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Let $V_X$ be a vector bundle of rank $r>1$ on a smooth (connected) projective variety of dimension $r$. Let s be the global holomorphic section, whose zero locus is a zero dimensional subscheme $Z\subset X$. Is there some result that the section is determined (up to scaling) by its zero locus? At least for some nice vector bundles on nice manifolds?

upd. I meant $X_\Bbb C$ and some conditions on some resolution of $V_X$ by some v.a. line bundles on $X$. jvp cites an excellent result below.

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The question is too general. For example, take $V=\mathcal{O}(1)^2$ on $\mathbb{P}^2$. Then $H^0(V)$ is six dimensional and an open set of a four dimensional subspace of this vector space has precisely just one fixed point as its zeroes. So, the zero does not determine the section. –  Mohan Jun 2 '11 at 21:56
    
Yes, certainly in general we do not have such a property. But maybe with some restrictions on the resolution of this bundle? –  Dmitry Kerner Jun 2 '11 at 21:59
    
Which resolution are you talking about qui-vadis? The structure sheaf of $Z$? –  J.C. Ottem Jun 2 '11 at 22:59
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I suppose you're looking for generalizations of the fact that a polynomial is determined up to scaling by its zeros? Note that this is very special to the algebraic situation, and also to the situation where the base field is algebraically closed, and also to the situation where you know the zero locus as a subscheme and not just a subset (so that you can distinguish $x$ from $x^2$ as sections of the trivial rank-1 bundle over the affine line). –  Theo Johnson-Freyd Jun 3 '11 at 0:54
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1 Answer

up vote 2 down vote accepted

Take a look at On sections with isolated singularities of twisted bundles and applications to foliations by curves by Campillo and Olivares.

There you will find the following result.

Theorem. Let $X$ be a projective manifold of dimension $n$, $\mathcal L$ be an ample >line-bundle on $X$, and $E$ a rank $n$ vector bundle over $X$. If $k\gg 0$ and $s, s'$ >are sections of $E \otimes \mathcal L^{\otimes k}$ such that

  • the zero scheme $Z(s)$ of $s$ has dimension zero, and
  • the zero scheme $Z(s')$ of $s'$ contains $Z(s)$

then there exists $\varphi \in H^0(X,\mathrm{End}(E))$ such that $s' = \varphi(s)$. In particular, if $E$ is simple ( $H^0(X,\mathrm{End}(E))=\mathbb C$ ) then $s = \lambda s'$ for a suitable complex number $\lambda$.

In the particular case $E=T\mathbb P^n$, it suffices to take $\mathcal L= \mathcal O_{\mathbb P^n}(1)$ and $k\ge 1$.

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