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Let $X_0$ be a smooth projective variety over $\mathbb{C}$ and let $\Theta_{X_0}$ be the locally free sheaf of $O_{X_0}$-module corresponding to tangent space of $X_0$.

Goal: To find a sufficient condition on $X_0$ so that it admits a model over $\overline{\mathbb{Q}}$ (the field of algebraic numbers).

By spreading out $X_0$ we may choose a proper morphism $$ f:X\rightarrow Spec(\overline{\mathbb{Q}}[T_1,\ldots,T_n])=:B, $$ where the $T_i$'s are "dependent variables" (i.e. they may satisfy some algebraic relations) such that when we specialize $T=(T_1,\ldots,T_n)$ to the point $P_0=(t_1,\ldots,t_n)\in\mathbb{C}^n$ we recover $X_0$. We may thus view $X$ via $f$ as a scheme over $Spec(\overline{\mathbb{Q}})$. Using sheaf cohomology, for every $\mathbb{C}$-valued point $p$ of $B$, we get a connecting homomorphism $$ \kappa:T_{B/Spec(\overline{\mathbb{Q}}),p} \rightarrow H^1(X_p,\Theta_{X_p}). $$ Note that an element $\partial\in T_{B/Spec(\overline{\mathbb{Q}}),p}$ may be viewed as a derivation of $\mathbb{C}$ over $\overline{\mathbb{Q}}$.

Now if we translate "naively" the Kodaira-Spencer deformation theory to our setting we should have a result which has the follwing flavor:

Tentative theorem: If for all $p\in B$ and all derivations $\partial\in T_{B/Spec(\overline{\mathbb{Q}}),p}$ one has that $\kappa(\partial)=0$ then $X_0$ admits a model over $\overline{\mathbb{Q}}$.

Question: Do we have such a result and if the answer is yes then what is a good reference where it is proved?

I would like a reference where the proof is as transparent as possible.

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  • $\begingroup$ Maybe my commnet is too naif and maybe I didn't get the point, but: isn't Lefschez principle sufficient to say that any complex variety has a "model" over any algebraically closed field of characteristic zero? What's the definition of "having a model over the algebraic numbers"? $\endgroup$
    – Qfwfq
    Dec 30, 2010 at 22:22
  • $\begingroup$ Also, what do you mean by "spreading out $X_0$" ? $\endgroup$
    – Qfwfq
    Dec 30, 2010 at 22:25
  • $\begingroup$ By a model of a variety $X$ over a field $K$ I simply mean that $X$ may be defined as the zero locus of a bunch of polynomials with coefficients in $K$. $\endgroup$ Dec 31, 2010 at 2:36
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    $\begingroup$ If $Y$ is a variety over a field $K$ of characteristic $0$ then by definition it is of finite type over $Spec(K)$. So any such variety may be viewed a scheme over $Spec(\mathbb{Q}[t_1,\ldots,t_n])$ where the $t_i$'s are the various coefficients which appear in a choice of a set of defining equations of $Y$. $\endgroup$ Dec 31, 2010 at 2:40
  • $\begingroup$ What if there are algebraic relations between the various coefficients $t_i$? $\endgroup$
    – Qfwfq
    Jan 2, 2011 at 17:06

3 Answers 3

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I think your question has a positive answer: The Kodaira-Spencer maps at each point $p \in B$ fit together to give a map of sheaves $\Theta_B \to R^1 f_* \Theta_{X/B}$ and your condition implies that this is map is zero. One may then base change to $C$ and apply Kuranishi's theorem (On the locally complete families of complex analytic structures. Ann. of Math. (2) 75 1962 536–577, in particular, Theorem 3) to deduce that all fibres are isomorphic as complex manifolds and hence, by GAGA, also as algebraic varieties. (The point is that one does not need a global moduli space, it suffices to have something that works analytically locally and this is supplied by Kuranishi's theorem.)

It is possible to give a purely algebraic proof of this by replacing the Kuranishi space by the Hilbert scheme of closed subschemes of $P^n$ for some large $n$ (depending on $X$) and analyzing the tangent map of the map from $B$ to such a scheme induced by choosing a projective embedding of $X$. This requires some more arguments since distinct points in the Hilbert scheme do not necessarily correspond to non-isomorphic subvarieties.

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I doubt you will get a result in the generality you want. First off, if $H^0(X_0,\Theta_{X_0}) \ne 0$, then Kodaira-Spencer is not well-behaved. This may not be a show-stopper. I think results like what you want are only possible when there is a coarse moduli space for a class of varieties containing your $X_0$. If there is such a moduli space $M$, then what you want should be true. Under your hypotheses, you get a map $f: B \to M$ and the vanishing of the Kodaira-Spencer classes amounts to $df_p = 0$ for all $p \in B$, so $f$ is constant and $f(B)$ gives your algebraic point. I can't think of a good reference offhand.

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  • $\begingroup$ So know very little about deformation theory but in my setting do we have the existence of a miniversal deformation $g:Y\rightarrow FormalSpec(R)$ which reduces modulo the maximal ideal of $R$ to $X_0\rightarrow Spec(\overline{\mathbb{Q}})$ Because if we have such a miniversal deformation then I think that this is good enough. So in otherwords, there is no need to have a moduli space! $\endgroup$ Jan 8, 2011 at 1:04
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The answer is yes, with the modification that the descent may not be from $B$ to $\overline{\mathbb Q}$ but from some etale cover of $B$. I.e. it really is a result about algebraically closed fields of definition. (The intuition being that Kodaira-Spencer trivial implies that a family is isotrivial).

The following was first proved by Buium:

Theorem Let $X$ be a variety, proper over an algebraically closed field $K$. Then $X$ is defined over the fixed field of the set of all derivations of $K$ which lift to derivations of the structure sheaf of $X$.

See:

Buium, Alexandru; Differential function fields and moduli of algebraic varieties. Lecture Notes n Mathematics, 1226. Springer-Verlag, Berlin, 1986.

Buium, Alexandru; Fields of definition of algebraic varieties in characteristic zero. Compositio Math. 61 (1987), no. 3, 339–352.

and also:

Gillet, Henri; "Differential algebra - A Scheme Theory Approach", in Differential algebra and related topics: proceedings of the International Workshop, Newark Campus of Rutgers, The State University of New Jersey, 2-3 November 2000, Editors Li Guo, William F. Keigher, World Scientific

The converse statement is of course trivial

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