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Consider a complex algebraic variety $X$ (namely a $\mathbb C$-scheme, of finite type, geometrically integral and separated); if $\sigma\in\textrm{aut}(\mathbb C)$, then is well defined the complex variety $X^\sigma$ in the following way:

  1. $X$ and $X^\sigma$ are equal as schemes.
  2. If $p:X\longrightarrow\textrm{Spec}\,\mathbb C$ is the structural morphism of $X$, then the structural morphism of $X^\sigma$ is given by $\left(\textrm{Spec}\,\sigma\right)\circ p$.

In this way, the group $\textrm{Aut}(\mathbb C)$ acts on the set of all complex varieties, but $X$ and $X^\sigma$ are in general not isomorphic (is some cases not homeomorphic).

Now suppose that $X$ is a complex surface, and consider the orbit set $$\Omega_X=\{X^\sigma\,:\sigma\in\textrm{Aut}(\mathbb C)\}$$ For which type of $X$ the set $\Omega_X$ contains only a countable number of isomorphism classes? Moreover, can you give me any reference/result that analyzes the orbit set $\Omega_X$?

Thanks in advance.

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This problem is of a more arithmetic nature, than geometric.

For example, since every automorphism of $\mathbb{C}$ preserves $\mathbb{Q}$, we see that if $X$ can be defined over $\mathbb{Q}$, then $\Omega_X$ consists of a single isomorphism class. Similarly if $X$ can be defined over a number field, then $\Omega_X$ consists of only finitely isomorphism classes.

Thus the size of $\Omega_X$ is telling you something about the size of the smallest field of definition of $X$, not really anything about the geometry of $X$.

On the other hand given any two algebraically independent transcendental numbers $\alpha$ and $\beta$, there is an automorphism of $\mathbb{C}$ which swaps $\alpha$ and $\beta$, and moreover there are uncountably many such numbers. Therefore, my guess is that $\Omega_X$ consists of countably many isomorphism classes, iff $\Omega_X$ consists of finitely many isomorphism classes, iff $X$ may be defined over a number field.

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    $\begingroup$ I think that a complex surface would always be definable over a finitely generated extension of $\mathbb{C}$, so your guess would imply that there are always only countably many isomorphism classes. $\endgroup$ – Matthias Wendt Jul 29 '14 at 15:30
  • $\begingroup$ Yes you are quite right... I shall edit my answer accordingly. $\endgroup$ – Daniel Loughran Jul 29 '14 at 16:28
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    $\begingroup$ @DanielLoughran Your guess is correct. See Criterion 1, page 3, of Gonzalez-Diez`s paper citeseerx.ist.psu.edu/viewdoc/… . $\endgroup$ – Ariyan Javanpeykar Jul 30 '14 at 7:18

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