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Let $X=\mathbb{CP}^n$. We proved using the hodge decomposition that $H^0(X,\Omega^p)=0$ if $p\neq 0$. But I do not understand why I cannot have global holomorphic differential p-forms not even constants.

  1. I want to understand why $H^0(X,\Omega^p)=0$ without using the Hodge decomposition.
  2. And without using GAGA I would like to see a proof of $H^0(Proj(\mathbb{C}[x_0,..,x_n]),\Omega_{X/\mathbb{C}}^p)=0$ for $p\neq 0$
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    Hint: Take a global section of $\Omega^p$ on $\mathbb{P}^{n}$ and pull it back to `$\mathbb{C}^{n+1} \setminus \{0\}$`. What do all forms on `$\mathbb{C}^{n+1} \setminus \{0\}$` look like? Which ones are pull backs from $\mathbb{P}^n$?2012-07-16
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    Let $\pi:\mathbb{C}^{n+1}\setminus \{0\}\rightarrow\mathbb{CP}^n $, all the forms in $\mathbb{C}^{n+1}\setminus \{0\}$ are give by $\sum f_i(z_0,...z_n)dz_i$, where the $f_i$ are holomorphic. When pulling back a differential form defined just in $U_0$ we get a one that is holomorphic in $\pi^{-1}(U_0)$ but it's not gonna be holomorphic when $z_0=0$. Then if we pull back a global holomorphic 1-form then it has to be also holomorphic in $\pi^{-1}(U_i)$, $i\neq0$. But there are points in $U_i$ with $z_0=0$ then there is not such a global holomorphic one form. Is this what you mean?2012-07-16
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    The first sentence is good, all forms on $\mathbb{C}^{n+1} \setminus \{ 0 \}$ are given by $\sum f_i (z_0, \ldots, z_n) d z_i$ (using Hartog's lemma). I think you are going in the wrong direction after that though. For example, is $z_0 dz_1$ pulled back from $\mathbb{P}^1$? Why not?2012-07-16

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