# How does a hyperplane become a linear bundle?

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?

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How a hyperplane can seem as a divisor? – user42912 May 27 '15 at 6:05

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 http://people.fas.harvard.edu/~amathew/linebund.pdf. 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).

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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 math.stackexchange.com/questions/1926/… (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