1
$\begingroup$

The situation is similar to this question

Let $D$ be a reduced projective scheme over $\mathbb{C}$ whose associated analytic space $D_{an}$ is simply connected as a topological space i.e. $\pi_1 (D_{an}) = 1$. Let $\Omega^1_D$ be its K\"{a}hler differential sheaf and $\hat{\Omega}_D^1$ be its double dual.

Question Is $H^0(D, \hat{\Omega}_D^1) = 0$? Is there some counterexample?

I'm considering how to deduce the above statement from the fact $\mathbb{H}^0 (D, \underline{\Omega}_D^1) = 0$ where $\underline{\Omega}_D^1$ is the graded piece of the Du Bois complex $\underline{\Omega}_D^{\bullet}$. But I don't know the way up to now.

$\endgroup$

1 Answer 1

4
$\begingroup$

Sorry, but it's not true. Let $D$ be a cone over an elliptic curve with vertex $p$. It is simply connected, but $H^0(D,(\Omega^1_D)^{**})= H^0(D-p, \Omega_{D-p}^1)\not=0$.

Added to address your question below: $H^0(D,\Omega_D^1)=0$ for any degree $d$ surface in $P=\mathbb{P}^3$, and so in particular for the example above. You can see this using the exact sequences $$0\to \mathcal{O}_D(-d)\to \Omega_P^1|_D\to \Omega_D^1\to 0$$ $$0\to \Omega_P^1(-d)\to \Omega_P^1\to \Omega_P^1|_D\to 0$$ and Bott's vanishing.


This goes to show that differentials can behave strangely on singular spaces. The Du Bois complex, exotic as it is, is thing that works best from certain points of view.

$\endgroup$
1
  • $\begingroup$ Thank you for the nice counterexample. Does it satisfy $H^0(D, \Omega^1_D) \neq 0$? $\endgroup$
    – tarosano
    Jul 8, 2011 at 11:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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