dimension 2 [Reference HarariHar20, Theorem 10.6], so there is nothing to prove. The case $k_v={\mathbb {R}}$ follows from Theorem 4.1 and Lemma 1.11 (ii). If k is a number field, we have an isomorphism $\mathrm {H}^3(k,\widehat T)\tilde \longrightarrow \prod \mathrm {H}^3(k_v,\widehat T)$ , where the product is over the real places of k, see [Reference HarariHar20, Exercise 18.1], so this case follows from the case of local fields.

Proof. Our goal is to compute the morphism

$$ \begin{align*}\delta_X:\mathrm{{Hom}}_{{\mathbb Z}/n}(A[n],{\mathbb Z}/n)\cong \mathrm{H}^1_{\mathrm{{\acute et}}}(X_{k_{\mathrm{s}}},{\mathbb Z}/n)\to \mathrm{H}^0_{\mathrm{{\acute et}}}(X_{k_{\mathrm{s}}},{\mathbb Z}/n)[2]\cong{\mathbb Z}/n[2]\end{align*} $$

in $D(\Gamma , {\mathbb Z}/n)$ corresponding to the complex $\tau ^{[0,1]}\mathrm {R}\Gamma _{\mathrm {{\acute et}}}(X_{k_{\mathrm {s}}},{\mathbb Z}/n)$ . The ${\mathbb Z}/n$ -linear dual of the map $\delta _X$ shifted by $2$ is a map $\delta _X^{\vee }[2]:{\mathbb Z}/n\to (A[n])[2]$ which is the following composition

$$ \begin{align*}{\mathbb Z}/n\xrightarrow{\Delta}\mathrm{{End}}_{{\mathbb Z}/n}(A[n])=\mathrm{{Hom}}_{{\mathbb Z}/n}(A[n],{\mathbb Z}/n)\otimes^L_{{\mathbb Z}/n}A[n]\xrightarrow{\delta_X\otimes\mathrm{id}}{\mathbb Z}/n[2]\otimes_{{\mathbb Z}/n}^L A[n], \end{align*} $$

where $\Delta $ sends $1\in {\mathbb Z}/n$ to $\mathrm {id}_{A[n]}\in \mathrm {{End}}_{{\mathbb Z}/n}(A[n])$ . After applying the derived functor of the functor of $\Gamma $ -invariants, we see that $\mathrm {id}_{A[n]}\in \mathrm {{End}}_{{\mathbb Z}/n}(A[n])^\Gamma $ goes to $\delta _X\in \mathrm {H}^2(k,A[n])$ .

The map $\delta _X\otimes \mathrm {id}_{A[n]}$ can be identified with the connecting map of the complex $\tau ^{[0,1]}\mathrm {R}\Gamma _{\mathrm {{\acute et}}}(X_{k_{\mathrm {s}}},{\mathbb Z}/n)\otimes ^L_{{\mathbb Z}/n}A[n]\simeq \tau ^{[0,1]}\mathrm {R}\Gamma _{\mathrm {{\acute et}}}(X_{k_{\mathrm {s}}},A[n])$ . Thus the differential

$$ \begin{align*}d\colon\mathrm{H}^1_{\mathrm{{\acute et}}}(X_{k_{\mathrm{s}}},A[n])^\Gamma \to\mathrm{H}^2_{\mathrm{{\acute et}}}(k,A[n])\end{align*} $$

of the spectral sequence

$$ \begin{align*}E_2^{p,q}=\mathrm{H}^p(k, \mathrm{H}^q_{\mathrm{{\acute et}}}(X_{k_{\mathrm{s}}},A[n]))\to\mathrm{H}^{p+q}_{\mathrm{{\acute et}}}(X,A[n])\end{align*} $$

is induced by $\delta _X\otimes \mathrm {id}_{A[n]}$ . There are natural isomorphisms of $\Gamma $ -modules

$$ \begin{align*}\mathrm{H}^1_{\mathrm{{\acute et}}}(X_{k_{\mathrm{s}}},A[n])\cong \mathrm{H}^1_{\mathrm{{\acute et}}}(A_{k_{\mathrm{s}}},A[n])\cong\mathrm{H}^1_{\mathrm{{\acute et}}}(A_{k_{\mathrm{s}}},{\mathbb Z}/n)\otimes_{{\mathbb Z}/n}A[n]\cong \mathrm{{End}}_{{\mathbb Z}/n}(A[n]).\end{align*} $$

