I've been trying to understand the following proof in Kobayashi's book. Let me state some relevent definitions.

Let $X$ be a complex manifold and $\mathscr{O}_X$ its structure sheaf. A coherent sheaf $\mathscr{E}$ is torsion-free if the map $\mathscr{E}\to\mathscr{E}^{**}$ is injective. It is a torsion sheaf if the above map is a zero map. A torsion-free sheaf $\mathscr{E}$ of rank $r$ has a determinant line bundle $\det\mathscr{E}=(\wedge^r\mathscr{E})^{**}$.

Every monomorphism $\mathscr{F}\to\mathscr{F'}$ between torsion-free coherent sheaves of the same rank induces a sheaf monomorphism $\det\mathscr{F}\to\det\mathscr{F'}$.

The proof goes like this:

Outside their singular sets (where coherent sheaves are not locally free), the map $\mathscr{F}\to\mathscr{F'}$ and the induced map $\det\mathscr{F}\to\det\mathscr{F'}$ are isomorphisms. Hence, the $\ker(\det\mathscr{F}\to\det{F'})$ is a torsion sheaf. Since it is a subsheaf of a torsion-free sheaf, it must be zero.

My questions are in the following:

  1. Outside their singular sets, $\mathscr{F}$ and $\mathscr{F'}$ are locally free. Since they have the same rank, the map $\mathscr{F}\to\mathscr{F'}$ has a cokernel which has rank 0 and hence is a torsion sheaf. But why is it necessarily zero so that $\mathscr{F}\to\mathscr{F'}$ is an isomorphism?

  2. If $\mathscr{F}\to\mathscr{F'}$ is an isomorphism outside the singular sets, then $\ker(\det\mathscr{F}\to\det{F'})=0$ outside the singular set. Then, how do I conclude that it must be a torsion sheaf? I don't know its stalks inside the singular set. To show it is a torsion sheaf, I need to make sure every stalk is a torsion module.



1 Answer 1

  1. I don't see that this is true without other assumptions. Let $X=\Bbb C$ and $\mathscr{O}$ be the structure sheaf. Then $\mathscr{O}\stackrel{\cdot z}{\to}\mathscr{O}$ is a perfectly good morphism of free sheaves with cokernel the skyscraper sheaf supported at $0$ with sections $\Bbb C$.

Perhaps there should be a modification of the proof here - since the cokernel is torsion, it's supported on a closed submanifold of strictly smaller dimension, and we can throw that in with the singular set as a locus we can ignore - we will still get that the map on determinant bundles is an isomorphism on a dense open subset, which is really the thing that makes the proof work.

  1. Any sheaf supported on a submanifold of strictly smaller dimension is torsion: consider the action given by multiplying by a nonzero function which vanishes on this submanifold.
  • $\begingroup$ Hi, Thank you for your reply. Your second bullet point is very useful. So, your definition of a torsion sheaf is that in every stalk every element is a torsion element, correct? $\endgroup$
    – YYF
    Oct 4, 2018 at 21:42
  • $\begingroup$ Yes. Did something I write suggest something else? $\endgroup$
    – KReiser
    Oct 4, 2018 at 22:10
  • $\begingroup$ Then, maybe I am just tired today but I don't follow your second point. Let's say $\mathscr{F}$ is supported in a proper analytic subvariety $A$ of the base manifold $X$ and $s_x\in\mathscr{F}$ where $x\in A$. Then, take some holomorphic function $f$ that vanishes on $A$, then why is $(fs)_x=0$? It is true that $(fs)_y=0$ for every $y\notin A$. It is also true that $(fs)|_A=0$. But I must've missed something obvious. $\endgroup$
    – YYF
    Oct 4, 2018 at 22:23
  • $\begingroup$ One more thing in my last comment: $(fs)_|A=0$ does not make much sense, since this is a section in an abstract sheaf and is not a function. $\endgroup$
    – YYF
    Oct 4, 2018 at 22:40
  • $\begingroup$ But the problem is $U\cap A$ is not open. Therefore, even if I am tempted to say $(fs)|_{U\cap A}=f|_{U\cap A}s|_{U\cap A}=0$, I couldn't. $\endgroup$
    – YYF
    Oct 5, 2018 at 1:57

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