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Let $C_1, C_2$ be two curves defined over a number field $K$. Suppose that $C_1, C_2$ are isomorphic over $\overline{\mathbb{Q}}$ but not over $K$ and that $C_1(K), C_2(K) \ne \emptyset$. Then the corresponding Jacobian varieties $J_1, J_2$ of $C_1, C_2$ respectively are also defined over $K$ and isomorphic over $\overline{\mathbb{Q}}$.

Fix a rational prime $p$, and consider the Tate modules $T_p(J_1), T_p(J_2)$. We know that since $J_1, J_2$ are isomorphic over $\overline{\mathbb{Q}}$ that there is a finite extension $L/K$ over which $J_1, J_2$ are isomorphic. It thus follows that the action of $\text{Gal}(\overline{\mathbb{Q}}/L)$ on $T_p(J_1), T_p(J_2)$ then give rise to the same Galois representation.

However, the action of $\text{Gal}(\overline{\mathbb{Q}}/K)$ on $T_p(J_1), T_p(J_2)$ need not look the same, since $J_1, J_2$ need not be isogenous over $K$. In general what can we say about the difference between the two Galois representations?

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  • $\begingroup$ I think that unless you try to make clear what "difference between two representations" means, your question runs the risk of being closed: you describe the situation pretty well, including that when you restrict the rep's to a finite-index subgroup they coincide (and have the same dimension, if $g(C_1)=g(C_2)$)... $\endgroup$ Commented Jun 27, 2019 at 8:34
  • $\begingroup$ @FilippoAlbertoEdoardo For example, how the local Euler factors differ $\endgroup$ Commented Jun 27, 2019 at 9:40

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