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The only interesting $\ell$-adic cohomology of a smooth cubic threefold $X$ is $H^3(X,\mathbb{Z}_{\ell}(2))$, which is isomorphic as a $\mathrm{Gal}_k$-module to $H^1(JX,\mathbb{Z}_{\ell}(1))^{\vee}$ when the intermediate Jacobian $JX$ exists. Moreover, the cup product $H^3(X,\mathbb{Z}_{\ell}(2)) \otimes H^3(X,\mathbb{Z}_{\ell}(2)) \to \mathbb{Z}_{\ell}(1)$ corresponds to $cl_1(\theta) \in H^2(JX,\mathbb{Z}_{\ell}(1))$ under this isomorphism, where $\theta$ is the canonical theta divisor on $JX$. It is known that the pair $(JX,\theta)$ determines $X$ up to isomorphism for arbitrary fields $k$ of characteristic not $2$.

So my question is, does the $\textrm{Gal}_k$-module $H^3(X,\mathbb{Z}_{\ell}(2))$ with its cup product determine the cubic threefold $X$ up to isomorphism? That is, if two such structures for cubic threefolds $X,X'$ are isomorphic, then are $X$ and $X'$ are isomorphic?

What other classes of varieties are distinguished by their $\ell$-adic cohomology rings?

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    $\begingroup$ Certainly not if $k$ is algebraically closed... $\endgroup$
    – abx
    Commented Apr 26 at 11:44
  • $\begingroup$ @abx Can it at least detect different twists of a cubic threefold? $\endgroup$
    – TCiur
    Commented Apr 26 at 15:55
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    $\begingroup$ For these kind of problems related to the Tate Conjecture, you usually need to assume that $k$ is finitely generated, for example a number field. $\endgroup$
    – Daniel Loughran
    Commented Apr 26 at 17:24

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