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Sep 29, 2011 at 19:34 comment added Damian Rössler Contained in this question is the question "Is a smooth projective variety determined by its motive" (because the motive will have the same Galois representations). For instance, for a curve, this is the question "Is the curve determined by its Jacobian ?". In this case (if one throws in polarisations...), Torelli's theorem gives the answer. There is also a Torelli-type theorem for K3 surfaces; but these are very special cases. In general, the motive loses a lot of the "non-linear" information contained in the variety.
Sep 29, 2011 at 17:11 comment added Joël One remark: suppose you replace étale cohomology (with its Galois action) by the étale fundamental group (with its outer Galois action), then in many cases (if your fundamental group is "anabelian") then by Grothendick's anabelian conjectures there should be only one variety having this $\pi_1$.
Sep 29, 2011 at 17:04 comment added Joël Interesting question, but which needs to be precised in several ways before admitting a satisfying answer. For example, if a Galois rep. $V$ occurs in the cohomology of $X$, then it occurs also in the cohomology of $X \times Y$ for any variety $Y$ so the set in question is very large.
Sep 29, 2011 at 17:00 comment added David Loeffler One example: any torsor for an abelian variety has the same $\ell$-adic cohomology as the abelian variety, so that gives examples where there are "many" varieties realizing the same Galois representation. This came up before in mathoverflow.net/questions/18006 (see the comment by Emerton and the linked notes by Dalawat).
Sep 29, 2011 at 16:36 history asked user18162 CC BY-SA 3.0