The hyperplane line bundle of $\mathbb P^n$ is just the dual bundle of the tautological line bundle $\mathcal{O}(1)$. It can also be defined through hyperplanes:
Let $[z_0: \ldots, z_n]$ be the homogeneous coordinates of $\mathbb P^n$. Write $U_i = {z_i \neq 0}$, a hyperplane is a set of the form $\{a \cdot z = a_0 z_0 + \ldots + a_n z_n\}$, where $a_i$ are constants and at least one of them is not zero. Let us denote it by $H_{\alpha}$. Since $\frac{a \cdot z}{b \cdot z}$ defines a global meromorphic function, $H_{a} = H_{b}$.
I am confused as to how does $H_{a}$ give a line bundle.
Best Answer
The fact that $\frac{a\cdot z}{b\cdot z}$ is meromorphic shows that $H_a$ and $H_b$ are linearly equivalent as divisors on $\mathbb{P}^n$ and that therefore every hyperplane divisor is linearly equivalent to $H = \{ z_0 = 0 \}$. Now, the hyperplane line bundle is the associated line bundle to the locally free sheaf $\mathcal{O}(H)$, which is defined by: $$ U \subset \mathbb{P}^n \mapsto \{ f \in K \ | \ (f) + H \geq 0 \ \text{on} \ U\} $$ where $K$ is the field of rational functions on $\mathbb{P}^n$. In other words, the hyperplane H (or any hyperplane $H_a$) defines a sheaf whose functions on an open set $U$ are those functions in $K$ that have at worst poles of order 1 on H. This sheaf has rank 1 and hence corresponds to a line bundle.