I was reading about the Hirzebruch surfaces construction to understand a few examples of Delzant construction, and need some help. I am using this reference Hirzebruck surfaces and can't understand the followinga few things about the final step: $P(L_{-n} \oplus \bar{\mathbb{C}}) $.

I understand the construction of $L_{-n}$ but is that a usual direct sum or what? Can't understand what is $\bar{\mathbb{C}}=X \times \mathbb{C}$ the "trivial complex line bundle", of what space? If someone can answer this questions perfect, or also a reference for me to read on my own would also be nice. Thanks!

  • $\begingroup$ $X \times \mathbb{C}$ is the trivial complex line bundle over $X$. $\endgroup$ – Max Apr 12 '17 at 17:22

$\overline{\mathbb C}$ is the trivial line bundle over $X$. It's a line bundle over $X$ (in your case, $X = \mathbb P^1$). So the total space of $\overline{\mathbb C}$ is $X\times \mathbb C$, with the projection given by $\pi(x, z) = x$.

In your case, $X = \mathbb P^1$ and the Hirzebruch surfaces is by definition $$P(L_{-n} \oplus \overline{\mathbb C}).$$ Here $L_{-n}$ is a line bundle on $\mathbb P^1$, so $L_{-n} \oplus \overline{\mathbb C}$ is a vector bundle of rank two over $\mathbb P^1$, and $P$ is the associated projective bundle: that is for each $\ell \in \mathbb P^1$ (the base), the fiber is $P((L_{-n})_\ell\oplus \mathbb C)$, all lines in the vector space $(L_{-n})_\ell\oplus \mathbb C$. Thus the fiber is again holomorphic to $\mathbb P^1$.

  • $\begingroup$ Hi, thank you very much. Just to make it sure, that means that a every point $p \in \mathbb{P}^1$ we have all lines of vector space $(L_{-n} \oplus \mathbb{C})$, isn't it? But here $\oplus$ means something like product ( I haven't studied yet vector bundle's or its direct sum but I can try to get the idea )? That would mean that a general point of our total space $P(L_{-n} \oplus \mathbb{C})$ is of the form $( p, [ z: w ] )$ with $p \in \mathbb{P}^1$, $z\in L_{-n}$ and $w \in \mathbb{C}$, am I correct? $\endgroup$ – Reb Apr 12 '17 at 18:16
  • 1
    $\begingroup$ @Rob, yes.${}{}{}$. $\endgroup$ – user99914 Apr 12 '17 at 18:31

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