In the notation of [Reference SkorobogatovSko01, Proposition 3.2.2], under these isomorphisms, $\mathrm {id}_{A[n]}\in \mathrm {{End}}_{{\mathbb Z}/n}(A[n])$ corresponds to the class $\tau \in \mathrm {H}^1_{\mathrm {{\acute et}}}(X_{k_{\mathrm {s}}},A[n])$ of the $X_{k_{\mathrm {s}}}$ -torsor for $A[n]$ given by (40) after an identification of $X_{k_{\mathrm {s}}}$ with $A_{k_{\mathrm {s}}}$ . (Note that $\tau $ does not depend on the choice of a $k_{\mathrm {s}}$ -point of X, thus $\tau \in \mathrm {H}^1_{\mathrm {{\acute et}}}(X_{k_{\mathrm {s}}},A[n])^\Gamma $ .) This gives $d(\tau )=\delta _X$ .

On the other hand, by [Reference SkorobogatovSko01, Proposition 3.2.2], which is a restatement of [Reference GiraudGir71, Proposition V.3.2.9], we have $d(\tau )=\beta _n([X])$ .

Proof. The canonical map $C\to X$ induces a map $\mathrm {R}\Gamma _{\mathrm {{\acute et}}}(X_{k_{\mathrm {s}}},\mu _n^{\otimes i})\to \mathrm {R}\Gamma _{\mathrm {{\acute et}}}(C_{k_{\mathrm {s}}},\mu _n^{\otimes i})$ in $D(\Gamma ,{\mathbb Z}/n)$ , which induces isomorphisms of cohomology groups in degrees 0 and 1. This gives (a).

Poincaré duality states that in $D(\Gamma ,{\mathbb Z}/n)$ the objects $\mathrm {{Hom}}_{{\mathbb Z}/n}(\mathrm {R}\Gamma _{\mathrm {{\acute et}}}(C_{k_{\mathrm { s}}},{\mathbb Z}/n),\mathbb{Z}/n)$ and $\mathrm {R}\Gamma _{\mathrm {{\acute et}}}(C_{k_{\mathrm {s}}},\mu_n)[2]$ are canonically isomorphic. Since each cohomology group of $\mathrm {R}\Gamma _{\mathrm {{\acute et}}}(C_{k_{\mathrm {s}}},{\mathbb Z}/n)$ is a free ${\mathbb Z}/n$ -module, this implies a duality on truncations: the objects $\mathrm {{Hom}}_{{\mathbb Z}/n}(\tau ^{[0,1]}\mathrm {R}\Gamma _{\mathrm {{\acute et}}}(C_{k_{\mathrm {s}}},{\mathbb Z}/n),{\mathbb Z}/n)$ and $\tau ^{[1,2]}\mathrm {R}\Gamma _{\mathrm {{\acute et}}}(C_{k_{\mathrm {s}}},\mu_n)[2]$ are canonically isomorphic. Therefore the differential $\delta ^{0,2}_{2,C}$ of the spectral sequence with coefficients ${\mathbb Z}/n$ is given by the class of a 2-extension in $\mathrm {Ext}^2_k({\mathbb Z}/n,\mathrm {H}^1_{\mathrm {{\acute et}}}(J^\vee _{k_{\mathrm {s}}},\mu _n))$ that is dual to the extension defining $\delta ^{0,1}_{2,C}$ . It follows that $\delta ^{0,2}_{2,C}$ sends the generator of $\mathrm {H}^2_{\mathrm {{\acute et}}}(C_{k_{\mathrm {s}}},\mu _n)^\Gamma \cong {\mathbb Z}/n$ to the image of $[X]$ in $\mathrm {H}^2(k,J[n])\cong \mathrm {H}^2(k,\mathrm {H}^1_{\mathrm {{\acute et}}}(J^\vee _{k_{\mathrm {s}}},\mu _n))$ .

