A Kodaira fibration is a compact complex surface X endowed with a holomorphic submersion onto a Riemann surface $\pi: X\to\Sigma$ which has connected fibers and is not isotrivial.


Is there an easy way to see why a compact complex surface that admits a Kodaira fibration is Kahler? I know for a complex compact surface is Kahler if and only if its first Betti number is even. I wonder it's possible to deduce that a compact complex surface that admits a Kodaira fibration has even Betti number?



Compact complex surface that admits a Kodaira fibration is Kahler
Let $f \colon S \longrightarrow B$ be a Kodaira fibration, and let $F$ be a general fibre. Then by [Kas68, Thm. 1.1] we have $g(B) \geq 2$ and $g(F) \geq 3$. 


In particular, $S$ contains no rational or elliptic curves: in fact, such curves cannot neither dominate the base (because $g(B) \geq 2$) nor be contained in fibres (because the fibration is by assumption smooth).


So every Kodaira fibred surface $S$ is minimal and, by the superadditivity of the Kodaira dimension, it is of general type. 


In particular, it is not only Kähler but actually *algebraic* (i.e., projective).


**References.**


[Kas68] A. Kas: [On deformations of a certain type of irregular algebraic surface](http://dx.doi.org/10.2307/2373484), *American J. Math.* **90**, 789-804 (1968). [ZBL0202.51702](https://zbmath.org/?q=an:0202.51702). 


