Let $f:X\to B$ a flat morphism of varieties over an algebraically closed field $k$. If $f$ is flat and with connected fibres we say that $f:X\to B$ is a family over $B$.

In literature you can find two definitions of "isotriviality" that seems to be not equivalent:

  1. (Mainly used when $X$ is a surface and $B$ is a curve). $f:X\to B$ is called isotrivial if the smooth fibres of $f$ are isomorphic.
  2. $f:X\to B$ is called isotrivial if there exists a dense open set $U\subseteq B$ such that $f^{-1}(x)\cong f^{-1}(y)$ for every $x,y\in U$.

Clearly $1)\Rightarrow 2)$ but the converse seems to be false. Am I right?

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    $\begingroup$ I would say that the right definition of isotrivial would be 2 in this case. It might actually even be better to take $U$ to be the complement of the union of a countable number of closed subsets in $B$. $\endgroup$ Commented Mar 19, 2016 at 11:20

1 Answer 1


Yes, 2 is a priori weaker than 1. And it is false in general.

There are examples of flag varieties specializing to smooth projective horospherical varieties; see Pasquier-Perrin

In particular, there exist a smooth projective morphism $X\to B$ with $B$ a smooth affine connected curve, and a closed point $b$ in $B$ such that $X_s \cong X_t$ for all $s,t\in B\setminus \{b\}$. However, $X_s \cong X_b$ if and only if $b=s$.

There are easier examples of this phenomenon, but this is the first one that came to mind.

  • $\begingroup$ Ok thank you. But, when we have a fibered surface over a smooth curve it seems that the authors tend to identify the two concepts... So, under which hypothesis $1)$ is equivalent to $2)$. $\endgroup$ Commented Mar 19, 2016 at 12:43
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    $\begingroup$ If the fibers of the curve are of genus at least two, then the two conditions are equivalent. It has to do with Isom-schemes of such curves being finite. To see that 1 and 2 are not equivalent in the case of a genus zero curve, try looking at some projective bundles. $\endgroup$ Commented Mar 19, 2016 at 18:25
  • $\begingroup$ Could you explain why the Isom-scheme $\operatorname{Isom}(C,D)$ for $C, D$ with genus $ \ge 2$ is finite and (assuming we know it) how does it imply the equivalence of 1 and 2 in case of fibered surfaces? $\endgroup$
    – user267839
    Commented Jun 2, 2021 at 22:48
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    $\begingroup$ @katalaveino The finiteness of the Isom scheme is proven in Deligne and Mumford's paper on the moduli of curves. First, you show that it is quasi-finite. This is because of the finiteness of Aut(C) for C of genus at least two. Then, the properness has to do with uniqueness and existence of minimal regular models. To show that 2 implies 1: Let F be a fibre of your fibred surface and consider $Isom_B(X, B\times F)$ over $B$. This is finite over $B$. It contains $U$ in its image, so it is dominant. Thus surjective (because finite morphisms are closed). So all fibres are isomorphic. $\endgroup$ Commented Jun 6, 2021 at 16:30
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    $\begingroup$ $X\to B$ in my comment above is a smooth proper curve of genus at least two. If $X\to B$ has some singular fibres, replace $B$ by $B^0$, where $B^0$ is the dense open over which $X\to B$ is smooth. $\endgroup$ Commented Jun 6, 2021 at 16:31

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