Let $K$ be a number field and let $A$ be an abelian variety over $K$ (I'm mostly interested in the case that $A$ is an elliptic curve). We use $v$ to denote places of $K$ and we write $H^i(k, A)$ for the Galois Cohomology $H^i(Gal(k^{al}/k), A(K^{al}))$ where $k=K$ or $K_v$ and $G=Gal(K^{al}/k)$.

The Tate-Shafarevich group $Ш(A/k)$ is defined to be the kernel of the map

\begin{align} H^1(K, A) \to \bigoplus_{v} H^1(K_v, A_v). \end{align}

There is a pairing \begin{align} Ш(A/k) \times Ш(A^{\vee}/k) \to \mathbb{Q}/\mathbb{Z} \end{align}

The construction (which can be found in detail in the reference) goes roughly as follows: For $a \in Ш(A/k) $ we get an $A$-torsor $V_a$ representing $a$. Then there is an isomorphism of $G$-modules $A^{\vee} \to \operatorname{Pic}^0(V_a)$. Let $Q_a=K(V_a)^{\ast}/K^{\ast}$ then there is an exact sequence \begin{align} 0 \to Q \to \operatorname{Div}^0(V_a) \to \operatorname{Pic}^0(A) \to 0 \end{align} which induces a map $Ш(A^{\vee}/k) \to H^2(G,Q_a)$. By some global class field theory there is a map $\phi_a:B \to \mathbb{Q}/\mathbb{Z}$ where $B \subset H^2(G,Q_a)$ and $\operatorname{im}(Ш(A^{\vee}/k)) \subset B$.

I havent been able to show that the pairing is additive in the first coordinate. In other words, if $a, b \in Ш(A/k)$ we get three maps \begin{align} \phi_{a,b,a+b}:Ш(A^{\vee}/k) \to \mathbb{Q}/\mathbb{Z}, \end{align} and I want to show that \begin{align} \phi_a+\phi_b=\phi_{a+b}. \end{align} I have tried looking at the product of torsors in the Weil–Châtelet group but I get stuck comparing $H^2(G, Q_a),H^2(G, Q_b),H^2(G, Q_{a+b})$.

Reference: Milne - 'Arithmetic Duality Theorems' pages 80-81 (http://math.stanford.edu/~conrad/BSDseminar/refs/MilneADT.pdf)


1 Answer 1


This is not quite complete in details, and there is some hand waving, some of which I can't justify (see added comment/note), but I obviously hope that all is correct, and that it in the meantime adds something...

Setting up (a lot of, and some of it rival) notation:

In the following $k$ is a number field.

Suppose $V$ is the $A$-torsor corresponding to the cocycle $\alpha$ belonging to a cohomology class in $ H^1(k, A)$: there is an action

$$\matrix{ V\times A & \to & V \\ (v,a)& \mapsto &v +a, }$$

the map $$\matrix { V \times A &\to &V \times V\\ (v,a)&\mapsto & (v,v+a) }$$ is an isomorphism, and $V(\bar k)$ is non-empty. Let $\phi:V\otimes \bar k \to A\otimes \bar k$ be an $A$-equivariant isomorphism, i.e., $$ \phi ( v + a ) = \phi(v) + a$$ for $v\in V(\bar k)$ and $a\in A(\bar k)$. Any other such isomorphism $\psi$ is of is of the form $ T_b\circ \phi$, for some $b\in A(\bar k)$: $$ \psi(v) = \phi (v) +b. $$ We may assume that $\sigma\phi = T_{\alpha_\sigma}\circ \phi$.

Now, there is a canonical isomorphism $$ \matrix{\text{Pic}^0A &\to &\text{Pic}^0V\\ L &\mapsto& \phi^* L}$$ which does not depend on the choice of $\phi$: if $L \in \text{Pic}^0A$, then, by definition, $T_b^*L \simeq L$, for $b\in A(\bar k)$. Hence, $$(T_b \circ \phi)^*L= \phi^*L,$$ where the equal sign is calculated in $\text{Pic}^0V$.

