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Suppose $X$ and $Y$ are two connected smooth projective varieties over $\mathbb{Q}$ (of the same dimension) that have the same $\ell$-adic Galois representations (up to semisimplification). What is the relation between $X(\mathbb{Q})$ and $Y(\mathbb{Q})$, even conjecturally?

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    $\begingroup$ (1) If $X$ and $Y$ are abelian varieties, then BSD should imply that the ranks are the same, (2) if you consider the action of ${\rm Gal}(\bar{ \mathbf{Q}}/\mathbf{Q})$ not on $\ell$-adic cohomology but on the arithmetic fundamental groups $\pi_1(X)$, $\pi_1(Y)$, then the section conjecture of anabelian geometry would imply that $X(\mathbf{Q}) = Y(\mathbf{Q})$ provided that $X$ and $Y$ are `anabelian'. $\endgroup$ – Piotr Achinger Jan 18 '17 at 6:19
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    $\begingroup$ @PiotrAchinger (1) already follows from the Tate conjecture for morphisms of abelian varieties, proved by Faltings, which implies that they are isogenous. Similarly for (2), by Grothendieck's anabelian conjecture, proved by Mochizuki, the condition on arithmetic fundamental groups implies the two curves are isomorphic. $\endgroup$ – Will Sawin Jan 18 '17 at 6:25
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In general one can say very little. There are some positive results (as indicated in the comments) in special cases, but the below example kills any hope that one can say something in general. NB "they have the same $\ell$-adic Galois representations" is vague -- I will interpret as "they have isomorphic $\ell$-adic etale cohomology in all degrees" which is the strongest reasonable interpretation I can think of.

So let $E$ be an elliptic curve over $\mathbb{Q}$ with positive rank and non-trivial Tate-Shaferevich group. Let $C$ be a torsor for $E$ corresponding to a non-trivial element of the group. Then $E(\mathbb{Q})$ is infinite and $C(\mathbb{Q})$ is empty.

However I think that the etale cohomology groups of $E$ and $C$ are all isomorphic. $H^0$ is trivial, $H^2$ is the cyclotomic character, and $H^1(C)$ is isomorphic to the dual of the Tate module of the Jacobian of $C$, and the Jacobian of $C$ is isomorphic to $E$ again, so $H^1(C)$ and $H^1(E)$ are both isomorphic to the dual of the Tate module of $E$.

In this example $E(\mathbb{Q})$ and $C(\mathbb{Q})$ are completely different, but at least $E(\mathbb{Q}_p)$ and $C(\mathbb{Q}_p)$ are the same for all $p$. So as another example, let $X$ be projective 1-space over $\mathbb{Q}$ and let $Y$ be a smooth plane conic over $\mathbb{Q}$ with no local points at some prime $p$. The problem here is that the Galois representations tell you very little about the geometry of the variety. $H^0$ is trivial, $H^2$ is cyclotomic, and $H^1$ is zero; again $X(\mathbb{Q})$ is infinite, $Y(\mathbb{Q})$ is empty, and even worse $X(\mathbb{Q}_p)$ is infinite and $Y(\mathbb{Q}_p)$ is empty. Note also that in this case $\pi_1(X)$ and $\pi_1(Y)$ are also isomorphic as Galois modules; the comments by Will Sawin and Piotr Achinger do not apply because $X$ and $Y$ do not have high enough genus to be anabelian.

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    $\begingroup$ Given your examples, maybe it makes sense to ask the question only for those projective varieties $X$ over a number field that have "large fundamental group" in the sense that for every normal subvariety $W$, the induced homomorphism $\pi_1^{alg}(W)\to \pi_1^{alg}(X)$ has infinite image. One could also ask for a comparison of rational points after a finite extension of the ground field. That would address the examples coming from torsors. $\endgroup$ – Jason Starr Jan 18 '17 at 12:09
  • $\begingroup$ What if we also have a morphism which induces the isomorphisms on cohomology? $\endgroup$ – user00000 Jan 19 '17 at 9:38
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    $\begingroup$ @JasonStarr: there is also Grothendieck's approach -- stick to the "anabelian" setting; then $\pi_1$ might determine the rational points purely group-theoretically. user00000: now there's a question! Let me show my ignorance by saying that of course one could ask this for e.g. compact complex manifolds first -- is it true then? Probably not -- you'll maybe just get something like a homotopy equivalence or something, rather than an isomorphism. It would not surprise me if one could then patch up an example in this more arithmetic setting. $\endgroup$ – Kevin Buzzard Jan 19 '17 at 11:51

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