Let $S$ be a smooth complex surface, If $c_1(S) \in N_1(S)$ is nef and non-torsion, then we know that this would imply some restrictions on the cone of effective curves (and surface itself)--see the discussion at the end of Borcea's paper on Horrocks-Mumford quintics--

In fact we will find that the cone of effective curves is generated by rational curves and possibly $c_1(S)$ itself, accumulating at most to $\mathbb{R}_{+} c_1(S)$.

My question is:

How much do we know about smooth surfaces with effective non-torion first Chen class?

Just remember that any effective curve has a unique Zariski decomposition

$$ H+N $$

where $H$ is nef, $N=\sum a_i \Gamma_i$, $a_i >0$, with neg-definite intersection matrix $(\Gamma_i\cdot \Gamma_j)$ and $H$ is orthogobal to all $\Gamma_i$.

  • $\begingroup$ Dear Mohammad, can you say a little more about what kind of information you would like about this class of surfaces? There are very obvious things one can say, for example all such surfaces are birationally ruled, but I guess you want something deeper than that. $\endgroup$ – user5117 Sep 19 '11 at 15:40
  • $\begingroup$ primarily,I am looking for a result similar to what I mentioned for the nef case on the structure of cone of effective curves. but any other result on the structure of such surfaces is welcome. please give a reference for the result you mentioned. $\endgroup$ – Mohammad F. Tehrani Sep 19 '11 at 15:54
  • $\begingroup$ The assertion in my first comment follows immediately from the Enriques--Kodaira classification of surfaces, explained for example in the book by Barth--Hulek--Peters--van de Ven. In brief: your assumption is that some power of $-K_S$ has a nonzero section, which implies that the Kodaira dimension $\kappa(S) = -\infty$. Kodaira dimension is a birational invariant, so your surface must be birational to a surface in the E--K classification with $\kappa = -\infty$. Those are exactly the birationally ruled surfaces. $\endgroup$ – user5117 Sep 19 '11 at 18:35
  • $\begingroup$ Oh, you are right. I was dumb. $\endgroup$ – Mohammad F. Tehrani Sep 19 '11 at 18:42
  • $\begingroup$ Regarding your main question: I don't know of any results about the cone of curves under your assumption. As a simple example of what can happen, blowing up an arbitrary number of points on a cubic in P^2 gives a surface with −K (equivalently c1(S)) effective. In this case there is only a single K-positive curve class, i.e. the cubic itself, but the $K^\perp$ part of the cone of curves could be a mess. In particular it must contain the intersection of K⊥ with the positive cone in N^1, giving a "round" part of the cone of curves in the hyperplane $K\perp$. $\endgroup$ – user5117 Sep 19 '11 at 18:59

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