Why $\mathcal{O}_{\mathbb{P}^n}(1)$ is a line bundle?

Question

Why $\mathcal{O}_{\mathbb{P}^n_{\mathbb{C}}}(1)$ is a line bundle? In the book of Hartshorne, $\mathcal{O}_{\mathbb{P}^n_{\mathbb{C}}}(1)$ is defined by $\mathcal{O}_{\mathbb{P}^n_{\mathbb{C}}}(1) = \tilde{S(1)}$, where $S=\mathbb{C}[x_0, \ldots, x_n]$.

The line bundle is defined as follows. Given a complex manifold $X$, a line bundle is given by the data of a cover $U_i$ ($i \in I$ some index set), and trivializations $U_i \times \mathbb{C}$, and transition functions $f_{ij}$ on $U_i \cap U_j$ that is an analytic function, nowhere zero, satisfying $f_{ij} f_{jk} = f_{ik}$ (the cocyle condition), from which $f_{ii} = 1$ and $f_{ij} = f_{ji}$. That is, if $u \in U_i \cap U_j$, then the point $(u, z) \in U_j \times \mathbb{C}$ is identified with the point $(u, f_{ij} z) \in U_i \times \mathbb{C}$.

How to show that $\mathcal{O}_{\mathbb{P}^n}(1)$ is a line bundle using this definition? Thank you very much.

Answer 1

Strictly speaking, $\mathcal{O}(1)$ is not a line bundle but a sheaf. You can construct a rank 1 vector bundle from it: locally, on $D_+(T_i)$, $\mathcal{O}(1)$ is isomorphic to $\mathcal{O}_{\mathbb{P}^n}$ via multiplication with $T_i^{-1}$ and these local isomorphisms and the cover $D_+(T_i)$ define a line bundle.

Answer 2

Question: "How to show that OPn(1) is a line bundle using this definition? Thank you very much."

Answer: Let $S:=k[x_0,..,x_n]:=\oplus_{d \geq 0} k[x_0,..,x_n]_d:=\oplus_{d \geq 0}S_d$ and let $S(l)$ be the twist of $S$ by an integer $l$. By definition

$$S(l)_d:=S_{l+d}:=k[x_0,..,x_n]_{l+d}.$$

When you sheafify $S(l)$ you get an invertible sheaf $\mathcal{O}(l)$ on $\mathbb{P}^n$. It has the property that

$$I1.\text{ }\mathcal{O}(l)(D(x_i)) \cong k[\frac{x_0}{x_i},. ,\overline{\frac{x_i}{x_i}}.,\frac{x_n}{x_i}]x_i^l:=A_i x_i^l$$

is a free $\mathcal{O}(D(x_i))$ module of rank one on the element $x_i^l$. Here $D(x_i) \subseteq \mathbb{P}^n$ is the open subscheme where "$x_i \neq 0$". There is an isomorphism $D(x_i) \cong \mathbb{A}^n$. Hence the sheaf of $\mathcal{O}$-modules $\mathcal{O}(l)$ is locally free of rank $1$.

Let $z_i:=x_i^{-l}$ and consider the "symmetric algebra"

$$1. Sym^*_{A_i}(A_i z_i) \cong A_i[z_i].$$

It follows $A_i[z_i]$ is a polynomial ring in $z_i$ over $A_i$ hence $Spec(A_i[z_i]) \cong D(x_i) \times \mathbb{A}^1$. If we consider the "linebundle"

$$\pi: \mathbb{V}(\mathcal{O}(l)^*) \rightarrow \mathbb{P}^n$$

it follows $\pi^{-1}(D(x_i)) \cong D(x_i) \times \mathbb{A}^1$

and on the open set $D(x_ix_j)$ it follows there is an equality

$$ z_i =(\frac{1}{x_i})^l =(\frac{x_j}{x_i})^l(\frac{1}{x_j})^l =(\frac{x_j}{x_i})^l z_j $$

hence the transition function $f_{ij}:=(\frac{x_j}{x_i})^l$ (which is rational) is analytic on the open set $D(x_ix_j)$.

A proof of $I1$ is found in Hartshorne, Prop.II.5.11b.

Note: If $X$ is a scheme and $E$ is a locally free finite rank sheaf on $X$, we define the "geometric vector bundle" of $E$ to be the scheme

$$ \mathbb{V}(E^*):= Spec(Sym^*_{\mathcal{O}_X}(E^*).$$

There is a canonical morphism of schemes

$$\pi: \mathbb{V}(E^*) \rightarrow X$$

and $E$ is canonically the sheaf of sections of $\pi$.