The base field is $\mathbb{C}$ throughout the question. Consider the projective space $\mathbb{P}^n$. It corresponds to a principal $\mathbb{C}^*$-bundle
$$ \mathbb{C}^* \to \mathbb{C}^{n+1}-0 \to \mathbb{P}^n,$$
where $\mathbb{C}^*$ acts on $\mathbb{C}^{n+1}-0$ by multiplication
$$ t\cdot(z_0,\dots,z_n) = (tz_0,\dots,tz_n).$$
Now fix an integer $d$. We can change the fiber $\mathbb{C}^*$ to $\mathbb{C}$ by considering the action of $\mathbb{C}^*$ on $\mathbb{C}$,
$$ t \cdot z = t^d z.$$
Then we obtain a line bundle $\left(\mathbb{C}^{n+1}-0\right) \times_{\mathbb{C}^*} \mathbb{C}$ over $\mathbb{P}^n$, where $\mathbb{C}^*$ acts on $\left(\mathbb{C}^{n+1}-0\right) \times \mathbb{C}$ by
$$t \cdot ((z_0, \dots, z_n),z) = ((tz_0, \dots, tz_n), t^d z). $$
Is the line bundle $\left(\mathbb{C}^{n+1}-0\right) \times_{\mathbb{C}^*} \mathbb{C} \cong \mathcal{O}(d)?$
The following is my attempt to identify $\left(\mathbb{C}^{n+1}-0\right) \times_{\mathbb{C}^*} \mathbb{C}$ with $\mathcal{O}(d)$. I use the standard open cover $\{U_i\}_{i=0,\dots,n}$ of $\mathbb{P}^n$ (i.e., $x_i \neq 0$ on $U_i$) and trivialize the line bundle over each $U_i$ as follows:
$$
\begin{aligned}
\left( \left(\mathbb{C}^{n+1}-0\right) \times_{\mathbb{C}^*} \mathbb{C} \right) |_{Ui} & \to U_i \times \mathbb{C} \\
[(z_0, \dots, z_n),z] & \mapsto ([z_0, \dots, z_n], z_i^{-d}z).
\end{aligned}
$$
This is well defined because another representative $((tz_0,\dots,tz_n),t^d z)$ maps to the same element.
The transition function from $U_i$ to $U_j$ is then
$$
\begin{aligned}
g_{ji}: U_i \cap U_j & \to \mathbb{C}^* \\
[z_0, \dots, z_n] & \mapsto (z_j/z_i)^{-d}.
\end{aligned}
$$
So these $g_{ji}$'s are transition functions for $\mathcal{O}(d)$.
Does everything above look right?
Best Answer
I made a sign error. The diagonal action of $\mathbb{C}^*$ on $\left(\mathbb{C}^n-0\right) \times \mathbb{C}$ should be $$ t \cdot ((z_0, \dots, z_n),z) = ((t^{-1}z_0,\cdots, t^{-1}z_n),t^{d} z) .$$
So the line bundle $\left(\mathbb{C}^n-0\right) \times_{\mathbb{C}^*} \mathbb{C}$ is actually $\mathcal{O}(-d)$. The original argument works with $d$ replaced by $-d$ everywhere.