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A Log del Pezzo surface is a normal complex surface with ample anti-canonical bundle and at worst quotient singularities.

It is known that such surfaces are rational. This is proven, for example, here (Lemma 1.3):

http://www.ms.u-tokyo.ac.jp/journal/pdf/jms010104.pdf

I would like to know if there is some place with a short and clean exposition of this result, or maybe you can explain an idea of a proof?

Note, that it is not hard to see that a Log del Pezzo has Kodaira dimension $-\infty$, so to prove that it is rational it is enough to prove that it has a trivial fundamental group.

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  • $\begingroup$ Did you try to look into Dolgachev's Classical algebraic geometry? $\endgroup$
    – Sasha
    Commented Sep 26, 2016 at 16:08
  • $\begingroup$ What's the relation with the title? $\endgroup$
    – Alex Degtyarev
    Commented Sep 26, 2016 at 16:24
  • $\begingroup$ Sasha, I had a look in a version of this book (maybe not the latest one), there is no mentioning of log Delpezzos there... Alex, you are right, I added one line at the end of the question - rationality of such surfaces is equivalent to triviality of their fundamental group. $\endgroup$
    – aglearner
    Commented Sep 26, 2016 at 20:05

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I would proceed as follows. Replacing $S$ by its minimal resolution I have a weak log Fano pair $(S,B)$ which is klt and $-(K_S+B)$ is semiample and big. In particular $-K_S\sim _Q -(K_S+B)+B$ is big and so $K_S$ is not pseudo-effective. Running the MMP $S\to S'$ I end up with either 1) a smooth del Pezzo surface of Picard number 1 (hence $\mathbb P ^2$) or 2) a Mori fiber space say $f:S'\to C$. We must then argue that $C$ is rational. There is probably a trivial way, but from the higher dimensional point of view, I would pick $H\sim _Q -(K_S+B)$ general so that $(S,B+H)$ is klt and then apply Kawamata's canonical bundle formula to $K_{S'}+B'+H'$ over $C$. Via some technical arguments, one can show that $K_{S'}+B'+H'\sim _Q f^*(K_C+G)$ where $G$ is big (we use the fact that $H'$ is big). But then as $\kappa (K_{S'}+B'+H')=0$, it follows that $\kappa (K_C+G)=0$ and hence $g(C)=0$.

One can generalize much of this argument to arbitrary dimension. In fact we can show that if $(X,B)$ is a klt log Fano then it is rationally connected see http://arxiv.org/pdf/math/0408301.pdf and http://arxiv.org/pdf/math/0504330v2.pdf (these proofs use Kawamata's canonical bundle formula: Y. Kawamata, Subadjunction of log canonical divisors. II, Amer.J. Math. 120 (5) (1998), 893-899). Of course, if $X$ is rationally connected and $X\to C$ is surjective, then $C$ is rationally connected, so if $\dim C=1$, then $C$ is a rational curve.

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  • $\begingroup$ Thanks for this answer! If you could provide the trivial reasoning I would be grateful. I am interested in the less high tech proof possible, because I want to push it to (real) symplectic geometry which only makes sense in real dimension 4 $\endgroup$
    – aglearner
    Commented Sep 26, 2016 at 20:03
  • $\begingroup$ One argument you could use is that if $C$ is not rational, it admits etale covers of arbitrarily large degree and hence so does $S$. But if $f:S_1\to S$ is etale of degree $k$, then for $K_{S_1}+B_1=f^*(K_S+B_S)$, $(S_1,B_1)$ is log Fano with $(K_{S_1}+B_1)^2=k(K_S+B)^2$. Suppose that $r(K_S+B)$ is Cartier, then $r(K_{S_1}+B_1)$ is Cartier and $(K_{S_1}+B_1)^2\geq k/r$ which goes to infinity. It is not hard to show that this can't happen (an instance of boundedness of log Fanos of bounded index). $\endgroup$
    – Hacon
    Commented Sep 26, 2016 at 22:36
  • $\begingroup$ Thanks! Would you be able to make a proof without using boundedness of log Fanos of bounded index? $\endgroup$
    – aglearner
    Commented Sep 27, 2016 at 14:26
  • $\begingroup$ Sure. Bounding the top self intersection (or volume) should be easier than bounding the actual pairs. For example one could use the connectedness lemma. If $(K_{S_1}+B_1)^2\gg 0$, then one should be able to produce a divisor $D\sim _Q t K_{S_1}+B_1$ where $t<1$ and $(S_1, B_1+D)$ has non-connected non-klt locus (eg. high multiplicity at two general points). The details should be similar to (but easier than) the proof of Angehrn-Siu. $\endgroup$
    – Hacon
    Commented Sep 27, 2016 at 16:50

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