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?

  • 1
    $\begingroup$ 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? $\endgroup$ – Mariano Suárez-Álvarez Jun 27 '17 at 8:53
  • $\begingroup$ Oh yes, that works for the tautological bundle. Sorry but how does one do this for the hyperplane bundle? $\endgroup$ – 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 $\sigma : \Bbb P^n \to \mathcal O(-1)$.

  • $\begingroup$ Thanks for taking the time, but I'm not familiar with cocyles. Isn't there an easy explanation for it? $\endgroup$ – abcdef Jun 27 '17 at 11:05
  • $\begingroup$ 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 $h_{ij} = g_{ij}^{-1}$. $\endgroup$ – user171326 Jun 27 '17 at 11:11
  • $\begingroup$ Thanks, maybe I didn't make it precise in my question. But could you also explain how a closed set in the hyperplane bundle looks like in terms of algebraic equations? $\endgroup$ – abcdef Jun 27 '17 at 11:18
  • $\begingroup$ @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. $\endgroup$ – user171326 Jun 27 '17 at 11:28
  • $\begingroup$ 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. $\endgroup$ – user171326 Jun 27 '17 at 11:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.