Timeline for Extending vector bundles from subvarieties
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Nov 16, 2017 at 13:18 | comment | added | user21574 | A theorem of Siu: Let $X$ is a Stein manifold of dimension at least $3$, $K$ is a holomorphically convex compact subset of $X$ with connected complement, and $E$ is a holomorphic vector bundle on $X\setminus K$, then there is a finite subset $P$ of $K$ such that $E$ extends to a holomorphic vector bundle on $X\setminus P$: See A Hartogs type extension theorem for coherent analytic sheaves. Ann. of Math. (2) 93 1971 166–188. 32.50 and also for details of proof see projecteuclid.org/download/pdf_1/euclid.afm/1485802742(Since Siu didn't prove it) | |
| Nov 15, 2017 at 14:50 | comment | added | user21574 | For topological condition to get extension theorems on bundles see Lemma 5 and Lemma 6 of the paper V. V. Shevchishin , The Ock–Grawert principle for the extension of holomorphic line bundles with integrable curvature V. V. Shevchishin , Mat. Zametki, 50:5 (1991), 109–119 link.springer.com/article/10.1007%2FBF01157706 | |
| Nov 14, 2017 at 14:50 | comment | added | user21574 | For a relative version of my previous comment on a surjective holomorphic fibre space see Theorem 2.2. of onlinelibrary.wiley.com/doi/10.1002/mana.19992040103/full | |
| Nov 14, 2017 at 14:00 | comment | added | user21574 | Let we have a complex manifold $X$, containing an analytic subset $A$ of complex codimension at least two. Let $(E,h) \to K\setminus A$ be a Hermitian-holomorphic vector bundle such that $F_h \in L^n(X\setminus A)$, then there exists a unique vector bundle $\hat E \to X$ such that $\hat E |_{X\setminus A}\cong E$. See Harris, A., and Tonegawa, Y. Analytic continuation of vector bundles with Lp-curvature, Int. J. Math. 11 No.l, (2000), 29-40. | |
| Jan 23, 2015 at 4:32 | vote | accept | Walter Neff | ||
| Jul 29, 2014 at 13:03 | comment | added | Vesselin Dimitrov | And preceding SGA2 there is Grauert's theorem that, in the case (over $\mathbb{C}$) that $\dim{X} > 2$ and $Y$ is an ample divisor, $E$ will so extend as soon as it extends to a vector bundle on a complex neighborhood of $Y$. A generalization in formal geometry appears in SGA2. | |
| Jul 28, 2014 at 11:50 | comment | added | Jason Starr | In addition to the excellent answers below, there is also SGA2. However, SGA2 will only extend your vector bundle to a reflexive coherent sheaf that is locally free on a Zariski open subset containing $Y$. If the rank is one, that forces the sheaf to be a line bundle. However, this can fail for higher rank. | |
| Jul 28, 2014 at 11:19 | answer | added | Misha Verbitsky | timeline score: 5 | |
| Jul 28, 2014 at 10:29 | answer | added | abx | timeline score: 7 | |
| Jul 28, 2014 at 10:23 | comment | added | Matthias Wendt | The first obstruction is that the determinant needs to be in the restriction on Picard groups. For the case $X=\mathbb{P}^2$ and $Y$ a smooth plane curve you can see that this is already a strong restriction. | |
| Jul 28, 2014 at 10:12 | review | First posts | |||
| Jul 28, 2014 at 10:13 | |||||
| Jul 28, 2014 at 10:10 | history | asked | Walter Neff | CC BY-SA 3.0 |