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?

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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.
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