Proof. Under the assumption that $[{\mathbf {Pic}}_{C/k}^1]$ is divisible by n, all differentials on the second page vanish by Proposition 6.1. To access the differentials on the next page, we relate the extensions between the cohomology modules of the complex $\mathrm {R}\Gamma _{\mathrm {{\acute et}}}(C_{k_{\mathrm {s}}},\mu _n)$ to those of $\mathrm {R}\Gamma _{\mathrm {{\acute et}}}(C_{k_{\mathrm {s}}},{\mathbb G}_m)$ .

The complex $\tau ^{\leq 2}\mathrm {R}\Gamma _{\mathrm {{\acute et}}}(C_{k_{\mathrm {s}}},{\mathbb G}_m)$ has non-zero cohomology modules only in degrees $0$ and $1$ , isomorphic to $k_{\mathrm {s}}^{\times }$ and ${\mathbf {Pic}}_{C/k}(k_{\mathrm {s}})$ , respectively, cf. [Reference Colliot-Thélène and SkorobogatovCTS21, Theorem 5.6.1 (iv)]. The degree $2$ extension between them is exactly the class $\delta $ . According to the Kummer sequence, applying the functor $-\otimes _{{\mathbb Z}}^L{\mathbb Z}/n:D(\Gamma ,{\mathbb Z})\to D(\Gamma ,{\mathbb Z}/n)$ to the object $\mathrm {R}\Gamma _{\mathrm {{\acute et}}}(C_{k_{\mathrm {s}}},{\mathbb G}_m)$ gives $\mathrm {R}\Gamma (C_{k_{\mathrm {s}}},\mu _n)[1]$ , and the extension

$$ \begin{align*}\mathrm{H}^0_{\mathrm{{\acute et}}}(C_{k_{\mathrm{s}}},{\mathbb G}_m)\to \tau^{\leq 2}\mathrm{R}\Gamma_{\mathrm{{\acute et}}}(C_{k_{\mathrm{ s}}},{\mathbb G}_m)\to \mathrm{H}^1_{\mathrm{{\acute et}}}(C_{k_{\mathrm{s}}},{\mathbb G}_m)[-1] \end{align*} $$

is carried to

$$ \begin{align*}\mathrm{H}^0_{\mathrm{{\acute et}}}(C_{k_{\mathrm{s}}},\mu_n)[1]\to \mathrm{R}\Gamma_{\mathrm{{\acute et}}}(C_{k_{\mathrm{s}}},\mu_n)[1]\to (\tau^{[1,2]}\mathrm{R}\Gamma_{\mathrm{{\acute et}}}(C_{k_{\mathrm{s}}},\mu_n))[1]. \end{align*} $$

Hence the extension

(51) $$ \begin{align} \tau^{[1,2]}\mathrm{R}\Gamma_{\mathrm{{\acute et}}}(C_{k_{\mathrm{s}}},\mu_n)[2]\to \mu_n[3] \end{align} $$

arising from the complex $\mathrm {R}\Gamma _{\mathrm {{\acute et}}}(C_{k_{\mathrm {s}}},\mu _n)$ is the result of applying the functor $-\otimes _{{\mathbb Z}}^L{\mathbb Z}/n$ to the map $\delta :{\mathbf {Pic}}_{C/k}(k_{\mathrm {s}})\to k_{\mathrm {s}}^{\times }[2]$ .

Under the assumption that $[{\mathbf {Pic}}_{C/k}^1]$ is divisible by n, the natural map

(52) $$ \begin{align} \tau^{[1,2]}\mathrm{R}\Gamma_{\mathrm{{\acute et}}}(C_{k_{\mathrm{s}}},\mu_n)[2]\simeq {\mathbf{Pic}}_{C/k}(k_{\mathrm{s}})\otimes^L_{{\mathbb Z}}{\mathbb Z}/n\to {\mathbb Z}/n \end{align} $$

has a section in $D(\Gamma ,{\mathbb Z}/n)$ and the task of computing the third differential is equivalent to computing the composition of this section with (51).

