Timeline for Really rigid varieties
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 3, 2011 at 21:34 | answer | added | Allen Knutson | timeline score: 13 | |
| Aug 3, 2011 at 21:14 | answer | added | Jason Starr | timeline score: 7 | |
| Aug 3, 2011 at 19:32 | history | edited | Mariano Suárez-Álvarez | CC BY-SA 3.0 |
added 117 characters in body
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| Aug 3, 2011 at 18:49 | comment | added | Anton Geraschenko | @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. | |
| Aug 3, 2011 at 17:28 | comment | added | Qfwfq | 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 $X_0$. | |
| Aug 3, 2011 at 17:11 | answer | added | Francesco Polizzi | timeline score: 7 | |
| Aug 3, 2011 at 16:43 | history | asked | Mariano Suárez-Álvarez | CC BY-SA 3.0 |