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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?

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    $\begingroup$ What is a Kodaira fibration? $\endgroup$
    – abx
    Feb 9, 2020 at 7:01
  • $\begingroup$ @abx I added a definition $\endgroup$
    – 6666
    Feb 9, 2020 at 17:42
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    $\begingroup$ It depends on what you call easy... A standard result in surface theory is that a surface with a line bundle $L$ such that $c_1(L)^2>0$ is projective. Consider a fiber $F$ of your fibration; its genus $g$ is $\geq 2$, hence by the genus formula $(K_X\cdot F)=2g-2>0$. Then $(K_X+nF)^2>0$ for $n\gg 0$, hence the result. $\endgroup$
    – abx
    Feb 9, 2020 at 18:01
  • $\begingroup$ @abx Sorry why is $g>1$? $\endgroup$
    – 6666
    Feb 9, 2020 at 18:41
  • $\begingroup$ If $g=1$ the $j$-invariant of the fibers must be constant, hence $X$ is an elliptic fiber bundle. I assumed that you exclude this case — otherwise, as observed by Nick L, you can get non Kähler surfaces. $\endgroup$
    – abx
    Feb 9, 2020 at 19:00

1 Answer 1

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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, American J. Math. 90, 789-804 (1968). ZBL0202.51702.

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  • $\begingroup$ Can you explain the theorem 1.1 of Kas' paper, why the map $U\to T$ is holomorphic? $\endgroup$
    – 6666
    Feb 16, 2020 at 0:57

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