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Oct 15, 2019 at 13:13 vote accept ssx
Oct 14, 2019 at 20:52 answer added Evgeny Shinder timeline score: 7
Oct 13, 2019 at 22:02 comment added Evgeny Shinder Actually, the argument above seems to be true for all cubics (odd or even-dimensional), as according to Totaro's paper arxiv.org/pdf/1506.00968.pdf, Thm 2.2, $Hilb_2(X)$ has torsion free cohomology as soon as this holds for $X$; in particular it applies to any hypersurface.
Oct 13, 2019 at 19:58 comment added Evgeny Shinder According to our work with Galkin, arxiv.org/pdf/1405.5154.pdf, to show that $H^*(F(X), \mathbb{Z})$ are torsion-free it suffices to show that $H^*(Hilb_2(X), \mathbb{Z})$ are torsion free. Indeed this follows from Thm 5.1 by taking the realization in the Grothendieck ring of integral Hodge structures (we do not do this in the paper; for rational Hodge structures this is explained on page 8 and goes back to Bittner). Now if $X$ is even-dimensional, Totaro's result in the post implies that this indeed is the case since cohomology of $X$ are even and torsion-free.
Oct 8, 2019 at 12:58 history edited ssx CC BY-SA 4.0
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Oct 8, 2019 at 12:52 history edited ssx CC BY-SA 4.0
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Oct 8, 2019 at 11:17 history asked ssx CC BY-SA 4.0