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Apr 18, 2016 at 9:53 comment added a_g Anycase, the refundation of Sabbah and Schnell are very useful, so thank you so much for the reference.
Apr 18, 2016 at 9:53 comment added a_g I totally agree: absolutely not since the graded pieces of $F$ are no longer $D$-modules. Indeed, if my question was true, then, using that shifting we'll get that the categories of pure Hodge modules over a fixed variety are all isomorphic, no matter the weight. But this is absolutely false as can be shown, for example, using the category of pure Hodge structures of weight 0 and 1, that have a different kind of symmetry in their decompositions. However, using Tate's twisting, we get all the pure Hodge modules with even weight are isomorphic, and odd-weighted too. Is it OK?
Apr 4, 2016 at 14:38 comment added Donu Arapura Oh yea good point. So no, in either case.
Apr 4, 2016 at 14:29 comment added Ben Webster But the associated graded for $F$ isn't a D-module, so it's obviously not a mixed Hodge module, right? Am I going crazy here?
Apr 4, 2016 at 14:07 comment added Donu Arapura Maybe yes, maybe no. No using Saito's original definition, because $F$ is almost never defined over $\mathbb{Q}$. On the other hand, Sabbah and Schnell have been reworking the foundations, so that it is no longer necessary to have a $\mathbb{Q}$ "lattice" in their version. See cmls.polytechnique.fr/perso/sabbah.claude/MHMProject/mhm.html
Apr 4, 2016 at 12:47 comment added Geordie Williamson It is probably helpful to think about a variation of pure Hodge structure. The associated graded for the Hodge filtration not a pure variation of Hodge structure. (Think about a non-trivial variation of Hodge structure of type (1,1) given by the $H^1$ of a family elliptic curves. One wants to think about this object as being "simple", so it shouldn't break up any further.)
Apr 4, 2016 at 10:25 history asked a_g CC BY-SA 3.0