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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.

Additional question. Are there such surfaces at all?

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?

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  • $\begingroup$ A trivial observation: singularities of such a surface cannot be rational. Do you have an example of rational normal Gorenstein surface with $H^0(\omega)\ne0$? $\endgroup$
    – Serge Lvovski
    Feb 22, 2013 at 20:33
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    $\begingroup$ Serge, I don't know if there is such an example (I have to add that when I was asking the question I was thinking that I have such examples. But analyzing them now more carefully I realised that they are birational to unirulled surfaces...) Still I don't see immediately why such examples can not exist. $\endgroup$
    – aglearner
    Feb 22, 2013 at 21:24
  • $\begingroup$ Since I am not really used to bad singularities, for me a rational surface cannot have an effective canonical divisor. In your case, maybe it is good to write down a resolution of singularities and compute the canonical divisor of the resolution (which is of course not effective) and compare with the canonical divisor of the singular surface, to see which kind of singularities you need, and in particular which kind of curve you want to contract in a smooth projective rational surface. When this is done, it should be easy to see if such curve exists or not. $\endgroup$
    – Jérémy Blanc
    Feb 24, 2013 at 13:29

2 Answers 2

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Here is an example (I hope!).

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.)

More generally, one can take a curve $B$ of degree $2r$ ($r\ge 3$) with a point $P$ of multiplicity $2r-2$.

I've no idea whether a classification exists, I had never considered this question before.

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  • $\begingroup$ Dear Rita, are you sure that your surface is rational and not ruled over an elliptic curve? In fact, could you please say what is "ordinary quadruple point"? (is this something like $x^4=y^4$?) Basically I am not sure that you get a pencil of lines on $Y$ -- you get one on $X$, but will they all pass through one point on $Y$? $\endgroup$
    – aglearner
    Feb 24, 2013 at 15:07
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    $\begingroup$ "ordinary quadruple point" means a point of multiplicity 4 whose tangent cone is made of 4 distinct lines, so $x^4+y^4=0$ is an example. The lines through $P$ give a base point free pencil $|f|$ of $\hat{\mathbb P}^2$. The preimage $F$ in $Y$ of a general $f$ is a double cover of a line branched over 2 points, so it is again a $\mathbb P^1$. This shows that $|F|$ is a base point free linear pencil of rational curves. Moreover, the computations for double covers give $\chi(\OO_Y)=1$, while a surface ruled over an elliptic curve has $\chi=0$. $\endgroup$
    – rita
    Feb 24, 2013 at 15:35
  • $\begingroup$ Thank you for these details. I still miss something. Consider the double cover of $\hat{\mathbb P^2}$ in $B'$. Then the preimage of the exceptional divisor in $\hat{\mathbb P^2}$ is an ellipitic curve. I think, that there is a projection from $Y$ to this curve collapsing all the rational curves from the "free rational pencil". Am I wrong? (if not, clearly $Y$ is not rational) $\endgroup$
    – aglearner
    Feb 24, 2013 at 15:46
  • $\begingroup$ I don't see that there is such a projection. The preimage of the exceptional curve is a BISECTION of the pencil (1 point of intersection downstairs gives 2 upstairs, since the exceptional curve is not in the branch locus). $\endgroup$
    – rita
    Feb 24, 2013 at 15:54
  • $\begingroup$ Thank you Rita and sorry for doubting the answer I finally got it:) I will accept it. $\endgroup$
    – aglearner
    Feb 24, 2013 at 15:58
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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)

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).

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 $H$.

Hence $\mu^*(K_X)=K_{X'}+E\leq E-H$ is not pseudoeffective, so that it cannot be effective, and the same holds for $K_X$.

Does it make sense?

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  • $\begingroup$ Sorry Gianny, this does not quite make sense for me. You seem to claim that on any rational surface anticanonial divisor is big. This is unfortunately wrong, it need not be effective even. Just take a blow up of $\mathbb P^2$ in $10$ generic points. $\endgroup$
    – aglearner
    Feb 24, 2013 at 14:06
  • $\begingroup$ Sorry, you are absolutely right about the first mistake, my proof is wrong. Anyway, just for the sake of truth, I didn't write that $-K_{X'}$ is ample; but it's not very important $\endgroup$
    – Gianni Bello
    Feb 24, 2013 at 14:43
  • $\begingroup$ Gianni, sorry my second comment is wrong indeed, I will delete it $\endgroup$
    – aglearner
    Feb 24, 2013 at 15:12

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