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A Hopf surface is a compact complex surface whose universal cover is complex analytically isomorphic to $\mathbb{C}^2 \setminus \{ 0 \}$. I would like to know whether anyone has any of the following examples:

(i) A non-compact complex surface whose universal cover is complex analytically isomorphic to $\mathbb{C}^2 \setminus \{ 0 \}$.

(ii) A compact complex surface whose universal cover is smoothly isomorphic to $\mathbb{C}^2 \setminus \{ 0 \}$, but not complex analytically isomorphic to $\mathbb{C}^2 \setminus \{ 0 \}$.

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    $\begingroup$ Isn't $\mathbb C^2 \setminus \{0\}$ a silly example of (i)? (Also, \backslash doesn't space well: $\mathbb C^2 \backslash \{0\}$. Prefer \setminus: $\mathbb C^2 \setminus \{0\}$. I have edited accordingly.) $\endgroup$
    – LSpice
    Commented May 12, 2020 at 1:25
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    $\begingroup$ Only marginally less silly: take a cyclic group acting linearly and freely on $C^2 \setminus \{0\}$. $\endgroup$
    – Danny Ruberman
    Commented May 12, 2020 at 1:27
  • $\begingroup$ @DannyRuberman Thank you, but now the question remains: Can we find non-silly examples? $\endgroup$
    – AmorFati
    Commented May 12, 2020 at 4:20
  • $\begingroup$ Look at Di Scala, Kasuya, Zuddas, Non-Kahler complex structures on $R^4$. These complex structures have embedded elliptic curves, so they won't be isomorphic to $\Bbb C^2 \setminus 0$ after removing a point. However, I doubt that they have compact quotients. $\endgroup$
    – Denis T
    Commented May 12, 2020 at 21:42

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(ii) I would like to prove that there are no complex surfaces that satisfy (ii).

Indeed, suppose that the universal cover $\widetilde X$ of a complex surface $X$ is diffemorphic to $\mathbb C^2\setminus 0$. Let's prove that $\widetilde X$ is biholomorphic to $\mathbb C^2\setminus 0$.

First, we note that $X$ has a finite cover that is diffeomorphic to $S^3\times S^1$. Indeed, take any

$$S^3\subset \widetilde X=\mathbb C^2\setminus 0$$

that encircles $0$. Then, since the action of $\pi_1(X)$ on $\mathbb C^2\setminus X$ is discreet, there exists only finite number of elements $g\in \pi_1(X)$ such that $g\cdot S^3$ intersects $S^3$ in $\mathbb C^2\setminus 0$. Let's take such $g_1\in\pi_1(X)$ that $g_1\cdot S^3$ is disjoint from $S^3$. Let $\mathbb Z=\langle g_1\rangle $ be the group generated by $g_1$. Then it is not hard to see that $(\mathbb C^2\setminus 0)/\langle g_1\rangle$ is diffeomorphic to $S^1\times S^3$. Clearly $S^1\times S^3$ is a finite cover of $X$.

It remains now to apply the result of Bogomolov that complex surfaces with $b_2=0$ are either Hopf surfaces or Inoue surfaces https://en.wikipedia.org/wiki/Surface_of_class_VII . Since $\pi_1(S^1\times S^3)=\mathbb Z$, the complex structure on $S^1\times S^3$ is that of a Hopf surface. We conclude that $\widetilde X$ is byholomorphic to $\mathbb C^2\setminus 0$.

(i) As for (i), you can take $(\mathbb C^2\setminus 0)/\Gamma$, where $\Gamma$ is any finite subgroup of $SU(2)$.

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    $\begingroup$ Instead of using Bogomolov's result, you can use the fact that any complex surface homeomorphic to $S^1\times S^3$ is biholomorphic to a Hopf surface. This was proved by Kodaira in Complex Structures on $S^1\times S^3$, Proc. Nat. Acad. Sci. U.S.A. 55 (1966), 240–243. $\endgroup$
    – Michael Albanese
    Commented Jun 20, 2021 at 20:20
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    $\begingroup$ Thanks Michael for giving this reference! I was not aware of this paper (regretfully) $\endgroup$
    – Dmitri Panov
    Commented Jun 20, 2021 at 22:27

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