In Hartshorne's Algebraic Geometry II.8.20.1 (page 182), he takes the dual of Euler sequence

$$0 \rightarrow \Omega_{X/k} \rightarrow \mathcal{O}_{X}(-1)^{n+1} \rightarrow \mathcal{O}_{X} \rightarrow 0,$$

where $X = \mathbb{P}_{k}^{n}$, and get

$$0 \rightarrow \mathcal{O}_{X} \rightarrow \mathcal{O}_{X}(1)^{n+1} \rightarrow \mathscr{T}_{X} \rightarrow 0,$$

where $\mathscr{T}_{X} = \mathcal{Hom}(\Omega_{X/k}, \mathcal{O}_{X})$ is the tangent sheaf of $X$. But, is the dualizing functor $\mathcal{Hom}( \cdot, \mathcal{O}_{X})$ exact? Is not it only a left exact contravariant functor? Why in this case we have exactness of the sequence?

  • 8
    $\begingroup$ The answer depends on how much you know. The next term in the sequence would be an Ext group, but $\mathcal{O}_X$ is free and hence it is $0$. $\endgroup$ – Matt Sep 8 '12 at 14:03
  • 2
    $\begingroup$ Dear @Matt, If you're thinking about $\mathrm{Ext}$ groups of, say, $R$-modules, then $\mathrm{Ext}_R^i(\cdot,\cdot)$ kills injective modules in the second variable, and $R$ is not necessarily self-injective (there are certain kinds of rings which are, such as fields, or the rings $\mathbf{Z}/p^n\mathbf{Z}$). Also, in the case of sheaves, while you do get a long exact sequence of $\mathcal{E}xt$ sheaves, the contravariant internal $\mathcal{H}om$ functor doesn't usually have derived functors because the category of $\mathscr{O}_X$-modules doesn't have enough projectives. $\endgroup$ – Keenan Kidwell Sep 8 '12 at 15:09
  • 1
    $\begingroup$ Proposition III.6.3, part (b) of Hartshorne's Algebraic Geometry states that $\mathscr{Ext}^i(\mathscr{O}_X,\mathscr{G})=0$ for $i>0$, which basically says that the sheaf Ext is acting like it 'should' even though we don't have projectives(that is, it effaces free sheaves) $\endgroup$ – John Stalfos Sep 8 '12 at 15:43
  • 2
    $\begingroup$ I'm sure the point Matt is making is that the next term in the sequence is $\mathscr{E}xt^1(\mathscr{O}_X, \mathscr{O}_X)$, which is zero by the observation. $\endgroup$ – Zhen Lin Sep 8 '12 at 16:04
  • 1
    $\begingroup$ Ah, of course! I didn't even notice that the last term was $\mathscr{O}_X$ when I wrote my answer\comment. I was just thinking about answering the question of whether or not $\mathcal{H}om_{\mathscr{O}_X}(-,\mathscr{O}_X)$ was always exact. Thank you for pointing this out, and to @Matt, my apologies. $\endgroup$ – Keenan Kidwell Sep 8 '12 at 16:14

If you have a short exact sequence

$0\rightarrow\mathscr{F}\rightarrow\mathscr{G}\rightarrow\mathscr{H}\rightarrow 0$

of finite locally free sheaves on a scheme $X$, then the sequence

$0\rightarrow\mathscr{H}^\vee\rightarrow\mathscr{G}^\vee\rightarrow\mathscr{F}^\vee\rightarrow 0$

is exact. The reason is that exactness can be checked on stalks, and because the sheaves in the original sequence are finitely presented, taking stalks commutes with taking $\mathcal{H}om$ sheaves, so the sequence of stalks of the second sequence is

$0\rightarrow\mathrm{Hom}_{\mathscr{O}_{X,x}}(\mathscr{H}_x,\mathscr{O}_{X,x}) \rightarrow\mathrm{Hom}_{\mathscr{O}_{X,x}}(\mathscr{G}_x,\mathscr{O}_{X,x}) \rightarrow\mathrm{Hom}_{\mathscr{O}_{X,x}}(\mathscr{F}_x,\mathscr{O}_{X,x})\rightarrow 0$

which is exact because the functor $\mathrm{Hom}_{\mathscr{O}_{X,x}}(-,\mathscr{O}_{X,x})$ is exact on short exact sequences of finite free $\mathscr{O}_{X,x}$-modules.

I'm not sure what happens when the sheaves in the sequence are not finite locally free. There is a long exact sequence of $\mathcal{E}xt$ sheaves. Note that the sheaves in the OP's original sequence are all finite locally free.

EDIT: Incidentally, this has nothing to do with schemes, and works for arbitrary locally ringed spaces.


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.