Timeline for Deformation of $\mathbb{P}^1 \times \mathbb{P}^1$
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 8, 2016 at 4:27 | comment | added | Allen Knutson | It's late to mention this, but I'm pretty sure the adjective Francesco wanted was "separated" not "separate". | |
| May 21, 2016 at 5:26 | comment | added | Francesco Polizzi | You can find all of this and much more in Sernesi and Hartshorne books on deformation theory. | |
| May 21, 2016 at 5:25 | comment | added | Francesco Polizzi | @Li Yutong: No, $H^1(X, \Theta_X)$ is the dimension of the tangent space at the point $[X]$ of the base $\textrm{Def}(X)$ of the semiuniversal deformation (=Kuranishi family). So the Kuranishi family has at most dimension $h^1(X, \, \Theta_X)$. If the strict inequality holds, than $\textrm{Def}(X)$ is singulat at $[X]$ and $X$ is said to be obstructed. | |
| May 21, 2016 at 0:48 | vote | accept | Li Yutong | ||
| May 21, 2016 at 0:45 | comment | added | Li Yutong | @FrancescoPolizzi Thank you for you explanation! Do we always have the dimension of Kuranishi space $= H^1(X, \Theta_X)$? I feel the dimension could greater than that (otherwise, it is too easy to see a variety is rigid). | |
| May 20, 2016 at 17:51 | answer | added | Allen Knutson | timeline score: 6 | |
| May 20, 2016 at 15:32 | comment | added | Sasha | Such deformation family is easy to construct, by considering a pencil of quadrics in $P^3$. An open subset in the base of this family corresponds to smooth quadrics (each isomorphic to $P^1\times P^1$), while finitely many points correspond to quadratic cones. If you consider a small resolution of singularities of the total family, the special fibers will turn into the Hirzebruch surface $F_2$. | |
| May 20, 2016 at 15:29 | comment | added | Francesco Polizzi | Note that local rigidity is actually sufficient to imply that the Kuranishi family of $\mathbb{P}^1 \times \mathbb{P}^1$ has $0$-dimensional basis. This gives in turn $$H^1(\Theta_{\mathbb{P}^1 \times \mathbb{P}^1})=0$$ (this last condition is called infinitesimal rigidity). | |
| May 20, 2016 at 15:22 | comment | added | Francesco Polizzi | The variety $\mathbb{P}^1 \times \mathbb{P}^1$ does not have a separate (=Hausdorff) moduli space. This allows the existence of a global deformation $\mathcal{X} \to \Delta$ such that $X_t \cong \mathbb{P}^1 \times \mathbb{P}^1$ for $t \neq 0$ and $X_0 \cong \mathbb{F}_2$. In other words, the rigidity in this example is only local, not in the large. | |
| May 20, 2016 at 14:54 | comment | added | Li Yutong | Which moduli space are you talking about? | |
| May 20, 2016 at 14:43 | comment | added | Francesco Polizzi | The point is that in this case you do not have a separate moduli space, so it can happen (and, in fact, it happens) that the variety is locally rigid but not globally rigid. The phenomenon you are talking about and the explicit description for the Kuranishi family of any Hirzebruch surface can be found in Catanese's paper moduli of algebraic surfaces, Theory of Moduli, Montecatini Terme, 1985, Springer LNM 1337 (1988). | |
| May 20, 2016 at 14:32 | history | asked | Li Yutong | CC BY-SA 3.0 |