3
$\begingroup$

Let $X$ be a rational surface, say $X$ is del Pezzo surface. Let $D$ be $(-1)$-class divisor, i.e: $D^2=-1$ and $D^2+D.K_X=-2$. It is easy to show that on del Pezzo surface any $(-1)$-class divisor is irreducible $(-1)$-rational curve. We know the intersection graph of $(-1)$-curve on $X$, say if $K_X^2=6$, it is a $6$-gon, if $K_X^2=5$, it is Perterson graph and so on. It is easy to show that the graph for any rational surface, say degree $\geq 1$ only depends on $deg(X)$, the differences are the irreducibility of vertices.

Now, assume that $X$ is smooth surface of general type with $p_g=q=0$,(maybe minimal or not)with $K_X^2>0$, is there any interesting example of surface such that the intersection graph of $(-1)$-class can be drawn? Or a simpler question, how many $(-1)$-classes can the surface have? Are there finitely many?

$\endgroup$

1 Answer 1

9
$\begingroup$

On any surface $X$ with non-negative Kodaira dimension the $(-1)$-curves (i.e, the smooth rational curves $D$ with $D^2=-1$) are isolated, in other words any two of them do not intersect. From this one can deduce the uniqueness of the minimal model of $X$, see the proof of Proposition 4.6, p. 79 in the book

W. Barth, C. Peters, A. Van de Ven: Compact Complex Surfaces.

In particular, the numerical classes of the $(-1)$-curves are linearly independent, so there are only finitely many of them because $\mathrm{NS}(X)$ is finitely generated.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .