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Consider a Kodaira fibration. i.e. a smooth non-isotrivial fibration $X\rightarrow C$ with $X$ a smooth complex surface and $C$ a smooth complex curve, such that both the genus of $C$ and genus of the fibers (which are complex curves) are at least $2$. By abuse of notation I call $X$ a Kodaira fibered surface. What is known about the structure of the universal cover of $X$? In particular, when is it a bounded symmetric domain? I know that $X$ is a minimal algebraic surface of general type and that the universal cover of $X$ can never be a ball in $\mathbb{C}^{2}$.

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    $\begingroup$ It is bounded domain (Bers) but is not symmetric: It is not a ball as you know and is not a polydisk. $\endgroup$ – Misha Aug 28 '13 at 8:10
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    $\begingroup$ Thank you very much for the answer! could you give a reference about your answer? also, do you mean that from the fact that it is not neither a ball nor a polydisk, follows that it is not a bounded symmetric domain? $\endgroup$ – Darius Math Aug 28 '13 at 8:18
  • $\begingroup$ Also, what is your argument that it is not a polydisk? $\endgroup$ – Darius Math Aug 28 '13 at 8:21
  • $\begingroup$ I think I got my first answer: since it is simply connected you are using the decomposition in to irreducible components: so it is either a ball or a polydisc. But still I don't know how you argue that it is not a polydisk? $\endgroup$ – Darius Math Aug 28 '13 at 9:15
  • $\begingroup$ OK, I think I got my second answer too! it is not a polydisc because otherwise $X$ will be a product of curves and this contradicts for example the positivity of $c_{2}$. Am I right? Is this also your argument? $\endgroup$ – Darius Math Aug 28 '13 at 9:39
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Here are the arguments to exclude polydisk and the ball (there are no other complex 2-dimensional bounded symmetric domains: In fact, one can do without this and argue that any domain other than the ball would have rank $\ge 2$ and, hence, Margulis superrigidity theorem would apply).

  1. Kefeng Liu ("Geometric height inequalities", Math. Research Letters, 3 (1996), 693–702) proved that a compact complex-hyperbolic surface cannot admit a holomorphic submersion to a Riemann surface. This excludes the complex ball.

  2. Consider the 2-dimensional polydisk $D^2$ and a group $\Gamma$ acting discretely, holomorphically and cocompactly on $D^2$. Then $\Gamma$ is a lattice in $Isom(D^2)$ and by Margulis' superrigidity theorem either $\Gamma$ is reducible or it is superrigid, and , ehnce, does not admit an epimorphism to a surface group. The second is impossible in the case of fibrations over Riemann surfaces, the first contradicts non-isotriviality assumption in the question.

On the other hand, the universal cover of your complex manifold $X$ is a certain bounded domain in ${\mathbb C}^2$: This result could be found in

P.A. Griffiths, Complex-analytic properties of certain Zariski open sets on algebraic varieties. Ann. of Math. (2) 94 (1971), 21–51.

who proves it using Bers' simultaneous uniformization.

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    $\begingroup$ @DariusMath: $IV(2)$ is the same as the bi-disk. I also explained in my answer (1st sentence) how you can avoid the classification. $\endgroup$ – Misha Aug 29 '13 at 9:04
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    $\begingroup$ Ah, I did not notice it! Thank you very much for your answers. It really helped me a lot. By the way, now that IV(2) is a polydisk, I think in addition to this argument, the argument that I gave in my comments above also work: It is not a ball by the Arakelov inequalities and not a polzdisk (hence a product of curves) since otherwise $c_{2}$ will vanish. $\endgroup$ – Darius Math Aug 29 '13 at 9:08
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    $\begingroup$ @DariusMath: Suppose you have an irreducible lattice $\Gamma$. Then Margulis' theorem implies that $\Gamma$ has finite abelianization. This implies that $\Gamma$ cannot have an epimorphism to a the fundamental group of a Riemann surface of genus $\ge 1$. But the long exact sequence of a fibration (applied to Kodaira's fibration) implies that $\pi_1(X)$ maps onto $\pi_1(C)$. $\endgroup$ – Misha Aug 29 '13 at 9:11
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    $\begingroup$ @DariusMath: Could you explain how to exclude the ball using Arakelov? The only proof that I know is by appealing to Liu's theorem. $\endgroup$ – Misha Aug 29 '13 at 9:12
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    $\begingroup$ @DariusMath: The proof of this (that I know) comes from algebra: If $F_1, F_2$ are two surface groups (of genus $\ge 2$) and $F_3< F_1\times F_2$ is a normal surface subgroup, then it has to be a (finite index) normal subgroup in one of the direct factors. Given this, in case when $F_3$ is $\pi_1$ of a fiber in a topological fibration, $F_3$ has to equal one of the factors. From there, you use the fact that two homotopic holomorphic maps from a compact Kahler manifold to the same hyperbolic Riemann surface have to be equal. You can probably fill in the details yourself now. $\endgroup$ – Misha Aug 30 '13 at 5:03
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If an algebraic surface has the bidisk as universal cover, then, by Hirzebruch's proportionality theorem, its topological index is 0 (the index is 1/3 (c_1^2 - 2 c_2)). But Kodaira proved that for Kodaira fibrations the index is strictly positive. To exclude the case that the universal covering is the ball, where c_1^2 = 3 c_2, again by Hirzebruch's theorem, is harder and was done by Kefeng Liu. You may look in my article with Rollenske for more results on Kodaira fibrations. Regards, Fabrizio Catanese.

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  • $\begingroup$ Very beautiful answer ! Thank you and I'm very glad to be the first one who receives an answer from you professor Catanese. $\endgroup$ – Darius Math Oct 22 '13 at 20:11
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    $\begingroup$ There is also another paper by A. Nadel which is linked with this topic (\emph{Semisimplicity of the group of biholomorphisms of the universal covering of a compact complex manifold with ample canonical bundle}). In this really beautiful paper Nadel shwos for instance that the automorphism groups of the universal cover a canonically polarized surface (it is the case for a Kodaira surface) are very constraints: if it's positive dimensional, it's the automorphism group of the ball or of the bidisk (else it's a discrete group and it reduces almost to the fundamental group of the surface). $\endgroup$ – Benoit Oct 25 '13 at 17:52

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