Timeline for Moduli space of motives vs moduli space of varieties
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| Nov 25, 2013 at 23:05 | comment | added | user25309 | If I want that my preceding comment makes sense, I should rather say : such that the family of Hodge structures over C admits a connection such that Griffits transversality is satisfied (i.e. one has a "variation of Hodge structures") | |
| Nov 25, 2013 at 22:36 | comment | added | user25309 | Let H in D be in the "image of motives". Let C a germ of curve in D through H such that the family of Hodge structures over C satisfies Griffits transversality. Is C in the "image of motives"? (In other words, is Griffiths transversality the only obstruction) | |
| Nov 25, 2013 at 11:54 | comment | added | Donu Arapura | Yes, I was taking $D$ to be Griffiths period domain, which is the moduli space of polarized Hodge structures with basis; it is still a complex manifold. | |
| Nov 25, 2013 at 10:58 | comment | added | eric | In the period domains Deligne uses in his "travaux de Shimura", Griffiths transversality is built into the axioms. So you're using a more general period domain probably -- it is still a complex manifold, right? Do you need some sort of algebraicity here to make this argument work though (you mention quasi-projective varieties and I am not sure why, or where they're coming from)? | |
| Nov 25, 2013 at 10:56 | vote | accept | eric | ||
| Nov 22, 2013 at 13:50 | comment | added | Donu Arapura | Indeed my answer was more of an extended comment. Perhaps later on if I have more time and energy... | |
| Nov 22, 2013 at 9:22 | comment | added | jmc | Thanks for this answer! You have not yet taken twisting into account: maybe $H$ is a summand of $H^{2+2j}(j)$ of some smooth projective variety. But I guess this does not change the rest of the argument, and you still find a countable union $\bigcup T_{i}$. There is one part that I do not understand. Why do we need to throw away some of the $T_{i}$? (Do all the Hodge structures parameterised by one $T_{i}$ have the same Hodge numbers?). And can we be sure that the non-typical cases, where the constraint $F^{p} \subset F^{p-1}$ is trivial, don't blow up the argument? | |
| Nov 21, 2013 at 22:36 | history | answered | Donu Arapura | CC BY-SA 3.0 |