Conjugate surfaces: informations about the orbits

Question

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.

Answer 1

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.