# Dolbeault cohomology of the complex projective space.

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 $\mathbb{P}^n$? – David Speyer Jul 16 '12 at 18:49
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? – Matilda Jul 16 '12 at 20:28
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? – David Speyer Jul 16 '12 at 22:33

I'll flesh out my hint. Suppose that $\alpha$ is a global holomorphic $p$-form on $\mathbb{P}^{n-1}$. Pullback $\alpha$ along the map $\mathbb{C}^{n} \setminus \{ 0 \} \to \mathbb{P}^{n-1}$; let the pulled back form be $\beta = \sum f_I(x_1, \ldots, x_n) d x_{i_1} \wedge \cdots d x_{i_p}$. The functions $f_I$ are holmorphic functions on $\mathbb{C}^{n} \setminus \{ 0 \}$; by Hartog's Lemma, they extend to holomorphic functions on $\mathbb{C}^n$.
The fact that $\beta$ is pulled back from $\mathbb{P}^{n-1}$ means that $\beta$ must be invariant under dilation. I.e. $\delta_t^{\ast} \beta = \beta$ where $\delta_t$ is the map $(x_1, \ldots, x_n) \mapsto (t x_1, \ldots, t x_n)$. Translating into specific equations, this shows that $f_I(t x_1, \ldots, t x_n) = t^{-p} f_I(x_1, \ldots, x_n)$.
If $f_I(x_1, \ldots, x_n)$ is nonzero, this implies that $\lim_{t \to 0} f_I(t x_1, \ldots, t x_n) = \infty$. This contradicts that $f_I$ extends to a homolorphic function at $0$.