The definition and basic properties of nef locally free sheaves appear for instance in the second volume of Lazarsfeld's book "Positivity in Algebraic Geometry" (beginning of chapter 6).

However, I am in a situation where some of the sheaves I deal with are not locally free, but only coherent; so I would like to know whether there is a well-behaved notion of nefness for coherent sheaves. The only mention of this that I found is at the end of section 1 of Kodaira Dimension of Subvarieties by Peternell-Schneider-Sommese, but it is just the definition with no references and no discussion of basic properties.

So my question is: is there a reference that gives an analogue of Theorem 6.2.12 in Lazarsfeld for nef coherent sheaves? (The results I'm mostly interested in are: (a) quotient of nef is nef; (b) pullback of nef is nef; and (c) extension of nef by nef is nef.)


Definition A coherent sheaf $\mathcal{F}$ on an algebraic variety $X$ is nef if the following condition holds: For every irreducible curve $C\subset X$, the line bundle $O(1)$ is nef on $\mathbb{P}(\mathcal{F}|_C)$.

I haven't seen this definition of a nef coherent sheaf either, but I think most of the properties you mention just follow formally from the properties of ordinary nefness. Here is a proof for a) and b):

a) A quotient of a nef sheaf is nef. Let $C$ be as in the definition. If $F\to E\to 0$ is a surjection, this restricts to a surjection $F|_C\to E|_C\to 0$ and hence gives an embedding $\mathbb{P}(\mathcal{F}|_C)\hookrightarrow \mathbb{P}(\mathcal{F}|_C)$ such that $O(1)$ on $\mathbb{P}(\mathcal{F}|_C)$ restricts to $O(1)$ on $\mathbb{P}(\mathcal{E}|_C)$. Since the restriction of a nef line bundle is nef, $E$ is also nef.

b) A pullback of nef is nef. Similarily, if $f:X\to Y$ is finite, then $f$ restricts to a finite map $f:C'=f^{-1}C\to C$ and hence there is a finite map $F:\mathbb{P}(f^*\mathcal{F}|_C)\to \mathbb{P}(\mathcal{F}|_C)$ such that $O(1)=F^*O(1)$. Now the claim just follows from the corresponding statement for line bundles.


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.