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According to this paper, Serre proved that there exists a pair $X, X'$ of smooth complex projective varieties, such that $X, X'$ are conjugate but not homeomorphic. Here, we say that $X, X'$ are conjugate if there exists a diagram

$\require{AMScd}$ \begin{CD} X @>{}>> X'\\ @VVV @VVV\\ \operatorname{Spec} \mathbb{C} @>{\sigma}>> \operatorname{Spec} \mathbb{C}, \end{CD} where $\sigma$ is complex conjugation.

I'm completely confused by this statement: complex conjugation induces a $\mathbb{Z}/2$ action on $\mathbb{CP}^n$ by homeomorphisms. By assumption, $X \subset \mathbb{CP}^n$, and the image of $X$ under this homeomorphism is precisely $X'$ (at least I think so...). Why does this not contradict Serre theorem?

user142700
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1 Answers1

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Your argument is correct, if we use complex conjugation, then these two varieties are isomorphic. Serre’s example however (and all such examples) use the base change along an exotic embedding of $\mathbb{C}$ into $\mathbb{C}$. In your linked paper it doesn’t mention $complex$ conjugation.

Chris H
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