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$\DeclareMathOperator\Aut{Aut}$Let $X$ and $Y$ be two complex manifolds of dimension $n$, $n\geq 2$. Denote by $\Aut(X)$ and $\Aut(Y)$ the group of bi-holomorphisms of $X$ and $Y$, respectively. Suppose the symmetric group on $n$-symbols $S_n$ is contained in both $\Aut(X)$ and $\Aut(Y)$ such that

  1. $X/S_n$ and $Y/S_n$ are complex manifolds of dimension $n$;

  2. $X/S_n$ and $Y/S_n$ are bi-holomorphic.

Question. Is it true that $X$ and $Y$ are bi-holomorphic?

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2 Answers 2

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Take two smooth hyperelliptic curves, which are not biholomorphic. Then, both have a degree $2$ morphism to $\mathbb{P}^1$. This is map may be seen as quotienting by involutions of the curves, I.e. an action of $S_{2}$.

Note that by taking different genus they don't have to be even homeomorphic.

Edit, I noticed you required that the manifolds have dimension at least 2,so taking product of the construction with a complex manifold should give the required example.

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For a series of examples in dimension $n=2$, take a degree $2k$ hypersurface $H_{2k} \subset \mathbb{P}^2$. Correspondingly, there is a double cover $X_{2k} \to \mathbb{P}^2$, branched over $k$, so $X_{2k}/\mathbb{Z}_2 \cong \mathbb{P}^2$.

If $k \neq h$ then $X_{2k}$ and $X_{2h}$ are not even homeomorphic, for instance because they have different topological Euler number.

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  • $\begingroup$ The $n$ in $S_n$ is the same as the dimensions of $X$ and $Y$ respectively. $\endgroup$
    – vikram
    Commented Jul 2, 2021 at 13:20
  • $\begingroup$ Ah, ok. Then just take $n=2$. I will edit the answer. $\endgroup$
    – Francesco Polizzi
    Commented Jul 2, 2021 at 13:24
  • $\begingroup$ In dimension $3$, I guess you can take the Galois closure of triple covers branched on hypersurfaces of the form $X_{3k} \subset \mathbb{P}^3$. $\endgroup$
    – Francesco Polizzi
    Commented Jul 2, 2021 at 13:28
  • $\begingroup$ I also would like to add that the case that I am in, the complex manifolds $X$ and $Y$ both admit the $n$-dimensional polydisc (the cartesian product of $n$-copies of the unit disc in the complex plane) as a holomorphic universal cover. Any ideas/suggestions in this case? $\endgroup$
    – vikram
    Commented Jul 2, 2021 at 13:28
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    $\begingroup$ @vikram Suitable product of curves (of genus $g \geq 2$) endowed with a diagonal $S_n$ action should do the job. You could look at the papers by Catanese and collaborators about varieties isogenous to a product. $\endgroup$
    – Francesco Polizzi
    Commented Jul 2, 2021 at 13:29

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