# Does smooth and proper over $\mathbb Z$ imply rational?

Does smooth and proper over $\mathbb Z$ imply rational?

I think someone told me that this is a standard conjecture. Is it a widely held? held at all? Did someone in particular make this conjecture? write it down somewhere? give evidence for it? Would you like to provide evidence? Why rational and not, say, unirational?

Fontaine's non-digitized letter to Messing shows that if $X$ is smooth and proper over $\mathbb Z$, then the low degree nondiagonal Dolbeault groups vanish $H^q(X_{\mathbb Q}; \Omega^p)=0$ for $p\ne q$, $p+q\leqslant 3$. An obvious conjecture is to remove the degree restriction. Is this expected? Does this conjecture imply the rationality one? (This hypothesis on a single space does not force rationality: there are fake projective planes that mimic the cohomology of the real projective plane, but they have finite covers that would again be smooth and proper, but which can't have such cohomology by the Atiyah-Bott fixed point formula. a little more here)

What I'm looking for: A citation of a survey would be ideal. Short of that, I would accept a citation of someone making this conjecture (which may well be the paper I cite above). In any event, I would be happy if people provide evidence for or against the conjecture.

Some definitions: propriety is a relative version of compactness; let's just say that the scheme is cut out of projective space by homogeneous equations with coefficients in $\mathbb Z$. Smoothness means that the fibers, that is, the schemes defined by the same equations considered over finite fields (or their algebraic closures) are smooth. This is stronger than regularity of the total space and more like a submersion, so by analogy with Ehresmann's theorem, all the fibers are supposed to be "the same." For example, $\operatorname{Spec}\mathbb Z[i]$ is regular, but the map to $\operatorname{Spec}\mathbb Z$ is not smooth, but ramifies at $(2)$. (Geometrically) rational means that that the field of rational functions with complex coefficients are the same as on $\mathbb CP^n$, namely $\mathbb C(x_1,\ldots,x_n)$. I think this question shows that strict rationality, a fraction field of $\mathbb Q(x_1,\ldots,x_n)$, is too much to ask for.

To complete the list of relevant MO questions, would (geometric) rationality imply a Hasse principle, answering this question?

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Interesting question. I think by the word "propriety" you mean properness. But Enriques surfaces have the Hodge diamond like the one you mentioned in the 2nd paragraph (And they are not simply connected, regarding the other question you just answered). I don't know if there are smooth Enriques surfaces over $\mathbb{Z}$ though. –  temp May 16 '12 at 4:57
I'm not so sure that it's reasonable to suggest that too many more of the cohomology groups vanish. One way to convince yourself that Fontaine's results might be true are that if you consider the etale cohomology of the variety in degree $i$ then there'll be a contribution from $H^q(\Omega^p)$ if $p+q=i$. But the associated $\ell$-adic Galois representation would be unramified away from $\ell$ and crystalline at $\ell$. Such representations, other than powers of the cyclotomic character (which contributes to the $p=q$ case) are hard to come by if $i$ is small, because the weights are too... –  Kevin Buzzard May 16 '12 at 6:58
...small. However the Galois representation attached to the Ramanujan Delta function works fine, which is perhaps some hint that if $p+q=11$ then there might be some non-trivial cohomology... –  Kevin Buzzard May 16 '12 at 6:59
Here is my summary of the discussion: (1) I hallucinated the conjecture. (2) The cohomological conjecture (and thus the birational conjecture) is almost certainly false in dimension 11+, but all known counterexamples are stacks. (3) In dimensions 3-10, the cohomological conjecture might be true, but no one offered an opinion on the birational conjecture. –  Ben Wieland May 17 '12 at 15:52
Why would the finite covers of fake projective planes be unramified over every prime? If they are ramified at some prime, why are they smooth? Enriques surfaces also have a degree 2 cover. It's defined canonically so it's unramified outside 2, but I don't see why it should be unramified at 2. –  Will Sawin Dec 9 '12 at 21:35

(This answer has been edited -- it used to say that a finite cover of $\overline M_{g,n}$ gives a counterexample, which no longer seems obvious.)

If you had written "Deligne-Mumford stack" instead of "scheme", then a counterexample would be given by the spaces $\overline M_{g,n}$, which are certainly smooth and proper but far from rational in the large $g$ limit (or, for $g > 0$, in the large $n$ limit). The original references here are, I guess, Deligne (for $\overline M_{1,11}$) and Harris-Mumford (for $\overline M_{25}$).

Kevin Buzzard's hint with the Ramanujan $\Delta$ function is relevant here; indeed, $H^{11,0}(\overline M_{1,11})$ is nonzero, and the $\ell$-adic Galois representation corresponding to $H^{11,0}\oplus H^{0,11}$ is the representation attached to $\Delta$.

My answer to the question Is the moduli space of curves defined over the field with one element? contains some more detailed information about these things.

The natural way to find a scheme instead of a stack would then be to consider finite or generically finite covers of $\overline M_{g,n}$ by smooth proper schemes. There are several closely related constructions of such covers in the literature by Looijenga, Boggi-Pikaart, Pikaart-de Jong, Abramovich-Corti-Vistoli, all using some kind of non-abelian level structure on curves, but as far as I can tell none of them work over the integers.

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@Dan, please can you expand your remark that ${\overline M}_{g,n}$ is a finite cover of a smooth proper scheme (besides projective space)? As you say, it is a smooth Deligne-Mumford stack, but the inertia is not always tame, which, maybe, makes it not morally equivalent to a smooth proper scheme. –  inkspot May 16 '12 at 10:16
@inkspot you're right, this seems much more delicate than I thought. Indeed every explicit construction of such a cover that I know of uses some kind of non-abelian level structure, which doesn't work over the integers. I have to run to a seminar but I'll edit the answer later. –  Dan Petersen May 16 '12 at 11:12
Ah, you intended (as I should have realized, sorry) "is finitely covered by" rather than "is a finite cover of". –  inkspot May 16 '12 at 11:25
Does the non-tame inertia mean that they could have more general cohomology? But in practice, known cohomology classes, like Delta, don't violate restrictions on cohomology that could appear in schemes? –  Ben Wieland May 16 '12 at 14:32
Let me move on from the issue of wild inertia (because I have no good response) and point out that smooth proper DM stacks with tame inertia are everywhere: take a general surface $X$ in $\mathbb P^3_{\mathbb Z}$. Then there are only finitely many bad fibers, all in odd characteristic, and all with one singularity, an ordinary node. In dimension $2$ such singularities have a $\mathbb Z/2$-uniformization, so that $X$ is the geometric quotient of a smooth proper stack over $\mathbb Z$. So stacks are not morally equivalent to schemes... . –  inkspot May 16 '12 at 18:58