MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 12 down vote accepted

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

share|cite|improve this answer
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? – Timo Schürg Oct 22 '09 at 8:37
@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. – David Lehavi Oct 22 '09 at 8:54

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.

share|cite|improve this answer
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 – Timo Schürg Oct 1 '10 at 6:59
Sorry, the rank of the vector bundle in the comment should be $rk(E^{-1})$. – Timo Schürg Oct 1 '10 at 7:00

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.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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