2
$\begingroup$

Let $\mathcal{F}_1, \mathcal{F}_2$ be coherent sheaves over $\mathbb{P}^n_{\mathbb{C}}$ for $n \ge 3$. Now, $\Gamma_*(\mathcal{O}_{\mathbb{P}^n})=\mathbb{C}[X_0,...X_n]$. Denote by $U_0$ the affine scheme $\{X_0 \not=0 \}$. Denote by $D_1$ (resp. $D_2$) the support of $\mathcal{F}_1$ (resp. $\mathcal{F}_2$). Assume that $D_i \cap U_0$ is dense in $D_i$, for $i=1, 2$. Let $\phi_0 \in H^0(U_0,\mathcal{H}om_{\mathbb{P}^n}(\mathcal{F}_1,\mathcal{F}_2)|_{U_0})$ be a morphism. Does there exist a morphism $\phi \in H^0(\mathbb{P}^n,\mathcal{H}om_{\mathbb{P}^n}(\mathcal{F}_1,\mathcal{F}_2))$ such that $\phi|_{U_0}=\phi_0$?

$\endgroup$
  • $\begingroup$ In your final sentence, what do you mean by $\textit{Hom}_{\mathbb{P}^n}(\mathcal{F}_1,\mathcal{F}_2)|_{U_0}$, as a sheaf on $\mathbb{P}^n$? $\endgroup$ – Jason Starr Feb 7 '14 at 18:41
  • 1
    $\begingroup$ Not in general, e.g. there is no nonzero map $\mathcal{O}\to\mathcal{O}(-1)$, but the restrictions of the two sheaves to $U_0$ are isomorphic. However, for large enough $N$, there always exists an extension of $X_0^n \varphi_0$ to a morphism $\mathcal{F}_1\to\mathcal{F}_2(N)$. $\endgroup$ – Piotr Achinger Feb 7 '14 at 18:42
  • $\begingroup$ @Starr: Sorry, that was a typo. $\endgroup$ – user46578 Feb 7 '14 at 18:47
  • $\begingroup$ @Achinger: Could you please give a counter-example where $\mathcal{F}_1$ or $\mathcal{F}_2$ are not the trivial sheaf? The example that I have in mind is when $\mathcal{F}_1$ is the ideal sheaf of a proper subscheme $Z$ and $\mathcal{F}_2$ is its structure sheaf. $\endgroup$ – user46578 Feb 7 '14 at 19:13
  • $\begingroup$ @Kovacs: I am sorry, I was a bit sloppy in my comment. I meant: For $i:Z \hookrightarrow \mathbb{P}^n$ the closed immersion, I am interested in $\mathcal{H}om_{\mathbb{P}^n}(\mathcal{I}_Z,i_*\mathcal{O}_Z)$ which is the same as the normal sheaf of $Z$ in $\mathbb{P}^n$. This has non-zero global sections most of the time. $\endgroup$ – user46578 Feb 8 '14 at 12:33
5
$\begingroup$

As long as the complement of your open set is "big" as in "codimension $1$", there is no hope to do anything like this. Here is a concrete example on $\mathbb P^n$: Let $\mathscr L_1$ and $\mathscr L_2$ be two arbitrary line bundles such that there exists no morphism $\phi:\mathscr L_1\to \mathscr L_2$. On $\mathbb P^n$ any line bundle restricts to the trivial line bundle on your $U_0$ so there will be a morphism there.

The point is that the morphism that you observe on $U_0$ may have poles along the complement if the complement is codimension $1$. This is exactly what happens in this example.

In the example of your interest, $\mathscr F_1=\mathscr I_Z$ and $\mathscr F_2=\mathscr O_Z$, pretty much the same thing is happening. Since here $\mathscr F_2$ is supported on $Z$, working on $\mathbb P^n$ is almost a red herring. Morphisms between these sheaves, as you remark in a comment, are nothing else but sections of the normal sheaf of $Z$. So in this case your question translates to asking whether a section of this sheaf on an open (in fact a concrete open) set extends to the entire $Z$. Since you are working in $\mathbb P^n$ you know that this sheaf is ample, but that still does not mean that sections from open affine sets extend to global sections. In fact, observe that this will never happen. The space of sections of a line bundle on an affine set tend to be huge. This one is likely infinite dimensional. In contrast, the space of global sections is always finite dimensional. I think based on this you will find plenty of counter-examples and will see that this extension property rarely happens and probably never with these particular open sets.

On the other hand, if you assume that $\mathscr F_1$ and $\mathscr F_2$ are reflexive sheaves on a normal variety (e.g., locally free, but typically ideal sheaves and structure sheaves are not reflexive), and that you have a morphism one an open subset $U$ whose complement has at least codimension two, then the morphism extends; it is simply the push-forward of the morphism on $U$.

$\endgroup$

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.