Question. Are there smooth complex surfaces of general type with an irregularity $q = 1$ and Euler characteristic $3$.
If the answer is yes, what is known about the geometry of such surfaces? Are there some explicit constructions?
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5$\begingroup$ Topological Euler characteristic or holomorphic Euler characteristic? $\endgroup$– Jason StarrJul 19, 2012 at 14:28
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$\begingroup$ Thanks for the remarks. I meant topological Euler characteristic. $\endgroup$– user25225Jul 19, 2012 at 14:34
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If you mean holomorphic Euler characteristic equal to $3$, i.e. $p_g=3$, then the answer is yes.
For some explicit constructions, look at the paper by Takahashi
Certain algebraic surfaces of general type with irregularity one and their canonical mappings, Tohoku Math. J. (2) Volume 50, Number 2 (1998), 261-290.
For any value of $p_g \geq 2$, the author shows the existence of minimal surfaces of general type with $K^2=3p_g$ and $q=1$ and studies their canonical mappings.
These surfaces are the minimal resolution of a relative quartic hypersurface (having at most rational double points as singularities) in a $\mathbf{P}^2$-bundle over an elliptic curve.
If you mean instead topological Euler characteristic, Noether formula gives $$K^2=12 \chi(\mathcal{O}) -3.$$
On the other hand, by Bogomolov-Miyaoka-Yau inequality one has $K^2 \leq 9 \chi(\mathcal{O})$, so the only possibility is $\chi(\mathcal{O})=1$, i.e. $p_g=q=1$, $K^2=9$.
As far as I know, the existence of such a surface was announced by Cartwright and Steger. They used a ball-quotient construction (based also on computer calculations), similar to the one used by Prasad and Yeung in order to classify the fake projective planes. I do not think that a more explicit geometric construction is currently known.
answered Jul 19, 2012 at 14:30
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$\begingroup$ Thank you Francesco for your remark and the reference. I will definitely look up Takahashi paper, but I meant topological Euler characteristic in my question above. $\endgroup$ Jul 19, 2012 at 14:39
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1$\begingroup$ In this case it seems to me that the only possibility is $p_g=q=1$. I edited the answer with some information about this case $\endgroup$ Jul 19, 2012 at 14:54