# complex automorphisms acting on a projective variety

Consider a complex projective variety $X=\operatorname{Proj}\frac{\mathbb C[T_1,\ldots,T_n]}{(f_1,\ldots,f_n)}$ with $f_1,\ldots,f_n$ homogeneous polynomials. If $\sigma\in\operatorname{Aut}(\mathbb C)$ one can define the variety

$$X^\sigma=\operatorname{Proj}\frac{\mathbb C[T_1,\ldots,T_n]}{(f^\sigma_1,\ldots,f^\sigma_n)}$$

where $\sigma$ acts on a polynomial in the obvious way i.e. changing the coefficients. Now this action on varieties is a bit subtle in fact $X$ and $X^\sigma$ can be not isomorphic or even not homeomorphic in the Euclidean topology, so I have two questions:

1. What about the dimensions of $X$ and $X^\sigma$? For example if $X$ is a surface can $X^\sigma$ be a curve or a threefold? I'd like some example because I can't imagine one.
2. When are $X$ and $X^\sigma$ biratonal? Is there some theorem about this fact?

• The dimension over $C$ is the trascendence degree minus one of the fraction field (let us suppose the ideal is prime) of the ring, and that does not change under the action of $\sigma$. – Mariano Suárez-Álvarez May 18 '14 at 8:23
As for (2), the answer is certainly almost never, unless $X$ comes from base change of a variety defined over the real numbers. If you replace "birational" by "isomorphic", then it is precisely when $X$ comes from base change of a variety defined over the real numbers (under the assumption that $X$ is projective).
• I don't understand why $X$ and $X^\sigma$ are isomorphic exactly when $X$ is defined over the real numbers (assuming $X$ projective). – Dubious May 18 '14 at 9:47
• If $X$ is defined over $\mathbb Q$ then $X$ and $X^\sigma$ are isomorphic (in particular equal), but why this argument holds for $\mathbb R$? – Dubious May 18 '14 at 9:58
• @fair-cointossing: Dear Fair, I think Bruno was thinking of the case when $\sigma \in Aut(\mathbb C/\mathbb R)$. Regards, – Matt E May 18 '14 at 11:07