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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!

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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... –  Hsueh-Yung Lin Aug 22 '12 at 6:08

1 Answer 1

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).

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