Extending vector bundles on a given open subscheme, reprise - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T09:13:12Z http://mathoverflow.net/feeds/question/35788 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/35788/extending-vector-bundles-on-a-given-open-subscheme-reprise Extending vector bundles on a given open subscheme, reprise Kevin Lin 2010-08-16T19:38:58Z 2012-06-05T19:51:09Z <p>In this <a href="http://mathoverflow.net/questions/22111/extending-vector-bundles-on-a-given-open-subscheme" rel="nofollow">question</a>, Ariyan asks about the question of uniqueness of extensions of vector bundles when they exist.</p> <p>Sasha's answer suggests that extensions of vector bundles don't always exist.</p> <p>More precisely, if $F$ is a vector bundle on an open subscheme $U$, there does not always exist a vector bundle $F'$ on the ambient space $X$ such that $F'|_U \cong F$.</p> <p>Can anyone give me a simple example of such an $F$?</p> <p>I am mainly interested in the case when $X$ is a variety (over $\mathbb{C}$), and $U$ is an open subvariety. Probably I want $X$ to be smooth.</p> http://mathoverflow.net/questions/35788/extending-vector-bundles-on-a-given-open-subscheme-reprise/35790#35790 Answer by Sasha for Extending vector bundles on a given open subscheme, reprise Sasha 2010-08-16T20:03:43Z 2012-06-05T19:51:09Z <p>The simplest example is the following. Take $X = A^3$ with coordinates $(x,y,z)$, and let $E = Ker(O_X \oplus O_X \oplus O_X \stackrel{(x,y,z)}\to O_X)$. Let $U$ be the complement of the point $(0,0,0) \in X$. Then $E_{|U}$ is a vector bundle. On the other hand, $E$ is not a vector bundle, but $E^{**} \cong E$, hence $E$ is the reflexive envelope of $i_*i^*E$, and thus there is no vector bundle on $X$ extending $E_{|U}$.</p> <hr> <p>[Edit by Anton: I just spent some time digesting some pieces of the above answer, so figured I'd include the results for future readers similar to me.]</p> <p> <b>("$E$ is not a vector bundle")</b> The sequence $O_X\xrightarrow{\pmatrix{z\\ y \\ x}}O_X^3\xrightarrow{\pmatrix{y & -z & 0\\ -x & 0 & z\\ 0 &x&-y}}O_X^3\xrightarrow{\pmatrix{x& y& z}}O_X$ is exact, so $E$ is the cokernel of the first map. Since taking fibers commutes with taking cokernels, we compute that $E$ has 2-dimensional fibers away from the origin, and 3-dimensional fiber at the origin. </p> <p> <b>("$E^{**}\cong E$")</b> Note that $E$ is $S_2$ (i.e. sections defined away from codimension 2 extend uniquely) since it is the kernel of a map from an $S_2$ sheaf to a torsion-free sheaf (the section of $O_X^3$ extends uniquely, and its image is zero away from codimension 2, so must be zero, so the extended section is in $E$). Note also that the dual of any sheaf is $S_2$ (if $\phi\colon F\to O_X$ is defined on an open set $V$ with codimension 2 complement and $s$ is a section, $\phi(s)$ must be the unique extension of $\phi(s|_V)$ as a section of $O_X$), so $E^{**}$ is $S_2$. The canonical map $E\to E^{**}$ is then a map of $S_2$ sheaves which is an isomorphism away from codimension 2, so it must be an isomorphism. </p> <p> <b>("and thus there is no vector bundle on $X$ extending $E|_U$")</b> If $F$ is an $S_2$ extension of $E|_U$ (i.e. $i^*F=i^*E$), then there is a map $F\to i_*i^*E\to (i_*i^*E)^{**}=E$ which is an isomorphism over $U$, so is an isomorphism by the argument in the previous paragraph. A vector bundle extension would be a different $S_2$ extension. </p> http://mathoverflow.net/questions/35788/extending-vector-bundles-on-a-given-open-subscheme-reprise/35819#35819 Answer by Olivier Benoist for Extending vector bundles on a given open subscheme, reprise Olivier Benoist 2010-08-17T00:44:08Z 2010-08-17T01:20:08Z <p>A vector bundle $E$ on an smooth integral curve $X$ that is not proper is trivial, so that it always extends. </p> <p>Indeed, such a curve is affine so that $E$ has a non trivial section $s$. Multiplying $s$ by a function constructed by Riemann Roch, you may assume that $s$ vanishes nowhere. This produces a short exact sequence $0\to\mathcal{O}_X\to E\to F\to0$, where $F$ is a vector bundle. By induction, $F$ is trivial. Since $X$ is affine, $H^0(E)\to H^0(F)$ is surjective, so that we can construct a splitting of this short exact sequence, which shows $E$ is trivial.