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Required fields*

2
  • $\begingroup$ Maybe the requirement $\mathfrak{X}\cong T\times X_0$ is too strong? I think when $X_0$ has a nontrivial automorphism one can often construct a nontrivial family (i.e. not $\cong$ to the product family) in which every fiber (over closed points) is $\cong$ to $X_0$. $\endgroup$ Commented Aug 3, 2011 at 17:28
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    $\begingroup$ @unknowngoogle: not any nontrivial automorphism will do. You need one which is "not homotopically trivial" in some sense. Your observation shows that $Aut(X_0)$ must be rationally connected. Given any automorphism $\phi$, you can form a family over the nodal cubic obtained by taking the trivial family over $\mathbb P^1$ and gluing two fibers together using $\phi$. A trivialization of this family produces a rational curve in $Aut(X)$ connecting $\phi$ to the identity. $\endgroup$ Commented Aug 3, 2011 at 18:49