Suppose $V$ and $W$ are $A$-torsors, corresponding to cocycles $\alpha$ and $\beta$, with corresponding $\phi$ and $\psi$ $A$-equivariant isomorphisms to $A$, normalized as above.

We consider the product space $ V \times W$, with the equivalence $ (v+a,w) \sim (v,w+a)$. We get a new torsor $Z=V\oplus W = V\times W / \sim$, and corresponding cocycle $\alpha + \beta$.

The map (over $\bar k$) $$\matrix { V \times W &\to& A \\ (v,w) &\mapsto &\phi(v) +\psi(w) } $$ descends to a $\bar k$- isomorphism $\gamma:Z \to A $, and $$\sigma\gamma (z) = \gamma(z) + \alpha_\sigma + \beta_\sigma.$$

As $V$ is an element of $Ш(A/k)$, $V_\nu=V\otimes k_\nu$ is a trivial $A_\nu$ torsor, and likewise for $W$, at every place $\nu$ of $k$.


  • $m: A\times A \to A$ for the addition morphism, and $p_A$ and $q_A$ for the projections onto the first and second factors;
  • $p_V$ and $q_W$ for the projections from $V\times W$ to $V$ and $W$;
  • $\pi$ for the quotient morphism $V\times W \to Z$.

Towards an answer of the question:

The main point is that if $L \in \text{Pic}^0 A$, then, almost by definition, $$m^*L \simeq p_A^* L\otimes q_A^*L. $$ Therefore, on $ V\times W$, we have isomorphic line bundles $$ M'= (\phi +\psi)^*L \text { and } N' = p_V^*\phi^* L \otimes q^*_W\psi^* L. $$ The line bundle $M'$ descends (hand-wave) to the line bundle $M=\gamma^*L$ on $Z$, and $$ \pi^* \gamma^* L\simeq p_V^*\phi^* L \otimes q^*_W\psi^* L.\tag{*}$$

The idea is to compare the various cohomological calculations on $V\times W$.

There is a $k_\nu$-rational point $(p,q)\in V_\nu\times W_\nu$. Therefore there is a section $ V_\nu \to V_\nu \times W_\nu$ to $p_V$, and likewise for $p_W$. Likewise, composing the isomorphism of $Z_\nu$ to $V_\nu$, there is a section of the quotient map $\pi$. There is therefore an injection on relevant cohomology groups everywhere possible - the only place still in question are the cohomology groups with coefficients in the function fields.

Consider the commutative diagram (exact rows) $$ \matrix{ 0 & \to & H^2(k_\nu, \bar k_\nu^*) & \to & H^2(k_\nu, \bar k_\nu(V)^*) & \to & H^2(k_\nu, Q_{V_\nu}) & \to & 0\\ & & \parallel & & \downarrow p^*_V & & \downarrow & & \\ 0 & \to & H^2(k_\nu, \bar k_\nu^*) & \to & H^2(k_\nu, \bar k_\nu(V\times W)^*) & \to & H^2(k_\nu, Q_{V_\nu\times W_\nu}) & \to & 0, }$$ where the subcripted $Q$'s denote the principal divisors.

The first and the third downward maps (equality and arrow) are injective. Therefore, so is the second.

Injectivity holds in the same manner for $p_W^*$ and $\pi^*$.

By running in parallel the calculation for $\gamma^*L$ using (*), and using the (local cohomological) injectivity, one can conclude that additivity holds on the $Ш(A/k)$ component of the Cassels-Tate pairing.

  • $\begingroup$ I obviously hope I haven't screwed up! But I am going to be on the road for the next while, so I wanted to post... But please comment esp. if there is something ludicrous - though it might take me a while to respond. $\endgroup$
    – peter a g
    Jun 25, 2016 at 17:14
  • $\begingroup$ My hand-waving caveat refers in particular to the 'descending to $Z$.' I am sure it's true (if it's not wrong!), but I don't have the scriptures on hand to quote - or the wherewithal to come up with a rigorous argument just like that. $\endgroup$
    – peter a g
    Jun 25, 2016 at 17:19

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