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.

  • 6
    $\begingroup$ I don't believe this is true. For example, Kodaira has constructed examples of nonalgebraic elliptic surfaces, which necessarily have disjoint fibres consisting of algebraic curves. I'm away from my books and papers at the moment, but I think you can find this in his papers, or perhaps in the book by Barth, Peters and Van de Ven. $\endgroup$ Oct 27 '11 at 0:08
  • $\begingroup$ @Donu: Right, they are the Hopf surfaces. $\endgroup$ Oct 27 '11 at 0:14
  • 2
    $\begingroup$ Depending on your interpretation of vacuous statements you might also be interested in complex tori of dimension 2. A general torus of dimension 2 (or higher) contains no closed subvarieties except disjoint unions of points. The condition $(*)$ is then vacuously satisfied for every point on such a torus. $\endgroup$ Oct 27 '11 at 5:26
  • 3
    $\begingroup$ these beautiful answers seem to suggest modifying the question to include the hypothesis your surface has two algebraically independent meromorphic functions. I believe it is then the case that it arises from blowing down a curve on an algebraic surface. you might ask, in order to obtain a non algebraic surface, whether that exceptional curve must meet all other curves on the algebraic resolution. $\endgroup$
    – roy smith
    Oct 27 '11 at 18:43
  • $\begingroup$ @roy smith: You are almost psychically (is that a word?) to the point - the examples I was considering have this property (with a minor correction in your last sentence - "... must meet all other non-exceptional curves on the ...") and I have been sort of debating with myself if I should ask another question. $\endgroup$
    – pinaki
    Oct 27 '11 at 19:47

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].

  • $\begingroup$ I am amazed at the elegance of your crisp yet thorough answer, Francesco: it is simply perfect. Bravo! $\endgroup$ Oct 27 '11 at 9:53
  • $\begingroup$ Dear Georges, thank you for your kind words. I just used the description of Hopf surfaces given in the beautiful book by Barth-Peters and Van de Ven. $\endgroup$ Oct 27 '11 at 10:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.