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A (projective) abelian variety $A$ over the complex numbers is determined by $H^1(A,\mathbb{Z})$ together with its Hodge structure and polarization. This miracle means that one can parametrise polarized abelian varieties (which sounds like a hard geometric problem) by parametrising their H^1's with the extra structure (which sounds like an easier algebraic problem, solved via the theory of Shimura varieties).

If we move away from abelian varieties, the miracle won't occur, and indeed attempting to understand a general variety (a complicated non-linear object) via its cohomology groups (linear objects) is presumably not going to work in general -- the linearization will lose information.

I have two related questions about what is going on in general. Let me talk about K3 surfaces below, although my real confusion has nothing to do with K3 surfaces in particular -- I could easily be saying "Calabi-Yau 8-folds" or "curves of genus 23" or indeed any random type of smooth projective variety.

Q1) I am confused about why the power of functors and representability theorems do not give me everything. This probably just reflects my lack of real understanding of what is going on. For example, let me just naively consider the functor sending a scheme S (over the complexes, if you like) to the set of isomorphism classes of polarized K3 surfaces over S (or the groupoid of polarized K3 surfaces). I now want to vaguely mutter that a big machine the likes of which I don't really understand says that this functor satisfies some basic continuity properties and hence (perhaps by some theorem of M. Artin) is representable by some algebraic stack. My understanding of the abelian variety situation is that this method is one way to prove the existence of things like Siegel modular varieties (perhaps even over $\mathbb{Z})$. Why does this method fail for more general families of varieties? I am guessing that I am perhaps being too sloppy with my polarizations but I don't really have a precise feeling for what goes wrong.

Q2) This general functor nonsense must surely fail, so here's a second approach to constructing moduli spaces of certain types of variety. Again let me stick to K3's for concreteness. I understand H^2 of a K3 surface quite well and presumably there is a Shimura variety parametrising the types of Hodge structures showing up as H^2 of a K3. I am wondering how far this Shimura variety would be from the "moduli space of K3 surfaces" which I am naively assuming exists. I can see the issue -- if the moduli space of K3's exists, there will be a map to the Shimura variety, but there is no reason to expect either injectivity or surjectivity on the face of it (I am losing information by linearising, and the linear stuff is too naive to know whether it comes from geometry). Let's forget about injectivity for a moment -- injectivity is an issue whose answer will depend on which type of variety I am trying to parametrise (e.g. for curves I have Torelli etc). But what about surjectivity? Because I'm not really interested in K3's, I'm interested in the general picture of "moduli spaces of varieties of type X" and understanding this "space" via Hodge structure, I am led to the following question, the answer to which is presumably well-known:

If $H$ is a polarizable $\mathbb{Z}$-Hodge structure of weight $n$, does one expect $H$ to be the singular cohomology of a pure motive? Perhaps more concretely, does one expect there to exist some smooth projective algebraic variety $X$ over the complexes such that $H$ is a subspace of $H^i(X,\mathbb{Z})(j)$ for some $i,j$, and even a subspace cut out by correspondences?

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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"). – ulrich Nov 21 '13 at 12:07
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. – John Salvatierrez Nov 21 '13 at 13:12
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. – John Salvatierrez Nov 21 '13 at 13:12
@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. – eric Nov 21 '13 at 13:59
@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. – ulrich Nov 22 '13 at 6:26
up vote 5 down vote accepted

Let me expand my (and ulrich's) comment slightly concerning your last question. Let $D$ be the period domain of all Hodge structures with fixed Hodge numbers and polarization. For the sake of simplicity, let's say the weight is $2$. Suppose that $H\in D$ is a summand of $H^2$ of some smooth projective variety, then by weak Lefschetz, it's a summand of $H^2$ of a smooth projective surface. All surfaces embed into $\mathbb{P}^5$. So using a Hilbert scheme argument, the set of surfaces is parametrized by a countable union of quasi projective varieties $\bigcup T_i$. With a bit more work, we can find $\bigcup T_i$, such that $t\in T_i$ parametrize pairs $(S_t, C_t)$, where $S_t$ is a surface and $C_t\subset S_t\times S_t$ is a correspondence yielding a motivic sub Hodge structure $$H^2([S_t,C_t]):=im(p_{1*}[C_t]\cup p_2^*-):[H^2(S_t)\to H^2(S_t)]$$ After throwing away some components $T_i$ we can assume that $H^2([S_t,C_t])\in D$ for all $i$ and $t\in T_i$. So we get holomorphic period maps $f_i:\tilde T_i\to D$ from the universal covers. Griffiths tranversality says that for any tangent vector $v$ of $\tilde T_i$, $df_i(v):F^p\subset F^{p-1}$, and this is typically (although not always) a nontrivial constraint. In such a typical case, this forces the image $f_i(T_i)\subset D$ to be proper. By Baire category $\cup f_i(T_i)\subset D$ is a proper subset as well. This shows that a generic element of such a typical $D$ does not come from an effective motive.

Of course this argument is highly nonconstructive. In fact, I don't know of a single explicit example of a polarizable Hodge structure which not motivic!

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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? – jmc Nov 22 '13 at 9:22
Indeed my answer was more of an extended comment. Perhaps later on if I have more time and energy... – Donu Arapura Nov 22 '13 at 13:50
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)? – eric Nov 25 '13 at 10:58
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. – Donu Arapura Nov 25 '13 at 11:54
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) – user25309 Nov 25 '13 at 22:36

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