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Surfaces can be classified using Kodaira dimensions and birational invariants. Denote by $\kappa(S)$ the Kodaira dimension of a smooth projective surface $S$. For a smooth projective surface, an equivalent characterization of Kodaira dimension uses the pluricanonical genus $P_{12}=\dim H^2(S, 12K_S)$ and $K_S^2$. When $K_S^2=0$ and $K_S$ is nef, we see that $0\leq \kappa(S)\leq 1$.

Is there an easier way (which means not using the classification of surfaces of $\kappa=0$) to proof that that $P_{12}=1$ and $K_S$ is nef implies $\kappa(S)=0$ or that $P_{12}\geq 2$, $K_S^2=0$ if $K_S$ is nef and $\kappa(S)=1$.

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  • $\begingroup$ The second question seems easier. The assertions $K$ nef and $K^2=0$ imply that if a multiple of $K$ moves, then it maps to a curve. Combining this with the inequality on the 12th plurigenus gives you the result. I am not sure how to proceed in the other case: you would need to know something similar to $K^2=0$ to proceed in a similar way. $\endgroup$
    – M P
    Commented Apr 29, 2013 at 8:04
  • $\begingroup$ Thanks, MP. You are right, the second one is easier. I should ask the other way around. $\endgroup$
    – Fei YE
    Commented Apr 30, 2013 at 9:30

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All of the following is from Beauville's wonderful, short, dense book: "Complex Algebraic Surfaces." I really recommend it if you want to learn about the classification or even general techniques in surface theory.

For a minimal surface, $P_{12}=1$ alone is equivalent to $\kappa=0$. It probably is necessary to go through the classification to get this result. The number $12$ is not random: It arises because of the bi-elliptic surfaces, which are quotients of a product of two elliptic curves by a finite group. These finite groups all have elements whose order divides $12$, which is why the number $12$ appears.

The key step is that $P_{12}=0$ implies $S$ is ruled. This follows from two cases:

(1) Suppose $q=0$. Because $P_{12}=0$ we also have $P_2=0$. Castelnuovo's rationality criterion then applies to show $S$ is rational.

(2) Suppose $q\geq 1$. Because $P_{12}=0$ we also have $p_g=0$. One can show relatively easily using the Albanese fibration that the only possibility for $S$ not to be ruled is when $q=1$ and $b_2=2$. Then lots of work shows that if $S$ were not ruled we would have $S=(B\times F)/G$ for curves $B$ and $F$ and a finite group $G$ (and a number of other technical restrictions). By analyzing the canonical bundle on the resulting surface, one can show that $P_{12}$ is never zero. Thus we have a contradiction, so $S$ is ruled. Furthermore, $P_{12}=1$ if and only if both $B$ and $F$ are elliptic curves.

Once this difficult step is out of the way, we know that $\kappa=0$ implies $P_{12}=1$. Conversely, if $P_{12}=1$ then $S$ is non-ruled, and one can show $\chi(O_S)\geq 0$. Since $p_g=0$ or $1$, the list of possibilities for $q$ is finite. After analyzing the five cases (one turns out to be impossible), dealing with the most difficult ones by invoking part (2) from above, one can show that $P_{12}=1$ does in fact imply $\kappa=0$.

Finally, the remaining case is $P_{12}\geq 2$. In this case, it is simple to show that $\kappa=2$ if and only if $K^2>0$. The backward direction is an application of Riemann-Roch, and the forward direction is proven by contrapositive. If $K^2=0$, then the mobile part $M$ of $|nK|$ satisfies $M^2=0$ and hence $\kappa=1$.

Hope this helps. - Phil

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  • $\begingroup$ Dear Philip, thanks for the answer. Maybe my question is not clear. The classification method is exactly what I want to avoid. I am looking for a direct way of proof. Thanks again. $\endgroup$
    – Fei YE
    Commented May 1, 2013 at 5:05
  • $\begingroup$ One direct way to show $\kappa(S)=-\infty$ implies $S$ is ruled is to use Iitaka's Conjecture: If $S$ is a surface and $S\rightarrow B$ is a fibration with generic fiber $F$ then $\kappa(S)\geq\kappa(B)+\kappa(F)$. Applying this to the Albanese fibration solves case (2) above instantaneously, because it implies that $F$ must be rational. This applies the stronger assumption $P_n=0$ for all $n$ rather than $P_{12}=0$ though. I think there really will be no way to get the specific number $12$ without some classification. $\endgroup$
    – Philip Engel
    Commented May 1, 2013 at 7:58

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