Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

As we know,a hyperplane can seem as a divisor,and a divisor can become a linear bundle,I want to know what the structure of linear bundle is. For example, the hyperplane is given by $a_0 z_0+a_1 z_1+\ldots+a_n z_n=0$,what is the projective map、transform function and so on?

share|cite|improve this question
How a hyperplane can seem as a divisor? – user42912 May 27 '15 at 6:05
up vote 6 down vote accepted

I'm going to assume your question is in the context of algebraic geometry. First, if we are working over affine space, then the hyperplane is cut out by a global function, so the divisor is principal; in particular, this line bundle is trivial. (In fact, any vector bundle over affine space is trivial, though this is a hard theorem, the Quillen-Suslin theorem.)

The question is more interesting when one is working with projective space $\mathbb{P}^n_k$ over a field $k$. In that case, one obtains the standard line bundle $\mathcal{O}(1)$. See section 4.3 of The basic idea is that given a line bundle, one obtains the associated Weil divisor by picking a rational section and taking its divisor. So there is a global section $x_0$ of $\mathcal{O}(1)$ (which is basically the structure sheaf twisted by a homogeneous degree), and where this section does not vanish (and does not vanish with multiplicity one) is just a hyperplane.

In fact, since one can show directly that the Weil class group of $\mathbb{P}^n_k$ is isomorphic to $\mathbb{Z}$ (any hypersurface of degree $d$ being equivalent to $d$ times the hyperplane $\{x_0 = 0\}$, as any homogeneous polynomial of degree $d$ divided by $x_0^d$ is a rational function on $\mathbb{P}^n$), so the Picard group of line bundles on $\mathbb{P}^n$ is precisely $\mathbb{Z}$, generated by a copy of $\mathcal{O}(1)$.

The line bundle $\mathcal{O}(1)$ is the dual of the so-called tautological line bundle over $\mathbb{P}^n$ consisting of pairs $(\ell, p)$ where $\ell$ is a line and $p$ is a point in that line (this is more basic in the topological category).

share|cite|improve this answer
Nice answer! I seem to remember someone asking, here or on MO, about how to go from a divisor to a bundle and back etc. Do you remember seeing this? if so could you point me in the direction or rather a source for the argument? – Sean Tilson Feb 25 '11 at 23:31
@Sean: Dear Sean, thanks! You may see… (this is in Hartshorne, among other places). – Akhil Mathew Feb 26 '11 at 2:25
Perfect!I know a little about algebraic geometry and get the question in the complex geometry,then I think if the bottem space has no hole just like a sphere,the linear bundle is always trivial.Maybe I ask a silly question. – Strongart Feb 26 '11 at 5:13
@Strongart: Dear Strongart, in fact, complex line bundles on $S^2$ even in the topological category are nontrivial (they are classified by $H^2(., \mathbb{Z})$). In the topological category, line (or more generally vector) bundles over contractible spaces are trivial. – Akhil Mathew Feb 26 '11 at 5:27
Thanks Akhil! For some reason I have not been able to find my copy of Hartshorne for months... :( (sorry for accidentally posting this as an answer) – Sean Tilson Feb 26 '11 at 20:25

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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