Pathologies of analytic (non-algebraic) varieties.

Question

Note: By an "analytic non-algebraic" surface below I mean a two dimensional compact analytic variety $X$ (over $\mathbb{C}$) which is not an algebraic variety.

A property of Nagata's example (see the end of the post for the construction) of a non-algebraic normal analytic surface $X$ is the following:

($\star$) $\quad$ There is a point $P$ on $X$ such that every (compact) algebraic curve $C$ on $X$ passes through $P$.

In a paper I am writing I also constructed (to my surprise) some examples of non-algebraic normal analytic surfaces which have this peculiar property.

Questions: Is this sort of behaviour "normal" for such surfaces? Or, more precisely, if an analytic surface does not satisfy ($\star$), is it necessarily algebraic? How about for higher dimensions?

Nagata's Construction (following Bădescu's book on surfaces): Start with a smooth plane cubic $C$ and a point $P$ on $\mathbb{P}^2$ such that $P - O$ is not a torsion point (where $O$ is any of the inflection points of $C$) on $C$. Let $X_1$ be the blow up of $\mathbb{P}^2$ at $P$, and for each $i \geq 1$, let $X_{i+1}$ be the blow up of $X_i$ at the point of intersection of the strict transform of $C$ and the exceptional divisor on $X_i$. Each blow up decreases the self-intersection number of the strict transform $C_i$ of $C$ by $1$, so that on $X_{10}$ the self-intersection number of $C_{10}$ is $-1$. $X$ is the blow down of $X_{10}$ along $C_{10}$. By some theorems of Grauer and Artin, $X$ is a normal analytic surface.

Answer 1

The answer is no.

A counterexample is provided by the so-called Hopf surfaces (they were actually constructed by Kodaira, see Donu Arapura's comment).

A Hopf surface of type $\alpha=(\alpha_1, \, \alpha_2)$, where $0 < |\alpha_1| \leq |\alpha_2| < 1$, is the compact complex surface $H_{\alpha}$ obtained as the quotient of $\mathbb{C}^2 \setminus (0,0)$ by the infinite cyclic group generated by the automorphism $$ (z_1, \, z_2) \to (\alpha_1 z_1, \, \alpha_2 z_2).$$

One can prove that $H_{\alpha}$ is a compact complex surface diffeomorphic to $S^1 \times S^3$, so it admits no Kähler metric. In particular, it is not algebraic.

However, $H_{\alpha}$ does not satisfy your property $(\star)$. In fact, there is the following result:

The Hopf surface $H_{\alpha}$ is an elliptic fibre space over $\mathbb{P}^1$ if and only if $\alpha_1^l=\alpha_2^k$ for some $l, \, k \in \mathbb{Z}$. Otherwise $H_{\alpha}$ contains exactly two compact curves, which are disjoint (they are the images of the $z_1$-axis and the $z_2$-axis).

So $H_{\alpha}$ contains either infinitely many or exactly two disjoint elliptic curves.

For more details, see [Barth-Peters-Van de Ven, Compact Complex Surfaces, Chapter V].