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?



2 Answers 2


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 $\mathbb{CP}^1$).

  • $\begingroup$ I must be getting confused with terminology here - I understand a line bundle to be a bundle of rank 1, i.e. each fibre has complex dimension one. In particular, a $\mathbb{CP}^1$ bundle has copies of $\mathbb{CP}^1$ as each fibre, which are of complex dimension one. So by my definition of a line bundle, these are line bundles. What's going wrong here? $\endgroup$
    – zmy
    Commented Aug 8, 2018 at 11:50
  • $\begingroup$ Ah, I think my problem is that a $\mathbb{CP}^1$ bundle is not a vector bundle. For some reason I've been thinking of these projective fibres as vector spaces, which they're obviously not. $\endgroup$
    – zmy
    Commented Aug 8, 2018 at 11:52
  • 2
    $\begingroup$ That's right. The fiber should be $\mathbb{C}$. $\endgroup$ Commented Aug 8, 2018 at 11:52

$\newcommand{\Proj}{\mathbf{P}}\newcommand{\Ocal}{\mathcal{O}}$For posterity, as a complement to Michael's (+1) answer:

If $p:E \to M$ is a holomorphic vector bundle of positive rank $k$, and if $\Ocal$ denotes the trivial line bundle over $M$, then

  • $\Proj(E \oplus \Ocal)$ is a holomorphic $\Proj^{k}$-bundle over $M$;
  • The total space of $E$ is naturally viewed as a dense open subset in $\Proj(E \oplus \Ocal)$, the image of $E \oplus \{1\}$ under projectivization. Geometrically, $\Proj(E \oplus \Ocal)$ is obtained from $E$ by "adding points at infinity in each fiber" in the same way $\Proj^{k}$ is obtained from affine $k$-space by adding a hyperplane at infinity.
  • The divisor at infinity is the image of $E \oplus \{0\}$ under projectivization, and is identified with the $\Proj^{k-1}$-bundle $\Proj(E)$.
  • If $E = L$ is a line bundle, as for the Hirzebruch surfaces, we've "completed" the total space by adding one point at infinity in each fiber, obtaining a $\Proj^{1}$-bundle. In this situation, the infinity section of $\Proj(L \oplus \Ocal)$ is also a copy of $M$. The normal bundle of the zero section is $L$, and the normal bundle of the infinity section is the dual/inverse $L^{-1}$. Formally, a point $\zeta$ in a fiber $L_{z}$ is identified with $1/\zeta$ in the fiber $L_{z}^{-1}$.
  • Blowing up the zero section of $E$ gives the total space of the tautological bundle $\tau_{E} \to \Proj(E)$, and commutes with completion ("because blowing up is local"). Whether or not $E$ is a line bundle, the normal bundle of the divisor at infinity in $\Proj(E \oplus \Ocal)$ is naturally viewed as the dual of the tautological line bundle $\tau_{E}$.

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