nt.number theory - Everywhere locally isomorphic abelian varieties

Here's a slight variant of Felipe Voloch's answer, for those who don't have a favorite group cohomology class. Let $C$ be an abelian variety over $mathbb{Q}$. Suppose that all the $overline{mathbb{Q}}$ automorphisms of $C$ are defined over $mathbb{Q}$ and let $P$ be this automorphism group.

Take two classes in $H^1(mathrm{Gal}(overline{mathbb{Q}}/mathbb{Q}), P)$ which are distinct, but become equal in $H^1(mathrm{Gal}(overline{mathbb{Q}_v}/mathbb{Q}_v), P)$ for every $v$. The corresponding twists of $C$ should give you the examples you want.

How have I made things easier? Because I made the action of $mathrm{Gal}(overline{mathbb{Q}}/mathbb{Q})$ on $P$ trivial, I can describe the group cohmology explicitly as
$$H^1(mathrm{Gal}(overline{mathbb{Q}}/mathbb{Q}), P) cong mathrm{Hom}(mathrm{Gal}(overline{mathbb{Q}}/mathbb{Q}), P)/P.$$
Here $P$ acts by conjugation on the target.

Since $P$ is finite, any of these Hom's factor through $mathrm{Gal}(K/mathbb{Q})$ for some finite extension $K/mathbb{Q}$.

So we are now reduced to the following: We must find finite groups $G$ and $P$, an extension $K/mathbb{Q}$ with Galois group $G$, an abelian variety with automorphism group $P$ and two maps $alpha$, $beta: G to P$ such that

• $alpha$ and $beta$ are not conjugated to each other by any element of $P$ but


• when we restrict to any decomposition subgroup group, $alpha$ and $beta$ become conjugate.

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Take $G=(mathbb{Z}/2)^2$ and $P=S_6 times (mathbb{Z}/2)$. We will not use the $(mathbb{Z}/2)$ factor at all in the following; the reason it is there is that the automorphism group of an abeliabn variety always contains a central involution, namely $-1$. Feel free to think of $P$ as $S_6$.

Take $K/mathbb{Q}$ to be any biquadratic extension in which no prime is completely ramified. This condition assures that no decomposition group is the whole of $G$. Let $alpha$ send the generators of $G$ to the elements $(12)(56)$ and $(34)(56)$ of $S_6$. Let $beta$ send the generators of $G$ to $(12)(34)$ and $(13)(24)$. Then $alpha$ and $beta$ are not conjugate in $S_6$, but they become conjugate when restricted to any of the three cyclic subgroups.

The one missing step is to construct an abelian variety with automorphism group $S_6 times (mathbb{Z}/2)$, and all automorphisms defined over $mathbb{Q}$. Dror Spieser, in the comments, points out that we can just take the restriction of scalars of an elliptic curve (without CM) defined over an $S_6$ extension of $mathbb{Q}$. I still don't have a good construction of this but, thanks to Bjorn's answer, I don't need one.