# Definition of the hyperplane bundle in algebraic geometry

Source: Smith et al., Invitation to Algebraic Geometry, Section 8.4 (pages 131 - 133).

I have a very limited background in algebraic geometry and I'm trying to understand line bundles. In the book they define the tautological bundle as follows:

The tautological bundle over $\mathbb{P}^n$ is constructed as follows. Consider the incidence correspondence of points in $\mathbb{C}^{n+1}$ lying on lines through the origin, $B = \{(x, \ell) \;|\; x \in \ell \} \subseteq \mathbb{C}^{n+1} \times \mathbb{P}^n$, together with the natural projection $\pi : B \rightarrow \mathbb{P}^n$. [...] The tautological bundle over the projective variety $X \subseteq \mathbb{P}^n$ is obtained by simply restricting the correspondence to the points of $X$...

Later on in the text they define the hyperplane bundle as

The hyperplane bundle $H$ on a quasi-projective variety is defined to be the dual of the tautological line bundle: The fiber $\pi^{-1}(p)$ over a point $p \in X \subset \mathbb{P}^n$ is the (one-dimensional) vector space of linear functionals on the line $\ell \subset \mathbb{C}^{n+1}$ that determines $p$ in $\mathbb{P}^n$. The formal construction of $H$ as a subvariety of $(\mathbb{C}^{n+1})^\ast \times \mathbb{P}^n$ parallels that of the tautological line bundle.

I can't see how to put a variety structure on the line bundle?

• Can't you find equations for a pair (x,L) to be in the set B in terms of the coordinates of x and if L? – Mariano Suárez-Álvarez Jun 27 '17 at 8:53
• Oh yes, that works for the tautological bundle. Sorry but how does one do this for the hyperplane bundle? – abcdef Jun 27 '17 at 9:24

The construction is exactly the same, taking the couples $(\phi,p)$ where $p \in \Bbb P(V)$ and $\phi : p \to \Bbb C$ is a linear functional.
If you know that line bundle are determined by cocycles, then compute the cocycles of $\mathcal O(-1)$ ( this is a notation for tautlogical line bundle), and the inverse cocycles will be the cocycles of H, this will gives you all the information you need about $H$.
Remark : this bundle is named like this because any $\mu \in V^* \backslash \{0\}$ create a section of our bundle, namely $\sigma : \Bbb P(V) \to H, L \mapsto \mu_{|L}$. Notice that the kernel of $\sigma$ is an hyperplane, hence the name. On the other hand, there is no algebraic regular section $\sigma : \Bbb P^n \to \mathcal O(-1)$.
• Sure, a vector bundle with fiber $V$ over $X$ is locally isomorphic to $U \times V$. We can cover $X$ by such open, and then we obtain isomorphism $h_i : \pi^{-1}(U_i) \to U_i \times V$. Composing such isomorphism on overlap $U_i \cap U_j$ gives a morphism $g_{ij} : U_i \cap U_j \to GL(V)$, indeed we are going from $U_i \cap U_j \times V \to U_i \cap U_j \times V$, nothing happens on the $X$ component, but the vector might change. This map $g_{ij}$ contains all the information about the vector bundle, and if $h_{ij}$ are the cocycles for the dual bundle then $h_{ij} = g_{ij}^{-1}$. – user171326 Jun 27 '17 at 11:11
• @abcdef : I don't know, because usually people consider it as a sheaf more than a space. and the tautological bundle has a very nice description since it's the blow-up of $\Bbb C^n$. I believe that $H$ is the complement of a point in $\Bbb P^{n+1}$ but I'm not sure. But don't try to think about it as a variety but more like a vector bundle, i.e the data of a vector space over each point + a way how they are moving. – user171326 Jun 27 '17 at 11:28
• Here is the proof : mathoverflow.net/questions/45116/… but again I think if you discover this for first time,maybe it's better just to accept that it's a variety and try to do some simple computations with $H$ and the tautological bundle. – user171326 Jun 27 '17 at 11:32