In Huybrechts' text Complex Geometry, I am told that any holomorphic line bundle over $\mathbb{CP}^n$ is of the form $\mathcal{O}(k)$ for some $k\in\mathbb{Z}$, where
\begin{equation}
\mathcal{O}(k)=\mathcal{O}(1)^{\otimes k}
\end{equation}
for $k>0$ and
\begin{equation}
\mathcal{O}(k)=\mathcal{O}(-1)^{\otimes -k}
\end{equation}
for $k<0$. Here $\mathcal{O}(1)$ is the hyperplane line bundle and $\mathcal{O}(-1)$ is its dual, the tautological line bundle (although I think Huybrechts calls these the other way round).
In my current work, I am looking at Hirzebruch surfaces, which are defined as holomorphic line bundles over $\mathbb{CP}^1$ in the following way: for the $n$'th Hirzebruch surface $H_n$, take the rank 2 bundle $\mathcal{O}(n)\oplus\mathcal{O}(0)$ over $\mathbb{CP}^1$, and projectivise it to obtain a rank 1 bundle
\begin{equation}
H_n=\mathbb{P}(\mathcal{O}(n)\oplus\mathcal{O}(0))
\end{equation}
over $\mathbb{CP}^1$. I am wondering why the Hirzebruch surfaces are defined in this way, if they are simply isomorphic to some $\mathcal{O}(k)$ by the above classification of holomorphic line bundles over complex projective space? Or am I missing something?
Thanks
Best Answer
The projectivisation of a complex rank $2$ vector bundle is not a complex line bundle, but rather a $\mathbb{CP}^1$ bundle (just as the projectivisation of $\mathbb{C}^2$ is not $\mathbb{C}$, but rather $\mathbb{CP}^1$).