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In this paper, Kodaira constructed a fiber bundle $\Phi:M_{m,n}\to S$ from a compact complex surface $M_{m,n}$ to a compact Rieman surface $S$ of genus $>0$. In particular, (on p.212) for any point $u\in S$, the fibre $C_u =\Phi^{-1}(u)$ is a compact Riemann surface which is an $m$-sheeted cyclic (branched) covering surface of $R$ with two branch points, where $R$ is a compact Riemann surface of genus $\not=0$. I want to know why this fiber bundle is not isotrivial.

Many thanks.

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Kodaira's examples have index $\tau>0$. If $M\to S$ were isotrivial, then it is not hard to see that after pulling back to a finite unramified cover of $S$, the surface becomes a product. But this would force $\tau(M)=0$ [See added note below]. You can look at the book Compact Complex Surfaces by Barth, (Hulek), Peters, and Van de Venn for further explanation.

There are examples of what are sometimes called Kodaira surfaces, where nonisotriviallity is essentially immediate. Namely, find a compact curve $S$ in $M_g$ (which exists once $g>2$), and pull back the "universal" curve.

Added Explanation The index is the signature of the intersection form. By a theorem of Hirzebruch, it can also be computed as $$\tau(M)= \frac{1}{3}(c_1^2(M)-2c_2(M))$$ It follows that if $M'\to M$ is a finite unramified cover, then $\tau(M')=0$ if and only if $\tau(M)=0$. In particular, if $M'$ can be chosen as a product of curves, then it can be checked that $\tau(M')=0$, so $\tau(M)=0$.

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  • $\begingroup$ Thanks, but how to see that Kodaira's examples have index $\tau>0$? $\endgroup$
    – 6666
    Jun 17, 2020 at 20:05
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    $\begingroup$ Kodaira's paper that you linked says "The purpose of this note is to exhibit a series of compact connected complex 4-manfolds ... with positive indices." You should read it. $\endgroup$ Jun 17, 2020 at 20:13
  • $\begingroup$ Sorry, I checked the book you mentioned but still one some questions, I can see the pullback as the product of curves has index $0$, but why would that imply $\tau(M)=0$? Is index invariant under base change? $\endgroup$
    – 6666
    Jun 18, 2020 at 19:14
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    $\begingroup$ OK, I'll expand my answer. $\endgroup$ Jun 18, 2020 at 20:05

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