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This question is about properties of Isom-schemes that are well-known over algebraically closed fields.

Let $K$ be a field of characteristic zero, let $C$ be a smooth projective geometrically connected curve over $K$ and let $P$ a $K$-rational point of $C$. Let $X$ be a smooth projective geometrically connected surface over $K$, and $X\to C$ a curve of genus at least two over $K$. Define $F:= X_P$. Then $F$ is a curve of genus at least two over $K$.

Consider the Isom-scheme:

$\underline{\mathrm{Isom}}_{C}(X, C\times_K F) \to C $.

This map is finite unramified. It is etale if $X\to C$ is "isotrivial", i.e., all geometric fibres of $X\to C$ are isomorphic. This is proven in Deligne-Mumford. (Are there any other references?)

Assume $X\to C$ is isotrivial, and non-trivial.

Is the Isom-scheme $\underline{\mathrm{Isom}}_{C}(X, C\times_K F) $ geometrically connected over the ground field $K$? What if $K$ is a number field?

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If $X\to C$ is trivial, i.e. $X=C\times_K F$, then $\underline{\mathrm{Isom}}_C(X,C\times_K F)$ is just $C\times_K \underline{\mathrm{Aut}}_K(F)$ which is geometrically connected if and only if $\underline{\mathrm{Aut}}_K(F)$ is trivial.

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  • $\begingroup$ Indeed. This has been confusing me for a while now. If $X\to C$ is isotrivial, then $X_D \to D$ is trivial, where $D$ is the Isom-scheme. People use this to show that $X\to C$ is trivial up to some finite etale cover $D\to C$ (by taking $D$ to be the Isom-scheme). Do I understand correctly that, if $X\to C$ is non-trivial and isotrivial, then the Isom-scheme is actually geometrically connected? $\endgroup$
    – Theaux G.
    Sep 13, 2013 at 12:57
  • $\begingroup$ No, because the Isom-scheme can be any $\underline{\mathrm{Aut}}_K (F)$-torsor over $C$. $\endgroup$
    – Laurent Moret-Bailly
    Sep 13, 2013 at 15:10
  • $\begingroup$ For an explicit example, take a curve with automorphism group $\mathbb Z/2 \times \mathbb Z/2$. $y^2= (x^2-1)(x^2-2)(x^2-3)$ should do. Then over an elliptic curve, the isotrivial families correspond to $Hom(\mathbb Z/2\times \mathbb Z/2, \mathbb Z/2 \times \mathbb Z/2)$. Only the surjective homomorphisms correspond to geometrically connected Isom-schemes. But there are non-trivial non-surjective homomorphisms. $\endgroup$
    – Will Sawin
    Sep 13, 2013 at 16:35

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