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Suppose I have a smooth projective family of varieties of varieties over $\mathcal M_g$ - i.e. a universal functor, commuting with deformations, from curves to smooth projective varieties. Can I compute its cohomology directly from the cohomology of the curves?

In some cases, like the functor associating a curve $C$ to the smooth projective variety $\operatorname{Sym}^n C$, I certainly can. Can I in general?

I think the right way to formalize this is to consider your favorite substitute for the conjectural category of relative motives - e.g. either $\ell$-adic sheaves or variations of Hodge structures. Consider the Tannakian category of relative motives on the stack $\mathcal M_g$ and take its Tannakian fundamental group $\pi_1^{mot} (\mathcal M_g)$. Then consider the interesting part, which I'm going to take to be the maximal reductive quotient of the identity component of $\pi_1^{mot} (\mathcal M_g)$. One should also take the kernel of the natural map tot he motivic Galois group of the base field.

Is this equal to $SP_{2g}$, the part coming from the universal family of curves, or is it larger?

One way to answer this would be to exhibit a smooth projective family over $\mathcal M_g$ with weird cohomology.

For $\mathcal A_g$, $g \geq 2$, I think you can show that the identity component of the motivic fundamental group (defined in an $\ell$-adic way) is exactly $SP_{2g}$ using the congruence subgroup property.

I'm also interested in the case where we restrict to motives with good reduction at every place.

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  • $\begingroup$ Nice question! In a weak sense, the answer is "yes" by Torelli, since the (polarized) cohomology knows the curve, hence whatever you construct from it. I don't see how to get at the formalization you ask about (which I would say is more like "can we compute the cohomology of the new variety from the cohomology of the curve, via linear algebra operations?"). $\endgroup$
    – Daniel Litt
    Commented Aug 14, 2015 at 17:52
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    $\begingroup$ Actually, I think that the reductive part of $\pi_1^{mot}(M_g)$ would bigger than $Sp_{2g}$. You can see my answer to mathoverflow.net/questions/186133/… for why I think so. $\endgroup$
    – Donu Arapura
    Commented Aug 14, 2015 at 17:57
  • $\begingroup$ @DanielLitt Good point. I guess I tend to think of the cohomology as a Galois representation rather than a Hodge structure. $\endgroup$
    – Will Sawin
    Commented Aug 14, 2015 at 22:25
  • $\begingroup$ @DonuArapura Yes, I agree. There should be no problem showing that these Prym-type families have big monodromy either by topology or moment methods. This is an example of what I asked for, but it still doesn't feel "exotic" to me, so probably I should have included this in the "expected" part. For the space of semistable vector bundles, is it the case that the irreducible factors come from SP_{2g} but the extension classes don't? $\endgroup$
    – Will Sawin
    Commented Aug 14, 2015 at 22:30
  • $\begingroup$ Will, no I would agree that these examples aren't exotic. The pure graded part of the moduli of semistable bundles come from $Sp_{2g}$ as you surmised. $\endgroup$
    – Donu Arapura
    Commented Aug 15, 2015 at 12:32

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