</p> <p>This implies that a rank $r$ vector bundle $E$ on an integral curve $X$ that is not proper always extends to any partial compactification $X'$. Indeed, $E$ is trivial on $X\setminus \rm{Sing}(X)$ so that it is possible to glue $E$ and $\mathcal{O}^r_{X'\setminus \rm{Sing}(X)}$ together.</p> <p>For surfaces, this is not true anymore. Here is a 2-dimensional example of a vector bundle that does not extend.</p> <p>Take $X=\mathbb{A}^2$ and $U$ be $X$ deprived from the origin. By Cech cohomology computation you can check that $Ext^1(\mathcal{O}_U,\mathcal{O}_U)=H^1(U,\mathcal{O}_U)\neq0$, so that there exists a non trivial extension $0\to \mathcal{O}_U\to E\to \mathcal{O}_U\to0$. If $E$ extended to a vector bundle $E_X$ on $X$, the section $\mathcal{O}_U\to E$ would extend to a section $\mathcal{O}_X\to E_X$ because $X$ is normal. This section vanishes nowhere because it can only vanish on the origin which is of codimension $2$, so that we get an extension $0\to \mathcal{O}_X\to E_X\to \mathcal{L}\to0$ with $\mathcal{L}$ a line bundle. Repeating the argument above, we see that the isomorphism $\mathcal{O}_U\to\mathcal{L}|_U$ extends to an isomorphism $\mathcal{O}_X\to\mathcal{L}$. Thus we extended our short exact sequence to a short exact sequence on $X$ : $0\to \mathcal{O}_X\to E_X\to \mathcal{O}_X\to0$. This extension is an element of $Ext^1(\mathcal{O}_X,\mathcal{O}_X)=H^1(X,\mathcal{O}_X)=0$, because $X$ is affine. Thus it is the trivial extension. It remains trivial when restricted to $U$ and this is a contradiction.</p> http://mathoverflow.net/questions/35788/extending-vector-bundles-on-a-given-open-subscheme-reprise/35829#35829 Answer by jvp for Extending vector bundles on a given open subscheme, reprise jvp 2010-08-17T02:29:03Z 2010-08-17T02:29:03Z <p>In the analytic category there are line-bundles over $X = \mathbb C^2 - { 0}$ which do not extend to $\mathbb C^2$. Since $X$ has the homotopy type of the sphere $S^3$, the exponential sequence $$0\to \mathbb Z \to \mathcal O_X \to \mathcal O_X^* \to 1$$ implies $H^1(X,\mathcal O_X) = H^1(X, \mathcal O_X^*)$. As $H^1(X,\mathcal O_X)$ is infinite dimensional, there are many non-zero elements in $H^1(X,\mathcal O_X^*)$. These define line-bundles which do not extend. </p> http://mathoverflow.net/questions/35788/extending-vector-bundles-on-a-given-open-subscheme-reprise/35873#35873 Answer by Donu Arapura for Extending vector bundles on a given open subscheme, reprise Donu Arapura 2010-08-17T14:37:47Z 2010-08-18T18:35:48Z <p>This is a little perverse, but rather than answering the question, I want to explain what can go wrong when attempting to construct an example. This is the sort of thing one never does normally so I think it's kind of interesting.</p> <ol> <li><p>If $X$ is a smooth curve, then any vector bundle $E$ on an open set $U$ extends. To see this, we can assume after shrinking $X$, that $E$ is trivial. Then it can be extended to a trivial bundle (the extension is not unique).</p></li> <li><p>If $X$ is smooth surface, then any vector bundle $E$ on an open set $U$ extends. (I think that Olivier Benoist's answer contains a very nice idea, but I don't think the conclusion is OK.) To simplify the argument, assume that <code>$X-U=\{p_1,p_2\ldots \}$</code> is zero dimensional. We can find finitely sections in a neigbourhood $V$ of $p_i$ which generate $E^*$. This yields an inclusion <code>$E|_V\hookrightarrow \oplus \mathcal{O}_V^n$</code>, and therefore <code>$j_*E|_V \hookrightarrow\mathcal{O}_X^n$</code>, where <code>$j:U\hookrightarrow X$</code> is the inclusion. It follows easily, that <code>$j_*E$</code> is coherent. Therefore <code>$F=(j_*E)^{**}$</code> is a reflexive extension of $E$. However, reflexive sheaves have depth 2. Since by Auslander-Buchsbaum-Serre depth+proj.dim=2 in $\mathcal{O}_{p_i}$, we can conclude that $F$ is in fact locally free.</p></li> <li><p>In view of jvp's answer, we see that 2 does not hold in the analytic category.</p></li> <li><p>One might seek a topological obstruction involving Chern classes as in David Treumann's comment, however: Claim: Any Chern class on $U$ extends to $X$, where $X$ is a smooth partial compactification. Proof: With a bit of fiddling one can see that $c_p(E)$ would lie in $W_{2p}H^{2p}(U,\mathbb{Q})=im H^{2p}(X,\mathbb{Q})$ by Deligne, Theorie de Hodge II, III</p></li> </ol>