Timeline for Moduli space of motives vs moduli space of varieties
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Nov 25, 2013 at 22:43 | comment | added | user25309 | related : mathoverflow.net/questions/114847/… | |
| Nov 25, 2013 at 10:56 | vote | accept | eric | ||
| Nov 22, 2013 at 6:26 | comment | added | naf | @eric: Donu has already given some more details. The key point is that in most cases the universal variation over the period domain does not satisy Griffiths transversality and there are only countably many family of smooth projective varieties. | |
| Nov 21, 2013 at 22:36 | answer | added | Donu Arapura | timeline score: 11 | |
| Nov 21, 2013 at 20:46 | comment | added | Donu Arapura | Eric: Suppose that every point of the period domain $D$ of polarized Hodge structures of a particular shape corresponds to a motive. Then (and this would take work show) a general point, and all nearby points, of $D$ would lie in the image of period map $f:T\to D$, where $T$ is the base of a family of motives. So on the one hand other $f$ is locally surjective, on the other hand Griffiths tranversality says that $f$ must be horizontal which may give an obstruction to subjectivity. | |
| Nov 21, 2013 at 13:59 | comment | added | eric | @ulrich: I thought Griffiths transversality was an assertion about how a Hodge structure is allowed to move in a family, so I don't see how to apply it to one Hodge structure to deduce what you say. A dimension count shows that for curves of big genus, their H^1 is a Hodge structure of the type which is not usually the H^1 of a curve. But it is the H^1 of an ab var. I am asking about whether in general a Hodge structure can show up as a sub of an H^i(X)(j) -- I don't know if this is a reasonable question or not, but I don't understand your comment. | |
| Nov 21, 2013 at 13:57 | comment | added | eric | @John Salvatierrez: Q1 is supposed to say "why isn't this a proof that there's a nice smooth moduli space of polarized Calabi-Yau 4-folds over the complexes: (1) write down the functor (2) check it has some nice local properties (3) big machine says it's representable (4) more local analysis [and adding some auxiliary structure to kill automorphisms] says it's representable by a smooth algebraic space, hence (5) what's all the fuss about constructing moduli spaces nowadays? A big machine always works." | |
| Nov 21, 2013 at 13:12 | comment | added | John Salvatierrez | Artin/Schlessinger gives you a way to tell whether the moduli functor is actually 'algebraic'. It doesn't come for free but the fact that you know some wild Artin stack parameterises the geometric problem you started with doesn't necessarily gain you the information you wanted. | |
| Nov 21, 2013 at 13:12 | comment | added | John Salvatierrez | I don't think I understand Q1. It reminds of fundamental groups, in a silly way. Saying 'compute the fundamental group' of some topological space isn't a really well-defined question. The group $\pi_1(X)$ has a definition and so is tautologically computed for any $X$. What we would really like is a (simple as possible) presentation of it. Similarly with moduli problems, the moduli functor is the tautological answer. | |
| Nov 21, 2013 at 12:09 | review | First posts | |||
| Nov 21, 2013 at 12:12 | |||||
| Nov 21, 2013 at 12:07 | comment | added | naf | For K3 surfaces there is also a Torelli theorem and the situation is as nice as you might want. For general varieties, the period map is usually far from being a surjection. Similarly, a general Hodge structure will not be the cohomology of a motive (by "Griffiths transversality"). | |
| Nov 21, 2013 at 11:51 | history | asked | eric | CC BY-SA 3.0 |