# How is the hyperplane bundle cut out of $(\mathbb{C}^{n+1})^\ast \times \mathbb{P}^n$?

[Question has been updated with more context and perhaps a better explanation of my question.]

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

First, a brief set-up, whose purpose will become obvious in a minute. Note that everything here is over $\mathbb{C}$.

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 $X$...

Next, it's shown that this bundle has no global sections, in the case $X = \mathbb{P}^1$:

A global section of the tautological bundle defines, for each point $p \in \mathbb{P}^1$,
a point $(a(p), b(p)) \in \mathbb{C}^2$ lying on the line through the origin corresponding to $p$. Since the assignment $p \mapsto (a(p), b(p))$ must be a morphism, we see that projecting onto either factor, we have morphisms $a, b : \mathbb{P}^1 \rightarrow \mathbb{C}$. But because $\mathbb{P}^1$ admits no nonconstant regular functions, both $a$ and $b$ are constant functions. But then both are zero...

So far, so good. Now:

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.

This, I don't understand. Specifically, whereas the set of $v \in \mathbb{C}^{n+1}$ such that $v \in \ell$ was a subspace of $\mathbb{C}^{n+1}$, the set of linear functionals $f : \ell \rightarrow \mathbb{C}$ appears to be a quotient of $(\mathbb{C}^{n+1})^\ast$, rather than a subspace. So I don't see how to perform the parallel construction here.

In particular, any method that cuts something out of $(\mathbb{C}^{n+1})^\ast \times \mathbb{P}^n$ would seemingly allow us to carry out the argument above and show that the hyperplane bundle has no global sections, which is false.

• er...couldn't you just take functionals which vanish on $l^\perp$ (I guess this involves choosing an inner product, but you're already working with a preferred basis..) – uncookedfalcon Jun 1 '12 at 1:53
• hmm, good point. – Daniel McLaury Jun 1 '12 at 2:02
• but there's probably a coordinate free way to do this...and that would probably tell us why $H$ is a dual more than point by point (i.e. in what category) – uncookedfalcon Jun 1 '12 at 2:07
• Well, the coordinate-free construction might not be a subvariety of $(\mathbb{C}^{n+1})^\ast \times \mathbb{P}^n$ in a natural way (whatever that means). – Daniel McLaury Jun 1 '12 at 2:11
• Dear Daniel, you are calling $\pi$ the projection morphism of the two different bundles $B,H$ . You must not do that: you might call $\rho$ that from $B$, if you don't want to change your highlighted quotation (from where, by the way?) – Georges Elencwajg Jun 1 '12 at 7:24

Let's be more canonical: call$V$ the vector space such that $\mathbb P^n=\mathbb P(V)$.
Given a point $p\in W$ (where $W\subset \mathbb P^n$ is your subvariety), you may consider the linear forms $\phi:l\to \mathbb C$ on the line $l$ corresponding to $p$.
These linear forms constitute the fiber at $p$ of the bundle $B^*$ dual to $B$: $\phi\in B^*[p]$.

The terminology hyperplane bundle is due to the fact that if you consider the same construction on $\mathbb P^n$ you get a bundle $H$ whose global sections are identified to $V^*$ i.e. $\Gamma (\mathbb P^n, H)=V^*$, so that the zero set of such a $\Phi\in V^*$ is a hyperplane $\mathbb P(Ker(\Phi))\subset \mathbb P^n$.
The restricted bundle $H\mid W$ is then the bundle you are interested in : $H\mid W=B^*$

Finally note that you have a canonical vector space morphism of global sections $\Gamma (\mathbb P^n, H)=V^*\to \Gamma (W, B^*)$ which is neither injective nor surjective in general.

Edit
Lat me address a subtle point raised by Daniel.
The trivial bundle $V^*\times \mathbb P^n$ and $H$ have the exact same vector space of global sections, namely $V^*$. But are they equal? Of course not: the first bundle has rank $n+1$ and the second one rank $1$.
All right, what then is the fiber of $H$ at $p$? Some dimension $1$ subspace of $V^*$ maybe?
Not at all! It is $l^*$, where $l$ is the line that $p$ represents and, as Daniel correctly notes, $l^*$ is a quotient of $V^*$, not a subspace.
Since we do not have $H\subset V^*\times \mathbb P^n$, we cannot apply the reasoning that led to the conclusion that the tautological vector bundle has only the zero section and the paradox vanishes.

To sum up, we have a canonical surjective (but not injective) morphism of vector bundles on $\mathbb P^n$ $$V^*\times \mathbb P^n\to H\to 0$$ which induces an isomorphism of ordinary $\mathbb C$-vector spaces $$\Gamma( \mathbb P^n,V^*\times \mathbb P^n)=V^*\stackrel{\cong}{\to} \Gamma(\mathbb P^n,H)$$

• I understand that this is the idea of the hyperplane bundle, but I don't understand how the text is suggesting I cut the hyperplane bundle out of $(\mathbb{C}^{n+1})^\ast \times \mathbb{P}^n$. I'll add a little more to the question to maybe make my confusion a little more clear. – Daniel McLaury Jun 2 '12 at 3:13
• I've updated the question now. In particular, it appears to me that if the total space of the hyperplane bundle could be cut out of $(\mathbb{C}^{n+1})^\ast \times \mathbb{P}^n$, then it would have no global sections, for the same reason that the tautological bundle does not. Can you help me out here? – Daniel McLaury Jun 2 '12 at 3:28
• Dear Daniel, I have tried to address your comments in an Edit. – Georges Elencwajg Jun 2 '12 at 8:32
• To clarify, is the textbook in fact incorrect when it says that $H$ can be constructed as a subvariety of $(\mathbb{C}^{n+1})^\ast \times \mathbb{P}^n$? – Daniel McLaury Jun 2 '12 at 9:05
• (This seems to be what you've said above, and is what I was beginning to suspect when I posted this question, but I just want to make sure that I'm not missing something here as this is all very new to me.) – Daniel McLaury Jun 2 '12 at 9:14