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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.

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.

2 deleted 65 characters in body

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.

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 $\mathbb{Q}$. This shouldn't be hardDror Spieser, but I don't have it yet. I originally wanted to in the comments, points out that we can just take $C=E^6$, with $E$ the restriction of scalars of an elliptic curve , but then $\mathrm{Aut}(C)$ contains $GL_6(\mathbb{Z})$, and $\alpha$ and (without CM) defined over an $\beta$ are conjugate as maps to S_6$extension of$GL_6(\mathbb{Z})$.\mathbb{Q}$.

1

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.

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 $\mathbb{Q}$. This shouldn't be hard, but I don't have it yet. I originally wanted to take $C=E^6$, with $E$ an elliptic curve, but then $\mathrm{Aut}(C)$ contains $GL_6(\mathbb{Z})$, and $\alpha$ and $\beta$ are conjugate as maps to $GL_6(\mathbb{Z})$.