Consider a Kodaira fibration. i.e. a smooth nonisotrivial 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 $\mathbb{C}^{2}$.

1$\begingroup$ 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

1$\begingroup$ 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
Here are the arguments to exclude polydisk and the ball (there are no other complex 2dimensional 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).
Kefeng Liu ("Geometric height inequalities", Math. Research Letters, 3 (1996), 693–702) proved that a compact complexhyperbolic surface cannot admit a holomorphic submersion to a Riemann surface. This excludes the complex ball.
Consider the 2dimensional 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 nonisotriviality 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, Complexanalytic properties of certain Zariski open sets on algebraic varieties. Ann. of Math. (2) 94 (1971), 21–51.
who proves it using Bers' simultaneous uniformization.

1$\begingroup$ @DariusMath: $IV(2)$ is the same as the bidisk. I also explained in my answer (1st sentence) how you can avoid the classification. $\endgroup$ – Misha Aug 29 '13 at 9:04

1$\begingroup$ 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

1$\begingroup$ @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

1$\begingroup$ @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

2$\begingroup$ @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
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.

$\begingroup$ 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

2$\begingroup$ 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