MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

Hi, i'm stuck on the following, please can someone help? Let $E$ be a complex holomorphic vector bundle of rank r over a compact kahler manifold $M$, let me indicate $\mathcal{E}$ the associated locally free sheaf of $E$, let $\mathcal{F}$ be a coherent subsheaf of $\mathcal{E}$ of rank $0< p < r$ such that $\frac{\mathcal{E}}{\mathcal{F}}$ is torsion free. The inclusion map $j:\mathcal{F}\rightarrow \mathcal{E}$ induces a homomorphism of sheaves $det(j): det(\mathcal{F})=({\Lambda}^p\mathcal{F})^{** } \rightarrow ({\Lambda}^p\mathcal{E})^{**}$ (with * i mean the dual sheaf), $det(j)$ is injective outside the singularity set $S_{n-1}(\mathcal{F})\subset M$ (the set of points in which $\mathcal{F}$ is not free), writing the sequence: $0 \rightarrow ker(det(j))\rightarrow det(\mathcal{F})\rightarrow (\Lambda^{p}\mathcal{E})^{ ** }$ why the sheaf $ker(det(j))$ is a torsion sheaf?

Thank you in advance

share|cite|improve this question

Let's assume for simplicity that $M$ is a smooth, complex, projective variety.

The set of points where the coherent subsheaf $\mathcal{F}$ is not locally free is a proper closed subset of $M$ (Hartshorne, Algebraic Geometry, Chapter II, ex. 5.8), so the stalk of $ker(det(j))$ at the generic point is zero, i.e. it is a torsion sheaf.

Moreover, you can say more. Indeed, since $\mathcal{E}$ is locally free and $\mathcal{E} /\mathcal{F}$ is torsion-free, it follows that $\mathcal{F}$ is a reflexive sheaf (Hartshorne, Stable Reflexive Sheaves, Theorem 1.1), so it is locally free except along a closed subset of codimension $\geq 3$ (same reference, Corollary 1.4).

In particular, if $M$ is a curve or a surface then $ker(det(j))$ is zero.

share|cite|improve this answer
Hi! Thank you for replying. I looked to hartshorne and it defines the torsion sheaf as the sheaf with the zero stalk at the generic point, i forgot to say that my definition (Kobayashi's) is: Let $\mathcal{F}$ be a coherent sheaf, $\sigma : \mathcal{F} \rightarrow \mathcal{F}^{ ** }$ the natural morphism into its double dual, $\mathcal{F}$ is a torsion sheaf if $ker(\sigma)=\mathcal{F}$, are the two definitions equivalent? – Italo Aug 16 '10 at 13:39
Yes, they are. More precisely, the kernel of your map $\sigma$ is equal to the torsion subsheaf of $\mathcal{F}$ in the sense of Hartshorne, see Stable Reflexive Sheaves p. 124. So $\sigma$ is the zero map if and only if $\mathcal{F}$ is a torsion sheaf (again, in the sense of Hartshorne). – Francesco Polizzi Aug 16 '10 at 21:42

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.