Timeline for Is there an Oka-Grauert principle for homogeneous spaces?
Current License: CC BY-SA 3.0
8 events
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| Nov 23, 2014 at 9:55 | comment | added | Matthias Wendt | Concerning the family of quadrics in your edit: it's not really a counterexample to Oka-Grauert, because it is topologically nontrivial. Interpreting the question - as the title suggests - as a question about Oka-Grauert, the topological triviality should be an assumption (which is missing in the actual question). | |
| Nov 23, 2014 at 5:41 | history | edited | abx | CC BY-SA 3.0 |
added 297 characters in body
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| Nov 22, 2014 at 21:41 | comment | added | Jason Starr | Oops, my second comment above was written before I saw abx's answer to the earlier question. I am just repeating what he already wrote. | |
| Nov 22, 2014 at 21:30 | comment | added | Jason Starr | Maybe to make more clear my concern, consider a deformation of the Hirzebruch surface $\Sigma_2$ to the Hirzebruch surface $\Sigma_0 = \mathbb{C}P^1\times \mathbb{C}P^1$ over the disk. This has the same basic flavor of the Pasquier-Perrin example, but the issue I raise is more evident in that example. Really we are just deforming $\mathcal{O}(-1)_{\mathbb{C}P^1}\oplus \mathcal{O}(1)_{\mathbb{C}P^1}$ over $\mathbb{C}P^1\times \Delta$. But the restriction to $\mathbb{C}P^1\times \Delta^*$ is just $\mathcal{O}_{\mathbb{C}P^1\times \Delta^*}\oplus \mathcal{O}_{\mathbb{C}P^1\times \Delta^*}$. | |
| Nov 22, 2014 at 21:21 | comment | added | Jason Starr | @abx: I am not sure that I agree with your comment. A single family over the punctured disk can, sometimes, have two different smooth, projective limits, e.g., flops. If the fibers are all orthogonal Grassmannians, then the family is a "torsor for the automorphism group scheme", which is itself some complex Lie group. So, I believe, Grauert-Oka implies that this torsor has a holomorphic section (possibly with an essential singularity at $t=0$), and thus the family over the punctured disk is a product family (as the OP hopes). | |
| Nov 22, 2014 at 21:14 | comment | added | abx | It is not, otherwise you could complete it by the trivial fibration on the full disk, so the fiber above 0 would not change. | |
| Nov 22, 2014 at 20:54 | comment | added | user42804 | I see. What if I am asking if it's trival over punctured disc? | |
| Nov 22, 2014 at 17:42 | history | answered | abx | CC BY-SA 3.0 |