In the survey paper


of Bauer-Catanese-Pignatelli, they mention a question of Mumford:

Can a computer classify all surfaces of general type with $p_g=0$?

I've been playing a bit with the Craighero-Gattazzo (CG) surface (a particular surface of this kind) using various computer algebra systems, and my life has been made difficult by the fact that the standard equations for this surface are defined over a cubic extension of $\mathbb{Q}$ rather than over $\mathbb{Q}$ itself (it seems to make computations run for longer and various algorithms are not implemented).

This made me worry about Mumford's question: because $\bar{\mathbb{Q}}$ is only countable, a generic complex surface in this moduli space will only be defined over some transcendental extension of $\mathbb{Q}$, which presumably makes Groebner basis calculations even less tractable. So my question is:

In the moduli space of general type surfaces, is anything known about the existence or density of surfaces defined over $\mathbb{Q}$ or $\bar{\mathbb{Q}}$? Should I be able to perturb the pluricanonical ring of the CG surface and find a "nearby" surface defined over $\mathbb{Q}$? Should every component of moduli space contain a surface defined over $\bar{\mathbb{Q}}$?

If this question is too general, I would be happy to know the answer to the following more concrete question.

The CG surface has an explicit birational model as a quintic in $\mathbb{P}^3$ with four simple elliptic singularities. The standard model is defined over $\mathbb{Q}[r]/(r^3+r^2-1)$. Is it known that it is necessary to work over this cubic extension, or could there be a similar quintic defined over $\mathbb{Q}$ with the requisite properties, i.e. whose minimal resolution is biholomorphic to (or at least deformation equivalent to) the CG surface?

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    $\begingroup$ If $M$ is a scheme of finite type defined over $\bar{\mathbb{Q}}$, then $M(\bar{\mathbb{Q}})$ is Zariski dense in $M$ (and $M_\mathbb{C}$). (Your questions about $\mathbb{Q}$ are way more difficult. Some hyperbolicity properties of the moduli space might imply that $M(\mathbb{Q})$ is not Zariski dense, perhaps even finite.) $\endgroup$ – Piotr Achinger Dec 6 '19 at 12:38
  • $\begingroup$ Thanks, Piotr. But is it clear that the moduli space of complex surfaces is defined over $\bar{\mathbb{Q}}$? $\endgroup$ – Jonny Evans Dec 6 '19 at 12:57
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    $\begingroup$ The moduli space is defined over $\mathbb{Q}$ as the moduli problem is over $\mathbb{Q}$. $\endgroup$ – Daniel Loughran Dec 6 '19 at 14:08
  • $\begingroup$ Sorry, I'm being dumb: why is the moduli problem over Q? My objects are complex surfaces with at worst canonical singularities and ample canonical bundle (and fixed Chern numbers). By Bombieri, the 5-canonical bundle is very ample, and these embed as projective varieties in some fixed projective space (i.e. they admit a description as Proj(R) for some graded ring R over C) but everything so far has been over C. From this point of view, how do you see that these are the complex points of a moduli space over Q? $\endgroup$ – Jonny Evans Dec 6 '19 at 14:23

Firstly, since you are interested in the field of definition of the surfaces, you should work with the moduli stack, rather than the coarse moduli space. A $k$-rational point on the moduli stack corresponds to a surface defined over $k$, but the same is not true for the coarse moduli space (this is the issue that ``Field of definition $\neq$ Field of moduli'' in general for coarse moduli spaces; though this issue only occurs for surfaces with non-trivial automorphism group).

So let $\mathcal{X}$ be moduli stack of interest. The functor of points is as follows: for a $\mathbb{Q}$-scheme $S$, we have that $\mathcal{X}(S)$ is the set of all smooth proper morphisms $Y \to S$ of relative dimensional two with connected fibres such that the relative canonical bundle $\omega_{Y/S}$ is relatively ample.

You seem to be also interested in singular surfaces, so you should adapt the definition as required; but this will correspond to some compactification of this stack. As you are interested in density of rational points, one can just work with the open subset corresponding to smooth surfaces. Moreover, if you want a stack of finite type, you should restrict the Hodge numbers or Hilbert polynomial, as you have done in the question.

Now note that the definition makes sense for any $\mathbb{Q}$-scheme $S$, so this means that $\mathcal{X}$ is defined over $\mathbb{Q}$. In fact, you can easily modify the functor of points definition for an arbitrary scheme $S$, which means that one can define a moduli stack over $\mathbb{Z}$, just as Deligne-Mumford defined the moduli stack over curves of given genus over $\mathbb{Z}$.

Next, in general, unless there is a good reason otherwise, one expects that moduli spaces are of general type. Therefore conjectures of Lang-Vojta predict that the $\mathbb{Q}$-rational points are not Zariski dense. Hence in general there is no reason to expect to be able to approximate a surface over $\mathbb{C}$ by a surface over $\mathbb{Q}$.

As for $\bar{\mathbb{Q}}$-points, as explained in the comments, for a non-empty finite type $\bar{\mathbb{Q}}$ scheme, the $\bar{\mathbb{Q}}$-points are Zariski dense.

  • $\begingroup$ Thanks for this answer. What's still bothering me is that I'm really interested in the Gieseker coarse moduli space (of canonically polarised complex surfaces). Presumably your moduli stack has an associated coarse moduli space. Is it clear that these coarse moduli spaces are the same, or is yours just a subset? Could there be components of the Gieseker moduli space of complex surfaces that just don't contain any surfaces which are biholomorphic to surfaces which are defined over Q? $\endgroup$ – Jonny Evans Dec 7 '19 at 9:29
  • $\begingroup$ Sorry I don't know what the Gieseker coarse moduli space is. The coarse moduli space of a stack satisfies a universal property so is unique. This is the right space to work with; I would guess that it agrees with your space on some dense open subset. If I understand you question correctly: yes a variety over $\mathbb{Q}$ need not have any $\mathbb{Q}$-rational points. $\endgroup$ – Daniel Loughran Dec 8 '19 at 20:54
  • $\begingroup$ It's the moduli space whose points correspond to isomorphism classes of complex surfaces with ample canonical bundle. Each of these is biholomorphic to some $Proj(R)$ for a finitely generated graded ring $R$ over $\mathbb{C}$, but I don't see why, for some dense set of surfaces in the moduli space, $R$ should be of the form $R'\otimes_{\bar{\mathbb{Q}}}\mathbb{C}$ for $R'$ defined over $\bar{\mathbb{Q}}$. Sorry, the Q in the question at the end of my comment should have been \bar{Q} not just Q. $\endgroup$ – Jonny Evans Dec 8 '19 at 21:17
  • $\begingroup$ For example, are you saying that if I have an isolated point in the moduli space of complex surfaces then it is biholomorphic to an algebraic surface defined over $\bar{\mathbb{Q}}$? This seems to follow from what you're saying (unless, as is very possible, there's something I'm missing). $\endgroup$ – Jonny Evans Dec 8 '19 at 22:14
  • $\begingroup$ I think providing your moduli stack is sufficiently nice (e.g Deligne-Mumford and quasi-compact and quasi-separated) then the answer is yes. I'm not sure about in general as there are some very bad stacks out there. $\endgroup$ – Daniel Loughran Dec 10 '19 at 10:36

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