15
$\begingroup$

Are there any general results on when a closed subscheme X of a quasi-projective smooth scheme M can be written as the zero-set of a section of a vector bundle E on M? To put it in a diagram: When is X the fiber product of M -> E <- M , where one arrow is the zero section and the other arrow is the section I'm looking for.

If this is not possible, can X be written as a degeneracy locus?

Improve this question
$\endgroup$

4 Answers 4

Reset to default
15
$\begingroup$

As for the first question, the class of X has to be the product of the Chern roots of the bundle, so in the Chow ring, it is the class of a complete intersection.

As for the second question, you would have to find classes that will solve the class of X in the Thom-Porteus formula, see Fulton's intersection theory 14.4

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thanks for the answer! Is it possible to recover the bundle from the Chern roots, or even better, to actually get your hands on the section? And how restrictive is the condition that it has to be the class of a complete intersection? A zero-set of a section can easily be a non-complete intersection, right? $\endgroup$
    – Timo Schürg
    Commented Oct 22, 2009 at 8:37
  • $\begingroup$ @Timo, I only have very partial knowledge on how to continue: You can build bundles as sub-bundles or quotient bundles of other bundles until you get the desired chern roots. A zero set can easily be non-complete intersection and be the same class of a complete intersection, but being the class of a complete intersection is a strong requirement. $\endgroup$
    – David Lehavi
    Commented Oct 22, 2009 at 8:54
8
$\begingroup$

Are you assuming that the rank of $E$ equals the codimension of the subscheme? You don't say so explicitly. If not, the answer is that every closed subscheme is a zero section, since it is the intersection of finitely many hypersurfaces.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thanks for the answer! The way I posed the question was really for arbitrary rank of $E$.I also figured out another way of proving your statement in the meantime: Just take the first step of a locally free resolution of the ideal sheaf of $X$ in $M$. That gives a surjective morphism $E \to I$, and thus a section. The case I really cared about was when $X$ has a perfect obstruction theory $E^{-1} \to E^{0}$. I wanted to fix $dim(M)=rk(E^{0})$, and the rank of the vector bundle to be $rk(E^{0})$. If $X$ is affine, that actually works! It's Appendix A of front.math.ucdavis.edu/1001.2719. $\endgroup$
    – Timo Schürg
    Commented Oct 1, 2010 at 6:59
  • $\begingroup$ Sorry, the rank of the vector bundle in the comment should be $rk(E^{-1})$. $\endgroup$
    – Timo Schürg
    Commented Oct 1, 2010 at 7:00
5
$\begingroup$

A necessary condition is that it be a locally complete intersection, since locally this is the same as asking that your scheme be the zero set of codimension many equations.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Do you happen to know when is this enough? $\endgroup$
    – Saal Hardali
    Commented Apr 9, 2017 at 15:11
  • $\begingroup$ I guess the question didn't require that the rank of the bundle is the codimension of the subscheme. $\endgroup$
    – Feng Hao
    Commented Feb 19, 2020 at 12:53
2
$\begingroup$

At least when the subvariety has codimension 2, this is known as "the Serre construction". There's a nice description of the case of points in a surface given in "Lectures on linear series" by Lazarsfeld. I'm sure there are many other excellent references too, but that's the first that comes to mind.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .