In [this paper](https://www.maths.ed.ac.uk/%7Ev1ranick/papers/kodaira1.pdf), 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.



Non-isotrival fiber bundle over compact Riemann surface
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 $\tau(M)=0$.


