Hermitian metric on $L=\mathcal{O}_{\mathbb{P^n}}(1)$ over the projective space

Question

Consider the line bundle $L=\mathcal{O}_{\mathbb{P^n}}(1)$ over the projective space. Locally is descriebd by $\{U_a,g_{ab}\}$ where $U_a=\{z_a\neq0\}$ is the standard covering of the projective space.

A Hermitian metric on $L$ can be given locally as a smooth function $h_a:U_a\rightarrow\mathbb{R_{>0}}$, satisfying $h_a =|g_{ba}|^2 h_b.$

My question is: how can we define $h_a$ locally? For example if we choose $h_a=\frac{1}{|z_a|^2}$ ,which satisfies $h_a =|g_{ba}|^2 h_b$, then $h_a$ is not well-defined on the homogeneous coordinates.

I would appreciate any help on this.

Answer 1

The line bundle $L$ is given by the data $\{U_{\alpha\beta}, g_{\alpha\beta}\}$, where $\alpha, \beta = 0, 1, \cdots, n$,

$$U_\alpha = \{ z_\alpha \neq 0\}\ \ \text{ and }\ \ g_{\alpha\beta} = \frac{z_\beta}{z_\alpha}.$$

If we set $h_\alpha : U_\alpha \to \mathbb R_{>0}$, where

$$h_\alpha = \frac{|z_\alpha|^2}{|z_0|^2 + \cdots + |z_n|^2},$$

then $h_\alpha$ is well defined and $h_\alpha |g_{\alpha\beta}|^2= h_\beta$. Thus it's a Hermitian metric on $L$.

Note that this metric is the most natural one on $L$: First we consider $L^*$, which can be described as the tautological line bundle

$$L^* = \{ (\ell, v) \in \mathbb P^n \times \mathbb C^{n+1} : v\in \ell\}.$$

Notice that $L^*$ has a natural Hermitian metric $\tilde h$: on each fiber $\ell \subset \mathbb C^{n+1}$, we restrict the Hermitian metric of $\mathbb C^{n+1}$ to $\ell$. So if on each $U_\alpha$,

$$e_\alpha = \left( \frac{z_0}{z_\alpha}, \frac{z_1}{z_\alpha}, \cdots, 1, \cdots \frac{z_n}{z_\alpha}\right)$$

is a basis for $L^*$ and

$$\tilde h_\alpha = \tilde h(e_\alpha, e_\alpha) = \frac{|z_0|^2 + \cdots + |z_n|^2}{|z_\alpha|^2}.$$

Thus $h_\alpha = \tilde h_\alpha^{-1}$ is a Hermitian metric on $L$.