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I am doing questions from [Huybrechts, Complex Geometry, An Introduction] page 143 questions 3.3.7 and 3.3.8. Basically the questions ask:

For question 7, for any complex line bundles on the projective space $\mathbb{P}^{n}$, we may endow it with a holomorphic structure, and it is unique.

For question 8, let $\mathbb{C}^{n}/\Gamma$ be a torus. Then any trivial complex line bundle over it can be given a holomorphic structure, and how many of them?

My question is how can I do so?

If I understand the meaning of holomorphic structure correctly, it means that while previously the transition functions on the line bundles are smooth, there is a way to make them holomorphic, but precisely I have no idea how can I do so. For example in the case for the 2ND question, there is no transition function to talk about...

How should I start?

Many thanks!

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Well I don't have the text here, so I don't know how Huybrechts wants you to prove the statements.

Let $X$ be a complex manifold. You probably know that any linebundle corresponds to an element of $H^{1}(X, \mathcal{O}^{*}_{X})$. Now you look at the exponential sequence. The interesting part of the associated long exact sequence of cohomology is

$\to H^{1}(X, \mathcal{O}_{X})\to H^{1}(X, \mathcal{O}^{*}_{X})\xrightarrow{c_{1}} H^{2}(X, \mathbb{Z})\to H^{2}(X, \mathcal{O}_{X})\to$

Here $c_{1}$ denotes the first chern class. Notice that $c_{1}$ is a complete invariant for topological line bundles, meaning that it gives a group isomorphism from the group of complex line bundles to $H^{2}(X, \mathbb{Z})$.

Now to your questions:

Let $X=\mathbb{P}^{n}$. Basically you have to show $Pic(\mathbb{P}^{n})\cong \mathbb{Z}$. There are several ways to do this, some of which work for fields other than $\mathbb{C}$ as well, but since Huybrechts book is all about Hodge-Theory, he probably wants you to use it here.

First note that $H^{2}(\mathbb{P}^{n}, \mathbb{Z})\cong \mathbb{Z}$. And by Serre duality we have $H^{1}(\mathbb{P}^{n}, \mathcal{O}_{\mathbb{P}^{n}})\cong H^{0, 1}(\mathbb{P}^{n})$ and $H^{2}(\mathbb{P}^{n}, \mathcal{O}_{\mathbb{P}^{n}})\cong H^{0, 2}(\mathbb{P}^{n})$.

Now use the Hodge decomposition theorem. We have $0=H^{1}(\mathbb{P}^{n}, \mathbb{C})= H^{1,0}(\mathbb{P}^{n})\oplus H^{0,1}(\mathbb{P}^{n})$, so $H^{1}(\mathbb{P},\mathcal{O}_{\mathbb{P}^{n}})=0$.

Also: $\mathbb{C}\cong H^{2}(\mathbb{P}^{n}, \mathbb{C})= H^{2,0}(\mathbb{P}^{n})\oplus H^{1,1}(\mathbb{P}^{n})\oplus H^{0,2}(\mathbb{P}^{n})$. But since $H^{1,1}(\mathbb{P^{n}})$ contains the Kähler-class it has $dim_{\mathbb{C}} > 0$, so $H^{2}(\mathbb{P}^{n}, \mathcal{O}_{\mathbb{P}^{n}})=H^{0,2}(\mathbb{P}^{n})=0$.

Together with the above exact sequence, we get $Pic(\mathbb{P}^{n})\cong H^{2}(\mathbb{P}^{n}, \mathbb{Z})\cong \mathbb{Z}$. So the only holomorphic linebundles on $\mathbb{P}^{n}$ are the bundles $\mathcal{O}_{\mathbb{P}^{n}}(k)$, which are also the only complex linebundles.

For the second question, note that the holomorphic structures on the trivial line bundle are classified by the Jacobian, which is simply the kernel of the map $H^{1}(X, \mathcal{O}^{*}_{X})\xrightarrow{c_{1}} H^{2}(X, \mathbb{Z})$. For any complex Kähler manifold $X$, the Jacobian is a complex torus of dimension $dimH^{1}(X)$. I know this is proved in Huybrechts book. So now you simply need to compute the first Betti-number of a complex torus.

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  • $\begingroup$ For the first problem, however I have a question. The fact the we are talking about elements in $H^{1}(X,\mathcal{O}_{X}^{*})$ already suggests that we are assuming the line bundles have holomorphic structures. Why should any complex line bundles (i.e. whose transition function is just smooth) belong to the cohomology group? Isn't this what we are suppose to prove? $\endgroup$ – enoughsaid05 Dec 7 '14 at 8:48
  • $\begingroup$ Like I said above. The group of complex line bundles is isomorphic to $H^{2}(X, \mathbb{Z})$. If $c_{1}$ is surjective, then every complex line bundle can be made holomorphic. But ok, it was somewhat misleading by me to say that we have to show $Pic(\mathbb{P}^{n})\cong \mathbb{Z}$. This is not enough of course. $\endgroup$ – Rieux Dec 7 '14 at 14:54

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