Computing the curvature of $\mathcal{O}(-1)$ from its pull back to $\mathbb{C}^{n+1}\setminus \{0\}$ I saw in different places a way to compute the curvature of $\mathcal{O}(-1)\to \mathbb{P}^n$ by pulling it back by the projection $\pi:\mathbb{C}^{n+1}\setminus \{0\}\to \mathbb{P}^n$.
As a matter of fact, the line bundle $L:=\pi^*\mathcal{O}(-1)\to\mathbb{C}^{n+1}\setminus \{0\}$ is trivial, with a trivializing section $\sigma:z\mapsto z$. We can compute easily its curvature
$$\Theta_L(z)=\bar{\partial}\partial\log||z||^2=\bar{\partial}\partial\log \sum_{i=0}^n|z|^2.$$
I do not understand now how to relate that (despite its obvious resemblance) to the curvature of $\mathcal{O}(-1)\to \mathbb{P}^n$ which is given for example in the chart $U_0=\{[z_0:...:z_n]|z_0\neq 0\}$ by
$$\Theta_{\mathcal{O}(-1)}(z)=\bar{\partial}\partial\log (1+\sum_{i=1}^n|z|^2).$$
There are formulae for pull back curvatures but not for "pushforward" curvatures. I feel that this is quite simple but do not get how exactly.

$\newcommand{\dd}{\partial}\newcommand{\ddbar}{\bar{\dd}}$Let $(Z^{0}, Z^{1}, \dots, Z^{n})$ denote Cartesian coordinates. For each $j$, if we write $z_{j}^{i} = Z^{i}/Z^{j}$ on the open set where $Z^{j} \neq 0$, we have
$$
\log \sum_{i=0}^{n} |Z^{i}|^{2}
= \log \biggl[|Z^{j}|^{2} \sum_{i=0}^{n} |z_{j}^{i}|^{2}\biggr]
= \log|Z^{j}|^{2} + \log \biggl[1 + \sum_{i\neq j} |z_{j}^{i}|^{2}\biggr].
$$
The function $\log|Z^{j}|^{2} = \log Z^{j} + \log \bar{Z}^{j}$ is annihilated by $\dd\ddbar$; taking $j = 0$ and writing $z^{i} = z_{0}^{i}$ we have
$$
\ddbar\dd\log \sum_{i=0}^{n} |Z^{i}|^{2}
= \ddbar\dd\log \biggl[1 + \sum_{i=1}^{n} |z^{i}|^{2}\biggr].
$$
