Let $X$ be a quasi-affine scheme; that is, the natural map $$X\rightarrow \overline{X}:=Spec(\Gamma(X,\mathcal{O}_X))$$ is an inclusion. Each scheme has a quasi-coherent sheaf of Kahler differentials $\Omega$, and the above open inclusion induces a $\Gamma(X,\mathcal{O}_X)$-module map of global Kahler differentials

$$ \Gamma(\Omega_{\overline{X}})\rightarrow \Gamma(\Omega_{X})$$

Is this map always an isomorphism?

Edit: Changed $\mathcal{O}_X$ to $\Gamma(X,\mathcal{O}_X)$.

  • 1
    $\begingroup$ You mean $Spec(\Gamma(\mathcal O_X))$ ? Otherwise the spec of the sheaf is just $X$... $\endgroup$ – Qing Liu Mar 5 '11 at 0:59
  • $\begingroup$ Yes, corrected. $\endgroup$ – Greg Muller Mar 5 '11 at 3:19
  • $\begingroup$ What is the meaning of "inclusion" in this context? (I'm not claiming the usage is not standard, only that I am not familiar with it.) $\endgroup$ – Charles Staats Mar 5 '11 at 16:35
  • $\begingroup$ Its an open inclusion, so it induces an isomorphism of schemes onto an open subscheme. $\endgroup$ – Greg Muller Mar 6 '11 at 3:05

The assumption implies that the natural embedding induces an isomorphism $\Gamma(X,\mathscr O_X)\simeq \Gamma(\overline X,\mathscr O_{\overline X})$. Then this means that the complement of $X$ has at least codimension $2$.

In addition assume that $X$ is noetherian and $S_2$ (for instance normal).

In this case if $\Omega_{\overline X}$ is a reflexive sheaf, then the restriction $\Gamma(\overline X,\Omega_{\overline X})\to \Gamma(X,\Omega_{\overline X})=\Gamma(X,\Omega_{X})$ is an isomorphism.

More generally, let $Z:=\overline X\setminus X$. If $\mathrm{depth}_Z\Omega_{\overline X}\geq 2$ then the restriction $\Gamma(\overline X,\Omega_{\overline X})\to \Gamma(X,\Omega_{X})$ is an isomorphism. This certainly holds if $\Omega_{\overline X}$ is a reflexive sheaf, but obviously it could hold "by accident" even if one of the above conditions fail, so I am not claiming that these are necessary conditions, but at least they sure seem to provide a natural set of conditions under which the required map is an isomorphism.

Sketch that if $X$ is $S_2$, then a reflexive coherent sheaf $\mathscr F$ is also $S_2$: First observe that by the argument in this answer to another MO question we may assume that $X$ is affine and it is enough to prove that $H^i_x(X,\mathscr F)=0$ for $i=0,1$ for all $x\in X$ and it also follows that $\mathrm{depth}_Z\mathscr F\geq 2$ even if $Z$ is not contained in an affine piece of $X$. To do that write $\mathscr F^\vee$ as the quotient of a (locally) free sheaf ($X$ is affine!). Then $\mathscr F$ is a submodule of the dual of this locally free sheaf, let's call it $\mathscr E$, and the quotient $\mathscr E/\mathscr F$ is torsion-free. Therefore none of them have torsion and so $H^0_x(X,\mathscr F)=0$ and $H^1_x(X,\mathscr F)$ embeds into $H^1_x(X,\mathscr E)$. But the latter is $0$ by the assumption that $X$ is $S_2$.

EDIT 1 removed intro paragraph about the starting assumption.

EDIT 2 added "more generally" paragraph.

EDIT 3 added noetherian assumption. this is probably not necessary but without this one should possibly be more careful about the other conditions.

EDIT 4 Added Sketch above.

  • $\begingroup$ Yes, Sandor, you are correct in that I explicitly want $\overline{X}=Spec(\mathcal{O}_X)$. And its true that, if we know $\Omega_{\overline{X}}$ is reflexive, then we are done. $\endgroup$ – Greg Muller Mar 4 '11 at 23:23
  • $\begingroup$ On the other hand, it is almost equivalent to being reflexive. $\endgroup$ – Sándor Kovács Mar 4 '11 at 23:24
  • $\begingroup$ To work correctly with the depth, do you have to assume $\overline{X}$ is noetherian ? $\endgroup$ – Qing Liu Mar 5 '11 at 1:00
  • $\begingroup$ @Qing Liu: I suppose I secretly assumed it was noetherian. $\endgroup$ – Sándor Kovács Mar 5 '11 at 6:48
  • $\begingroup$ Sandor, on an S2 scheme, is it true that every reflexive sheaf is S2? I thought you needed S2 + G1 (do you know a reference?) $\endgroup$ – Karl Schwede Mar 5 '11 at 13:01

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.