Let $X$ be a smooth projective variety and let $Y\subset X$ be a smooth subvariety. Given a vector bundle $E$ on $Y$, when can $E$ be extended to a vector bundle $\tilde E$ on $X$? I.e., are there cohomology groups containing the obstructions to extending $E$?
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2$\begingroup$ 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. $\endgroup$– Vesselin DimitrovCommented Jul 29, 2014 at 13:03
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4$\begingroup$ 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. $\endgroup$– user21574Commented Nov 14, 2017 at 14:00
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2$\begingroup$ 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 $\endgroup$– user21574Commented Nov 14, 2017 at 14:50
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3$\begingroup$ 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 $\endgroup$– user21574Commented Nov 15, 2017 at 14:50
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3$\begingroup$ 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) $\endgroup$– user21574Commented Nov 16, 2017 at 13:18
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2 Answers
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Without hypotheses on $Y$ there is no hope to define such obstructions, already for line bundles. A natural hypothesis is to take for $Y$ a (smooth) ample divisor in $X$, of dimension $\geq 2$. In this paper, Fujita gives some cohomological conditions which imply that $E$ extends : $H^2(Y, \mathcal{E}nd(E)(-tY))=0$ for all $t\geq 1$ and $H^p(Y, E(tY))=0$ for all $t\in\mathbb{Z}$ and $0<p<\dim Y$. Be aware that these conditions are extremely strong.
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$\begingroup$ This has been a while. You mention that $Y$ needs to be a smooth ample divisor. Fujita only mentions non singularity of $X$. Is the assumption of smoothness of the divisor implicit throughout the paper? $\endgroup$ Commented Dec 11, 2020 at 16:35
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$\begingroup$ No, I guess I put "smooth" because the OP did so. You are right, Fujita makes no smoothness assumption on $Y$. $\endgroup$– abxCommented Dec 11, 2020 at 17:27
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$\begingroup$ Does Fujita assume the divisor is irreducible or maybe a different definition of ampleness? Because I'm considering the case when everything is affine, the cohomologies vanish because cohomology of affine schemes vanish. By Hartshorne's definition any divisor on affine scheme is ample. Then it implies you can extend any vector bundle from any effective divisor. Which is not correct. $\endgroup$ Commented Dec 11, 2020 at 20:05
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$\begingroup$ I think he is implicitly assuming that $X$ is a projective variety. His Lemma 3.2 is obviously false in the affine case. $\endgroup$– abxCommented Dec 11, 2020 at 20:43
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An obvious obstruction comes from topology: the Chern classes of your bundle should be obtained from restriction of Hodge classes on an ambient variety. This is (more or less) enough to extend a smooth bundle $B$ from $Y$ to $X$. To be precise, you need the classifying map from $Y$ to the space $BU(r)$ to be extendable to a continuous map from $X\supset Y$ to $BU(r)$, where $r$ is rank $B$. This is not the only obstruction, because a way to find a holomorphic bundle with prescribed $(p,p)$-Chern classes on $X$ amounts to a result which is much stronger than the Hodge conjecture (and false, generally speaking). The easiest obstruction to finding a bundle with prescribed $(p,p)$-Chern classes comes from the Bogomolov inequality: for any stable $B$, one has $$\int_M [2rc_2(B) - (r - 1)c_1(B)^2]\wedge \omega^{n-2}> 0,$$ where $n$ is dimension of your manifold $M$, and $\omega$ its Kahler form (case of unstable bundles is considered separately using the Jordan-Holder filtration). Also, if this inequality is non-strict, $B$ admits a projectively flat connection, and therefore $c_2$ and the rest of Chern classes are powers of $c_1$.
answered Jul 28, 2014 at 11:19