The situation is similar to [this question](https://mathoverflow.net/questions/69643/question-about-arguments-using-du-bois-complex)


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.



Vanishing of $H^0( D, \hat{\Omega}^1_D)$ on simply connected surface $D$?
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. 




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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.


