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Let $F$ and $G$ be coprime complex homogeneous polynomials in three variables of the same degree $d\geq 4$. Suppose that a general member of the pencil $\{F+tG=0\}\subset \mathbb{P}^2$ is smooth.

Which are the tools one can use to verify whether this pencil is isotrivial?

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    $\begingroup$ The Kodaira-Spencer map comes to mind. $\endgroup$
    – meh
    Commented May 31, 2018 at 19:04

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This is not an answer but rather a lengthy comment.

A necessary condition for the pencil to be isotrivial is that a smooth member in that pencil has a non-trivial automorphism: By blowing-up the base points, $\mathbf{P}^2$ is birational to the total space of an isotrivial family $f : X \to \mathbf{P}^1$ of curves of genus $\ge 2$. Let $C$ denote a smooth fiber of $f$. As $f$ is isotrivial, there exists a Galois cover $B \to \mathbf{P}^1$ (with the Galois group denoted by $G$) and a $G$-action on $C$ such that $f$ is birational to the projection $(C \times B)/G \to \mathbf{P}^1$. If $G$ acts trivially on $C$, then $\mathbf{P}^2$ would be birational to $C \times \mathbf{P}^1$, which is not possible.

This implies in particular that isotrivial pencils on $\mathbf{P}^2$ cannot be Lefschetz, because the automorphism group of a general member of a non-hyperelliptic Lefschetz pencil on a simply connected surface is trivial (see 10.6.18 in the book of Katz and Sarnak for a monodromy argument proving this fact).

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    $\begingroup$ Isotrivial pencils are everywhere non-transversal to the discriminant divisor, which is much stronger than being non-Lefschetz (i.e., that the pencil is somewhere non-transversal). $\endgroup$
    – Jason Starr
    Commented Jun 2, 2018 at 9:35
  • $\begingroup$ @JasonStarr, could you please elaborate this or point out a reference? $\endgroup$
    – Alan Muniz
    Commented Jun 2, 2018 at 12:57
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    $\begingroup$ For a transverse crossing at $t_0\in C$, the first-order deformation to the versal deformation space of the node $x_0\in f^{-1}(t_0)$ is a nonzero quotient of the Kodaira-Spencer map $T_B\to h^1(R\pi_*\textit{RHom}(L_f,\mathcal{O}_X))$. So the Kodaira-Spencer map is nonzero. For a family of hypersurfaces, that target is locally free (on a neighborhood of $t_0$), so the family is nonisotrivial. $\endgroup$
    – Jason Starr
    Commented Jun 3, 2018 at 15:11
  • $\begingroup$ @JasonStarr Then a pencil with a node not in the base locus will give a nonisotrivial fibration, right? Could you give any good reference to study the Kodaira-Spencer map with reasonable examples? $\endgroup$
    – Alan Muniz
    Commented Jun 25, 2018 at 12:11

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