Suppose we have a fibration over the punctured disc (i.e., a deformation of complex manifolds) such that each fiber is a homogeneous space. Is the total space a product of a fiber with the punctured disc? What about the case where the fiber is Fano?
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$\begingroup$ Ngaiming Mok has done some relevant work. There are nontrivial examples where all fibers away from the central fiber are isomorphic and the central fiber is something topologically different, even for homogeneous Fano manifolds. $\endgroup$– Ben McKayCommented Nov 22, 2014 at 17:01
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2$\begingroup$ The question does not seem to have anything to do with the Oka-Grauert principle mentioned in the title... If you are trying to get a holomorphic section by applying the Oka-Grauert principle to a topological section, then you probably want "fibration" to mean subelliptic submersion and consult Forstneric's book "Stein manifolds and holomorphic mappings". $\endgroup$– Matthias WendtCommented Nov 22, 2014 at 18:28
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$\begingroup$ I believe Grauert-Oka does apply to a holomorphic fibration by projective homogeneous spaces over the punctured disk. The family is equivalent to a holomorphic principal bundle for complex Lie group of biholomorphisms of the projective homogeneous space, and you can apply Grauert-Oka to this principal bundle. For a Fano fibration, the same holds if you know that every fiber is biholomorphic (again reduce to a principal bundle for the biholomorphism group of the Fano). Some of this is discussed in my paper, "Discriminant Avoidance..." with de Jong. $\endgroup$– Jason StarrCommented Nov 22, 2014 at 21:24
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2 Answers
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To elaborate on my comment: yes, there is an Oka-Grauert principle for homogeneous spaces. The definite reference (besides the earlier papers of Grauert and Gromov) is the book "Stein manifolds and holomorphic mappings" by Forstneric. In there, you find Corollary 5.4.8 telling you that for a holomorphic fiber bundle $\pi:Z\to X$ over a reduced Stein space $X$ with Oka manifold fibers, the inclusion of holomorphic into continuous sections is a weak equivalence. Proposition 5.5.1 tells you that every complex homogeneous manifold is an Oka manifold (via the exponential spray from the corresponding complex Lie group). Combining these two, any continuous section of a holomorphic fiber bundle with homogeneous space fibers can be deformed to a holomorphic section. Therefore, a holomorphic fiber bundle over the punctured disc with homogeneous space fibers is holomorphically trivial if it is topologically trivial.
Now that we have translated the problem into algebraic topology, we can get rather complete information. The fiber bundle with fiber $G/H$ over the punctured disc $\Delta^\ast$ is completely classified by the homotopy class of the monodromy map $\pi_1(\Delta^\ast)\to\pi_0(\operatorname{Aut}(G/H))$, i.e., by a connected component of the holomorphic automorphisms. Therefore, any homogeneous space with non-connected automorphism group can be used to manufacture non-trivial families. In abx's answer, the quadric $Q\cong\mathbb{P}^1\times\mathbb{P}^1$ has an automorphism - the map which switches the two factors - which is not homotopic to the identity, giving rise to a topologically non-trivial family. On the other hand, there are two cases where the bundle is topologically (and by the above also holomorphically) trivial: 1) $\operatorname{Aut}(G/H)$ is connected, or 2) the fiber bundle with fiber $G/H$ is the associated bundle for a principal $G$-bundle, and $G$ is connected.
answered Nov 23, 2014 at 10:32
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The relevant paper is "Local rigidity of quasi-regular varieties" by Pasquier and Perrin, Math. Z. 265 (2010), no. 3, 589–600. They construct a smooth fibration over $\mathbb{C}$ such that the fiber over $t\neq 0$ is a orthogonal grassmannian $\mathbb{G}_q(2,7)$, but the fiber over 0 is not homogeneous.
Edit: Actually I overestimated the question. As it stands, just take the family of quadrics given by $X^2+Y^2+Z^2+tT^2=0$ in $\mathbb{P}^3\times \mathbb{C}$. The family cannot be trivial over the punctured disk because the monodromy exchanges the two generators of $H^2$ of a smooth fiber
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$\begingroup$ I see. What if I am asking if it's trival over punctured disc? $\endgroup$ Commented Nov 22, 2014 at 20:54
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$\begingroup$ It is not, otherwise you could complete it by the trivial fibration on the full disk, so the fiber above 0 would not change. $\endgroup$– abxCommented Nov 22, 2014 at 21:14
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$\begingroup$ @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). $\endgroup$ Commented Nov 22, 2014 at 21:21
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$\begingroup$ 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 $\mathcal{O}_{\mathbb{C}P^1\times \Delta^*}\oplus \mathcal{O}_{\mathbb{C}P^1\times \Delta^*}$. $\endgroup$ Commented Nov 22, 2014 at 21:30
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1$\begingroup$ 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). $\endgroup$ Commented Nov 23, 2014 at 9:55