Timeline for When a proper smooth fibration is isotrivial?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| May 11, 2016 at 2:25 | comment | added | Jorge Vitório Pereira | I think it works. After base change we get a map to an appropriate Hilbert scheme of a certain projective space. On the Hilbert scheme we consider the foliation induced by automorphism of the ambient projective space. We pullback this foliation to the relative Picard. It is a foliation by algebraic leaves. Since its leaves dominates the leaves of the original foliation on $Y$ we have that its leaves cannot be Zariski dense unless the fibration is isotrivial. | |
| May 11, 2016 at 0:09 | comment | added | Jason Starr | I am still checking the details . . . | |
| May 10, 2016 at 20:26 | comment | added | Jason Starr | This should be true. Here is a sketch. First, let $\widetilde{X}\to \text{Pic}^0_{X/Y}$ be the pullback of $X$ to the relative Picard scheme. Now pullback $\mathcal{O}_X(1)$ to $\widetilde{X}$, then twist that by the Poincare bundle. Now, instead of considering isomorphisms of "bare" varieties, consider isomorphisms of the polarized varieties coming from this invertible sheaf. For a fixed pair $(X_0,\mathcal{L}_0)$, the corresponding subset $\widetilde{Z}(X_0,\mathcal{L}_0) \subset \text{Pic}^0_{X/Y}$ is "constructible". Using connectedness, $\widetilde{Z}\to Z$ should be surjective. | |
| May 10, 2016 at 13:50 | comment | added | Jorge Vitório Pereira | Thanks Jason. You are of course right. The key hypothesis is connectedness. The real question is about the nature of the leaves of the foliation defined by the saturation inside $TY$ of the image of the natural map $\pi_* TX \to TY$. | |
| May 10, 2016 at 13:45 | history | edited | Jorge Vitório Pereira | CC BY-SA 3.0 |
Corrected grammar.
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| May 10, 2016 at 13:44 | comment | added | Jason Starr | Silly observation: I think I can choose a countable subset of the Zariski dense subset above so that the only accumulation point is $(0,0)$, and of course we removed this from $X_0$ to form $Y$. So that countable subset of $Y$ would be an analytic subvariety. So probably the important hypothesis is connectedness. | |
| May 10, 2016 at 13:34 | comment | added | Jason Starr | If you drop the connectedness and analyticity hypotheses, it is false (as you probably already know). You can begin with $X_0=E\times E$, for $E$ an elliptic curve, you let $Y=X_0\setminus\{(0,0)\}$, and you let $X$ be the blowing up of $X_0\times Y$ along the union of $\{(0,0)\}\times Y$ and the diagonal. Fixing a general point $(y,z)\in Y$, the Zariski dense subset $\{(ay+bz,cy+dz)\in Y: (a,b,c,d)\in \mathbb{Z}^4, ad-bc=1\}$ has your property, but it is neither connected nor analytic (since it accumulates). | |
| May 10, 2016 at 13:17 | history | asked | Jorge Vitório Pereira | CC BY-SA 3.0 |