A classification of rational surfaces with effective $K$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T13:36:20Z http://mathoverflow.net/feeds/question/122650 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/122650/a-classification-of-rational-surfaces-with-effective-k A classification of rational surfaces with effective $K$ aglearner 2013-02-22T18:30:41Z 2013-02-24T14:40:53Z <p>I would like to know if there can be some kind of classification of normal rational surfaces with Gorenstein singularities, such that their canonical divisor is effective. </p> <p><strong>Additional question.</strong> Are there such surfaces at all?</p> <p>I could imagine constructing such a surface by blowing up several points on an elliptic curve in $\mathbb CP^2$ and then contracting the proper transform of the curve, but will this give an example? </p> http://mathoverflow.net/questions/122650/a-classification-of-rational-surfaces-with-effective-k/122797#122797 Answer by Gianni Bello for A classification of rational surfaces with effective $K$ Gianni Bello 2013-02-24T13:55:52Z 2013-02-24T14:40:53Z <p>EDIT: The proof below is wrong, because it is false that $h^0(X',-mK_X')=h^0(\mathbb{P}^2,-mK_{\mathbb{P}^2})$ (see comments)</p> <p>Consider a normal rational surface $X$ with Gorenstein (or $\mathbb{Q}$-Gorenstein) singularities. Let $\mu:X'\to X$ be a resolution. Then $\mu^*(K_X)=K_{X'}+E$, where $E$ is a $\mu$-exceptioanl divisor. If $X$ is very singular it can happen that $E$ is effective, but it is an exceptional divisor, so that it is not big, or, in other words, it cannot be in the interior of the pseudoeffective cone, or, in other words, given any $A$ is an ample ($\mathbb{Q}$-)divisor, for sure $E-A$ is not effective (it is not pseudoeffective in fact).</p> <p>On the other hand, as $X'$ is a smooth rational surfaces it should be easy to see that, for all $m\in \mathbb{N}$, $h^0(X', -mK_X')=h^0(\mathbb{P}^2,-mK_{\mathbb{P}^2})$, so that $-K_{X'}$ is in fact big, that is $-K_{X'}\geq H$, for some ample $H$.</p> <p>Hence $\mu^*(K_X)=K_{X'}+E\leq E-H$ is not pseudoeffective, so that it cannot be effective, and the same holds for $K_X$.</p> <p>Does it make sense?</p> http://mathoverflow.net/questions/122650/a-classification-of-rational-surfaces-with-effective-k/122799#122799 Answer by rita for A classification of rational surfaces with effective $K$ rita 2013-02-24T14:37:45Z 2013-02-24T14:37:45Z <p>Here is an example (I hope!).</p> <p>Take $X$ a double cover of $\mathbb P^2$ branched over a normal sextic $B$. It is a normal Gorenstein surface and the standard formulae for double covers give $K_X=0$. Now assume that $B$ has an ordinary quadruple point $P$ and is smooth elsewhere, so that $X$ has an elliptic Gorenstein singularity at the point $Q$ lying over $P$. The minimal desingularization $Y$ of $X$ can be obtained by blowing up the plane at $P$ and taking base change + normalization of the cover. So $Y$ is a double cover of $\hat{\mathbb P}^2$ branched on the strict transform $B'$ of $B$. The pencil of lines through $P$ induces on $Y$ a linear pencil of rational curves, so $Y$ is rational. (Indeed, by the usual formulae for double covers it is easy to show that $\chi(Y)=1$ and that $K_Y$ is the pull back of $-E$, where $E$ is the exceptional curve of $\hat{\mathbb P}^2\to \mathbb P^2$, and this gives a different proof of the fact that $Y$ is ruled.)</p> <p>More generally, one can take a curve $B$ of degree $2r$ ($r\ge 3$) with a point $P$ of multiplicity $2r-2$. </p> <p>I've no idea whether a classification exists, I had never considered this question before. </p>