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I am reading the paper "Canonical models of surfaces of general type" by E. Bombieri. In the last section of this paper, there is a statement saying that surfaces with $K^2=1$ and $p_g=0$ do not have pencils of genus $2$, and there is no proof. Is there a proof of this statement?

edited Nov 11 2011 at 12:51

Francesco Polizzi
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asked Nov 11 2011 at 1:16

Michael Zhang
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Perhaps it is worth including Bombieri's footnote from the paper: "Our proof is too long to be inserted here; we hope to return to this argument in another paper." – Jack Huizenga Nov 11 2011 at 2:06

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You are right. But I really want to see this proof and I can not find the "another paper". – Michael Zhang Nov 11 2011 at 4:11

I think you should ask this question to Torsten Ekedahl (because of his paper "Canonical models of surfaces of general type in positive characteristic"). He is also active in MO. – Damian Rössler Nov 11 2011 at 7:12

rita · Accepted Answer · 2011-11-11 14:53:14Z UTC

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In fact it seems that the statement is not correct.

The paper [Calabri, Ciliberto, Mendes Lopes, Numerical Godeaux surfaces with an involution. Trans. Amer. Math. Soc. 359 (2007), no. 4] contains the classification of numerical Godeaux surfaces (i.e., minimal surfaces of general type with $K^2=1$ and $p_g=0$) that have an automorphism of order 2. The examples described in section 6 have a pencil of curves of genus 2 (cf. Remark 6.3).

answered Nov 11 2011 at 14:53

rita
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	Thanks, rita. I remember that minimal surfaces of general type with $p_g=0$ and $K^2 \ge 3$ do not have genus two pencil. I just want to know the cases that when $K^2$ is smaller, which correspond the Numerical Godeaux or Campedelli surfaces. I will check the paper you mentioned. – Michael Zhang Nov 11 2011 at 17:41
	For numerical Campedelli, you can look at [Calabri, Mendes Lopes, Pardini, Involutions on numerical Campedelli surfaces. Tohoku Math. J. (2) 60 (2008), no. 1, 1–22]. This is essentially a continuation of the paper I referred to in the answer and in 3.5 you can find examples of (numerical) Campedelli surfaces with a genus 2 pencil. – rita Nov 11 2011 at 18:07
	Thanks again, rita! – Michael Zhang Nov 12 2011 at 8:20
	I have another quick question, rita. It is about what I said before. I said that minimal surfaces of general type with $p_g=0$ and $K^2 \ge 3$ have no pencils of genus 2. Actually I do not know what the pencil means here. Does it mean a \emph{morphism} of fibration or just a \emph{rational map}? – Michael Zhang Feb 17 2012 at 19:17
	If the surfaces is minimal of general type and $\|C\|$ is a pencil with base points such the general $C$ is smooth of genus 2 then the index theorem gives $K^2= 1$. On the other hand, I do not know whether one could have a pencil with base points and such that the general $C$ is singular but has geometric genus 2. – rita Feb 17 2012 at 20:56

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Francesco Polizzi · Answer 2 · 2011-11-11 15:46:04Z UTC

I do not know whether this result is true.

In fact G. Xiao in his book Surfaces fibrees en courbes de genre deux, page 66 claims that a surface of general type with $p_g=q=0$, $K^2=1$ and a pencil of curves of genus $2$ has been constructed by Oort and Peters in their paper A Campedelli surface with torsiongroup $\mathbb{Z}/2$. You should check their construction, since unfortunately I have not time to do it now.

Notice that Oort and Peter call "Campedelli surface" what is nowadays called "Numerical Godeaux surface" (i.e., a surface of general type with $p_g=q=0$, $K^2=1$); in fact, the name "Campedelli surface" is currently used for surfaces with $p_g=q=0$, $K^2=2$.

At any rate, the following is surely true:

if $S$ is a minimal surface of general type with $p_g=q=0$ and $K^2=1$, then the bicanonical pencil $|2K_S|$ cannot be composed with a pencil of curves of genus $2$.

See the paper by Catanese and Pignatelli Fibrations of low genus I, Section 5. The proof given there is not based on Bombieri's paper, but it uses a structure theorem for genus $2$ fibrations which involves vector bundles techniques.

On a result about genus two pencils

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