Extending vector bundles on a given open subscheme, reprise - MathOverflow

Extending vector bundles on a given open subscheme, reprise

2010-08-16T19:38:58Z

In this question, Ariyan asks about the question of uniqueness of extensions of vector bundles when they exist.

Sasha's answer suggests that extensions of vector bundles don't always exist.

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 $F'|_U \cong F$.

Can anyone give me a simple example of such an $F$?

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.

Answer by Sasha for Extending vector bundles on a given open subscheme, reprise

2010-08-16T20:03:43Z

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 $E_{|U}$.

[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.]

("$E$ is not a vector bundle") 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.

("$E^{**}\cong E$") 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.

("and thus there is no vector bundle on $X$ extending $E|_U$") 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.

Answer by Olivier Benoist for Extending vector bundles on a given open subscheme, reprise

2010-08-17T00:44:08Z

A vector bundle $E$ on an smooth integral curve $X$ that is not proper is trivial, so that it always extends.

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.

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.

For surfaces, this is not true anymore. Here is a 2-dimensional example of a vector bundle that does not extend.

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.

Answer by jvp for Extending vector bundles on a given open subscheme, reprise

2010-08-17T02:29:03Z

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.

Answer by Donu Arapura for Extending vector bundles on a given open subscheme, reprise

2010-08-17T14:37:47Z

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.

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).
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 $X-U=\{p_1,p_2\ldots \}$ is zero dimensional. We can find finitely sections in a neigbourhood $V$ of $p_i$ which generate $E^*$. This yields an inclusion $E|_V\hookrightarrow \oplus \mathcal{O}_V^n$ , and therefore $j_*E|_V \hookrightarrow\mathcal{O}_X^n$ , where $j:U\hookrightarrow X$ is the inclusion. It follows easily, that $j_*E$ is coherent. Therefore $F=(j_*E)^{**}$ 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.
In view of jvp's answer, we see that 2 does not hold in the analytic category.
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