Let $X$ be a projective variety. Suppose we have computed the graded modules corresponding to $\Omega_{\mathbb P^n}$ (the cotangent sheaf) and $\mathcal O_X$. One way to get a representation for $\Omega_{\mathbb P^n} \otimes \mathcal O_X$ is to tensor the two corresponding modules.

In section 3 of the paper "Projective Geometry and Homological Algebra" by Eisenbud, contained in this freely available book, the author objects, saying

The result would represent the right sheaf, but would not be the module of twisted global sections of $\Omega_{\mathbb P^n}\otimes \mathcal O_X$ (the unique module of depth two representing the sheaf).

I have never seen this terminology before. What is the "module of twisted global sections," and what is the connection with depth? I suppose it is the module in the definition at the top of page 106 in Hartshorne (the "graded $S$-module associated to $\mathcal F$") but I have no idea how depth is relevant here.

After that passage, Eisenbud uses the Euler exact sequence to compute a description for $\Omega_{\mathbb P^n}$, and remarks that since this method produces a module with depth at least 2, we have found the module of twisted global sections. Again, I wonder about this depth statement.

  • $\begingroup$ nice ............+1 @patrick $\endgroup$ – Bhaskara-III Dec 28 '15 at 0:14
  • 1
    $\begingroup$ I haven't read the article, but my guess is that he refers to the $\Gamma_*$ construction as in Hartshorne [AG], the definition before Prop. II.5.13 (p.118). While it is always true that $\Gamma_*(\mathscr F)^{\tilde{}} = \mathscr F$, it may not be true that $\Gamma_*(\tilde{M}) = M$. In fact, $\tilde M \cong \tilde N$ if and only if $M_n \cong N_n$ for $n \gg 0$, so you couldn't possibly reconstruct $M$ from $\tilde M$; in particular you can't reconstruct $\Gamma(X,\tilde M)$ from knowing a (not the) presentation of $\tilde M$ as graded module. $\endgroup$ – Remy Dec 28 '15 at 8:16
  • $\begingroup$ See also exercise II.5.9 of Hartshorne. $\endgroup$ – Remy Dec 28 '15 at 8:23
  • $\begingroup$ @Remy Thanks. I figured it out. One needs local cohomology. See below. $\endgroup$ – user4571 Dec 28 '15 at 9:23

I found the result Eisenbud is alluding to in Appendix 1 of his book The Geometry of Syzygies. The result uses local cohomology.

Corollary A1.13. Let $M$ be a finitely generated graded $S$-module. The natural map $$M\mapsto \bigoplus_d \Gamma(\tilde M(d))$$ is an isomorphism if an only if $\operatorname{depth}(M)\ge 2$.

This implies the second claim.

If I am interpreting this corollary correctly, it gives a one to one correspondence between modules of depth at least two and sheaves induced by modules. So there is one such module representing $\Omega \otimes \mathcal O_X$. The claim that this module has has depth exactly two should fall out of the later analysis in the paper, where he computes this module explicitly in Macaulay 2. One could then use Macaulay 2 to compute the depth and show it is exactly two. This settles the first claim.


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.