Recall from (50) that for any Galois module M there is a natural isomorphism $\mathrm {Ext}^i_{\Gamma }({\mathbb Z}/n, M[1])\cong \mathrm {H}^i(k, M\otimes ^L_{{\mathbb Z}}{\mathbb Z}/n)$ . The morphism $h:{\mathbb Z}/n\to {\mathbf {Pic}}_{C/k}(k_{\mathrm {s}})[1]$ corresponding to the extension $0\to {\mathbf {Pic}}_{C/k}(k_{\mathrm {s}})\to A(k_{\mathrm {s}})\to {\mathbb Z}/n\to 0$ is carried to a class in $\mathrm {H}^0(k, {\mathbf {Pic}}_{C/k}(k_{\mathrm {s}})\otimes _{{\mathbb Z}}^L{\mathbb Z}/n)$ which defines a splitting of (52) because the composition

$$ \begin{align*}{\mathbb Z}/n\xrightarrow{h}{\mathbf{Pic}}_{C/k}(k_{\mathrm{s}})[1]\xrightarrow{\mathrm{{deg}}[1]}{\mathbb Z}[1]\end{align*} $$

is the Bockstein map.

The composition of this splitting with (51) is then the map ${\mathbb Z}/n\to \mu _n[3]$ whose class in $\mathrm {H}^3(k,\mu _n)$ corresponds to the composition ${\mathbb Z}/n\xrightarrow {h} {\mathbf {Pic}}_{C/k}(k_{\mathrm {s}})[1]\xrightarrow {\delta [1]}k_{\mathrm {s}}^{\times }[3]$ , and we arrive at the definition of the class $c_{1/n}$ , as desired.

Proof. Since n is invertible modulo m, the class $[{\mathbf {Pic}}^1_{C/k}]$ is divisible by n, so we are in the setup of Proposition 6.5. Specifically, choose $m'$ to be any integer such that $m'\cdot n\equiv 1\bmod m$ and let ${\mathbf {Pic}}_{C/k}^{1/n}:=[m']_*{\mathbf {Pic}}^1_{C/k}$ . The point x defines a map of Galois modules $f_x:{\mathbb Z}\to {\mathbf {Pic}}_{C/k}(k_{\mathrm {s}})$ sending $1\in {\mathbb Z}$ to x. The composition

$$ \begin{align*}{\mathbb Z}\xrightarrow{f_x}{\mathbf{Pic}}_{C/k}(k_{\mathrm{s}})\xrightarrow{\delta}k_{\mathrm{s}}^{\times}[2]\xrightarrow{\mathrm{Bock}_{{\mathbb G}_m}}\mu_n[3] \end{align*} $$

corresponds under the isomorphism $\mathrm {H}^3(k,\mu _n)\simeq \mathrm {Ext}^3_{\Gamma }({\mathbb Z}/n,k_{\mathrm {s}}^{\times })$ to the composition

(53) $$ \begin{align} {\mathbb Z}/n\to{\mathbb Z}[1]\xrightarrow{f_x[1]}{\mathbf{Pic}}_{C/k}(k_{\mathrm{s}})[1]\xrightarrow{\delta[1]}k_{\mathrm{s}}^{\times}[3]. \end{align} $$

The composition of the first two maps in (53) equals

$$ \begin{align*}{\mathbb Z}/n\xrightarrow{f_x\otimes^L_{{\mathbb Z}}{\mathbb Z}/n}{\mathbf{Pic}}_C(k_{\mathrm{s}})\otimes^L_{{\mathbb Z}}{\mathbb Z}/n\xrightarrow{\mathrm{ id}\otimes\mathrm{Bock}}{\mathbf{Pic}}_C(k_{\mathrm{s}})[1].\end{align*} $$

Since $f_x\otimes ^L_{{\mathbb Z}}{\mathbb Z}/n$ is a section of the natural map ${\mathbf {Pic}}_C(k_{\mathrm {s}})\otimes ^L_{{\mathbb Z}}{\mathbb Z}/n\to {\mathbb Z}/n$ , the above composition (53) is the map inducing the differentials on the third page of our spectral sequence, as in the proof of Proposition 6.5.