24
$\begingroup$

Let $U$ be a dense open subscheme of an integral noetherian scheme $X$ and let $E$ be a vector bundle on $U$. Suppose that the complement $Y$ of $U$ has codimension $\textrm{codim}(Y,X) \geq 2$. Let $F$ be a vector bundle on $X$ extending $E$, i.e., $F|_{U} = E$.

Is any extension of $E$ to $X$ isomorphic to $F$?

$\endgroup$

4 Answers 4

36
$\begingroup$

This is true if $X$ satisfies Serre's condition $S_2$, i.e. $\mathcal O_X$ is $S_2$. Then a vector bundle is $S_2$ since locally it is isomorphic to $\mathcal O_X^n$.

More generally, a coherent sheaf $F$ on a Japanese scheme (for example: $X$ is of finite type over a field) which is $S_2$ has a unique extension from an open subset $U$ with $\operatorname{codim} (X\setminus U)\ge 2$. This follows at once from the cohomological characterization of $S_2$.

Thus, another name for the $S_2$-sheaves: they are sheaves which are saturated in codimension 2, and another name for the $S_2$-fication: saturation in codimension 2.

P.S. Of course, by Serre's criterion, normal = $S_2+R_1$. So the above statement is true for any normal (e.g. smooth) variety.

P.P.S. And of course, Gorenstein implies Cohen-Macaulay implies $S_2$. So the statement is also true for hypersurfaces and complete intersections, which could be very singular and non-reduced.


Edit to define some terms:

  1. A Japanese (or Nagata) ring is a ring obtained from a ring finitely generated over a field or $\mathbb Z$ by optionally applying localizations and completions. The property used here is that for a Japanese ring $R$, its integral closure (normalization) $\tilde R$ is a finitely generated $R$-module. This is important because the $S_2$-fication $S_2(R)$ lies between $R$ and $\tilde R$.

  2. A coherent sheaf $F$ satisfies $S_n$ if for any point $x\in Supp(F)$, one has $$ depth_x (F) \ge \min(\dim_x Supp(F),n) $$ If $F$ locally corresponds to an $R$-module $M$, and $x$ to a prime ideal $p$, then the depth is the length of a maximal regular sequence $(f_1,\dots, f_k)$ of elements of $R_p$ for $M_p$ (so, $f_1$ is a nonzerodivisor in $M_p$, etc.).

$\endgroup$
12
  • $\begingroup$ "Japanese scheme"?? :D :D $\endgroup$
    – Qfwfq
    Apr 22, 2010 at 12:43
  • 1
    $\begingroup$ Very satisfactory answer! $\endgroup$
    – Qfwfq
    Apr 22, 2010 at 12:52
  • $\begingroup$ Dear VA, sorry for my ignorance: what is the definition of (S_2) for sheaves? $\endgroup$ Apr 22, 2010 at 15:15
  • $\begingroup$ Hi! I wonder if I understand correctly: Could you replace the Japaneseness hypothesis by requiring O_X and F both to satisfy S2? $\endgroup$
    – Lars
    Apr 23, 2010 at 15:03
  • $\begingroup$ Lars: No. What is needed is this property: the normalization of $R$ is a finite $R$-module. Japanese implies that. $\endgroup$
    – VA.
    Apr 23, 2010 at 18:05
16
$\begingroup$

Let $i:U \to X$ be the embedding. Assume that $i^*F = E$. Then by the adjunction we have a map $F \to i_*E$ which is an embedding, since the kernel is zero on $U$, so it is a torsion sheaf, and a vector bundle doesn't have torsion subsheaves (if $X$ doesn't have embedded components). So, we have an exact triple $0 \to F \to i_*E \to G \to 0$ for some $G$, which is supported on $X \setminus U$. If $X$ satisfies and $codim_X (X\setminus U) \ge 2$, then (provided $S_2$ condition) we have $G^* = {\mathcal Ext}^1(G,O_X) = 0$, so dualizing the sequence we see that $F^* = (i_* E)^\ast$, hence $F = (i_* E)^{\ast\ast}$, so $F$ has to be the reflexive envelope of $i_* E$. This proves the uniquencess. This also allows to construct an example of a vector bundle on $U$ which does not extend to a vector bundle on $X$.

$\endgroup$
1
  • 1
    $\begingroup$ Hi, Sasha, can you explain how to construct a an example of a vector bundle on $U$ which does not extend to a vector bundle on $X$. Thanks! $\endgroup$
    – Fei YE
    Apr 26, 2010 at 19:59
12
$\begingroup$

This is false as stated; for example, if $X$ is obtained from a projective geometrically connected smooth surface over a field $k$ by gluing two points together and $U$ is the complement of the singular point, then the kernel of the restriction map from the Picard group of $X$ to the Picard group of $U$ is easily seen to be $k^*$. You need to assume that the depth of all the components of the complement of $U$ in $X$ is at least 2; then the statement is correct. If $j\colon U \to X$ is the embedding, then $j_*j^*F = F$.

$\endgroup$
5
$\begingroup$

Edit: The following argument cannot work always, given the counterexample by Angelo, but maybe it works if $X$ is smooth? Edit2: Yes, this argument should work for $W$ smooth ("smooth" being, as noted in another answer, a particular case of "Serre S2").

The transition functions of an extension $F$ of $E$ are morphisms on open (affine) subsets $V$ of $X$ with values in $GL_n$ where $n=rank(F)$. Their restrictions to $W:=V\cap U$ are the transition functions of $E$. So, the problem reduces to proving that a morphism from a the complement of a codimension $\geq 2$ subvariety $Y$ of a quasi-affine variety $W$, with values in $GL_n$, extends at most in one way to the whole variety $W$. This is true, since you can assume $W$ is a subset of affine space, and each component $W-Y \rightarrow GL_n \hookrightarrow \mathbb{A}^{n\times n} \rightarrow \mathbb{A}^1$ is a regular function, and polynomials extend uniquely through codimension $\geq 2$ subvarieties. So, if you have an extension, it has to be unique.

$\endgroup$
2
  • 1
    $\begingroup$ The fact that regular functions extend over to codim $\geq 2$ subvarieties doesn't mean that the morphism to $GL_n$ will extend; you can only claim that a morphism to $\mathbb{A}^{n^2+1}$ will extend, but the extension might well wander off the embedded $GL_n$. (I'll just leave this comment here in case someone gets confused while reading the answers, as I did) $\endgroup$ Mar 7, 2012 at 20:45
  • $\begingroup$ @DimaSustretov: that is not the problem. If a dense subset lands in a closed subvariety, then everything does (note that $\operatorname{GL}_n$ is affine, by the closed embedding into $\mathbb A^{n^2+1}$). The problem is that Angelo's example is not $S_2$, hence not normal. "Hartog's theorem" holds for normal schemes. $\endgroup$ May 30, 2017 at 12:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.