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Let $\mathcal{O}(-1)$ be the tautological bundle over $\mathbb{P}^n$ and $\mathcal{O}(1)$ its dual bundle, also known as the hyperplane bundle. I know that there is a bijection between $\Gamma(\mathbb{P}^n,\mathcal{O}(1))$ and $\mathbb{C}[z_0,\dots,z_n]_1$, the set of homogeneus degree $1$ polynomials in $n+1$ variables. In fact, I think that given any two sections defined over two different affine open sets of $\mathbb{P}^n$ that glue are also defined by one of these polynomials.

So my question is, what can we say about a section only defined in one of the affine open sets? Can these sections be arbitrary or are they also determined by a linear functional?

Edit: MooS already provided a great answer for the algebraic geometric case. Yet, I forgot to mention in the OP that I'm looking for a holomorphic approach to the problem.

Edit 2: In fact, given any affine open set $U_i$ of $\mathbb{P}^n$ it would be enough for me to know how the sections $\mathcal{O}(1)(U_i)$ look in the holomorphic setting. MooS' answer makes me think that these sections should be of the form $\frac{f(z_0,\dots,z_n)}{z_i^{\operatorname{deg}f-1}}$, for $f$ an homogeneous polynomial, just as in the algebraic setting, yet I don't know how to show it.

For example, in $U_i$ a section $s$ is defined by a function $\tilde{s}:\mathbb{C}^n\to\mathbb{C}$ so that $$s([z])=([z],\tilde{s}(z_0/z_i,\dots,z_n/z_i)).$$

Taking the series expansion of $\tilde{s}$ naturally yields a homogeneous polynomial as long as it's finite. And even in this case we end up with something of the form $\frac{g(z_0,\dots,z_n)}{z_i^{\operatorname{deg g}}}$ for $g$ a homogeneous polynomial. Any ideas on how to proceed from here?

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    $\begingroup$ The standard affine cover is a trivialization for the hyperplane bundle, i.e. the sections on $\{ z_i \neq 0 \}$ are polynomials in the $z_j$'s, $j \neq i$. $\endgroup$ Commented Jun 29, 2016 at 9:19
  • $\begingroup$ Hi @MooS, thanks for your reply. How do you know that's the case? Along the lines of my question, does this imply that the gluing condition is the one imposing these polynomials to become linear? $\endgroup$ Commented Jun 29, 2016 at 9:24
  • $\begingroup$ You should have a look at Hartshorne, Propositions 2.5.11-12. $\endgroup$ Commented Jun 29, 2016 at 9:32
  • $\begingroup$ They are also determined a linear functional. This is a standard fact. For any vector bundle $E$ the dual $E^{*}$ gives a perfect pairing of vector bundles, this means that you can restrict to any open set. $\endgroup$ Commented Jul 1, 2016 at 23:52
  • $\begingroup$ If you want to know what are the holomorphic sections you must first give the explicit trivialization of your line bundle which is given by $(l, z) \mapsto (l, z_i)$ when $z_i \neq 0$ ($z_i$ is the coordinate of $z$). $\endgroup$ Commented Jul 1, 2016 at 23:54

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$\newcommand{\Cpx}{\mathbf{C}}\newcommand{\Proj}{\mathbf{P}}$If $V$ is a finite-dimensional complex vector space, a section of $\mathcal{O}(1) \to \Proj(V)$ is a linear functional $\lambda:V \to \Cpx$. Fixing an identification $V \simeq \Cpx^{n+1}$, a section of $\mathcal{O}(1) \to \Cpx\Proj^{n}$ is a homogeneous linear polynomial $$ \lambda(Z^{0}, Z^{1}, \dots, Z^{n}) = \lambda^{0}Z_{0} + \lambda^{1}Z_{1} + \dots + \lambda^{n}Z_{n} $$ in the homogeneous coordinates on $\Cpx\Proj^{n}$.

In the affine chart $U_{j} = \{Z^{j} \neq 0\}$ with affine coordinates $z_{j}^{i} = Z^{i}/Z^{j}$, this becomes the complex-valued function $$ \lambda_{j}(z_{j}^{0}, z_{j}^{1}, \dots, \widehat{z_{j}^{j}}, \dots, z_{j}^{n}) = \frac{\lambda(Z^{0}, Z^{1}, \dots, Z^{n})}{Z^{j}}. $$ Particularly, a collection $(\lambda_{j})$ of $n + 1$ affine functions on $\Cpx^{n}$ glues into a global section if and only if it satisfies the transition formula $$ \lambda_{i} = \frac{Z^{j}}{Z^{i}} \lambda_{j},\quad 0 \leq i, j \leq n. $$ (To me, the transition function $g_{ij} = Z^{j}/Z^{i}:U_{i} \cap U_{j} \to \Cpx^{\times}$ defines the holomorphic line bundle $\mathcal{O}(1)$.)

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  • $\begingroup$ Dear @Andrew D. Hwang, at the beginning of your answer you claim that a global section of $\mathcal{O}(1)$ is the same data as a linear functional $\lambda: V\rightarrow\mathbb{C}$. I understand that such a linear functional yields a global section, but could I ask about the converse? Of course, I am guessing that holomorphicity of the section must be used somehow, because for smooth, complex bundles, it is false that the $\mathbb{C}$-vector space of global sections is finitely generated, right? $\endgroup$ Commented Nov 4 at 15:58
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Let $U_i = \{ z_i \neq 0 \} = \{ z_i=1 \}$. By defintion we have that $\Gamma(U_i, \mathcal O_{\mathbb P^n}(1))$ is the degree $1$-part of the localization $k[z_0, \dotsc, z_n]_{z_i}$, i.e. the sections are of the form $$\frac{f(z_0, \dotsc, z_n)}{z_i^{\deg f -1}}.$$

After letting $z_i=1$, this is nothing else but an arbitrary polynomial in the $z_j$'s with $j \neq i$.

If you want to glue such sections to a global section, you need to make sure that there are no denominators, i.e. $\deg f=1$. This - along the lines of your question - explains why the global sections are given by linear polynomials.

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