MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ be a complex compact Kähler minimal surface with zero algebraic dimension and $H^2(X,\mathcal{O}_X) \ne 0$. We know that according to Enriques–Kodaira classification, $X$ is either a torus or a K3 surface. In particular, its canonical bundle is trivial.

My question is, are there are any direct ways to show that the canonical bundle $K_X$ of $X$ is trivial without using the classification of complex compact surfaces? Thank you!

share|cite|improve this question
Because it's Kahler we have $H^0(X,K_X)\neq 0$, so we have a section. If we have two linearly independent sections then their ratio is a meromorphic function, so $H^0(X,K_X)$ has dimension $1$. So we are interested in showing that the vanishing set of a generator of this is empty, which I guess you do using the nonexistence of $-1$ curves. – Will Sawin Aug 22 '12 at 5:01
Thanks Will, this is what I was thinking. By minimality and the fact that $K_X$ is not ample, one has $K_X \cdot K_X = 0$; and since $K_X$ is nef, we have $K_X \cdot C = 0$ where $C$ is an irreducible curve contained in the support of the canonical divisor (which I suppose exist). So by the genus–degree formula, we have $2g-2 = C^2 \le 0$, then I stuck here... – HYL Aug 22 '12 at 6:08

You can extract a proof from the classical book "Compact complex surfaces" by Barth-Hulek-Peters-Van de Ven, in chapter VI, section 6 "The Case $a(X)=0$". Here are some details. The proof is quite direct, and does not require quoting the classification of surfaces.

First, one shows that $h^{1,0}(X)\leq 2$ (Proposition 8.1 in Chapter IV of that book).

Assume that $h^{1,0}(X)=0$. From this (and the fact that $K_X\cdot K_X=0$) you deduce immediately from Riemann-Roch that $H^0(X,-K_X)\neq 0$, which together with $H^0(X,K_X)\neq 0$ implies that $K_X$ is trivial.

The case when $h^{1,0}(X)=1$ cannot happen since it would imply on the one hand that $\chi(\mathcal{O}_X)=1$, while on the other hand $\chi(\mathcal{O}_X)=0$ by the "unbranched covering trick", Proposition 18.1 in Chapter I (plus the fact that $\chi(\mathcal{O}_X)$ behaves multiplicatively under finite unramified covers).

Finally, in the case when $h^{1,0}(X)=2$ one can show that $X$ is a torus by proving that the Albanese map of $X$ is a finite unramified cover. The details are on page 258 of that book (case e) in their proof).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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