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I always confused by whether tautological bundle is $\mathcal{O}(1)$ or $\mathcal{O}(-1)$, and definitions from different sources tangled in my brain. However, I thought this might not be simply a matter of convention.

Let $\mathbb{P}^n$ be a projective space of dimensional $n$, if we realize a point $[l]$ in $\mathbb{P}^n$ as a line $l \subset \mathbb{C}^{n+1}$ passing through origin. Then the tautological bundle $S$ of $\mathbb{P}^n$ is defined as a subbundle of $\mathbb{P}^n\times \mathbb{C}^{n+1}$ by

$$[l] \times l \subset [l] \times \mathbb{C}^{n+1}$$

In many books, the convention is $S \cong \mathcal{O}(-1)$ on $\mathbb{P}^n$. Here $\mathcal{O}(-1)$ is defined as it is in Hartshorne which is the sheaf of modules associated to $\mathbb{C}[x_0,\dots,x_n](-1)$ (for example $x_i^{-1}$ is degree $0$ element in $\mathbb{C}[x_0,\dots,x_n](-1)$). I know something need to be clarified in $S \cong \mathcal{O}(-1)$: for $S$, I mean the sheaf associated to the tautological line bundle S. So it seems the problem becomes to show $S$ does not have global sections (because $\mathcal{O}(-1)$ is different from $\mathcal{O}(1)$ by does not have global sections)?

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2 Answers 2

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As a complement to Matt's fine answer let me explain why $\mathcal O(-1)$ has only zero as global section.

A section $s\in \Gamma(\mathbb P^n,\mathcal O(-1))$ is in particular a section of the trivial bundle $\mathbb P^n \times \mathbb C^{n+1}$, so that it is of the form $s(x)=(x,\sigma (x)) $ with $\sigma:\mathbb P^n \to \mathbb C^{n+1}$ a regular map.
But such a map $\sigma$ is a constant, since any regular map $\mathbb P^n \to \mathbb C$ is constant by completeness of $\mathbb P^n$.
So $\sigma (x)=v\in \mathbb C^{n+1}$, a fixed vector independent of $x$.
However for $x=[l]$, we must have $\sigma (x)=v\in l$.
In other words, that constant vector $v\in \mathbb C^{n+1}$ must lie on all lines $l\subset \mathbb C^{n+1}$, which forces $v=0$ .
We have thus proved that $$\Gamma(\mathbb P^n,\mathcal O(-1))=0$$

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    $\begingroup$ I know this answer is very old now, but I've got a question. A map $\mathbb{P}^n \rightarrow \mathbb{C}$ is constant because $\mathbb{P}^n$ is compact right? But I don't see why this would mean $\sigma(x)$ is a fixed vector in $\mathbb{C}^{n+1}$. What I understood is that $\sigma(x)$ where $x = [l]$ must be parallel to the vector $l$, so we can write $\sigma([l]) = s([l])l$ for some $s:\mathbb{P}^n \rightarrow \mathbb{C}$. Hence $\sigma$ will varies in $\mathbb{C}^{n+1}$ from point to point. $\endgroup$
    – user113988
    Oct 21, 2016 at 1:43
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    $\begingroup$ Then for the proof, can I just say that $-s([l])l = \sigma([-l]) = \sigma([l]) = s([l])l$ hence $s([l]) = -s([l]), \forall [l] \in \mathbb{P}^n \implies s([l]) = 0$ so there is no non-zero global section? Sorry for picking up on such an old answer (that maybe clear to most people already) but I always feel like the proof suggests that we can't have a non-zero global section for a line bundle over any compact space. I tried follow the proof with $\mathbb{S}^n$ instead and $\mathbb{S}^n$ normal line bundle should have a non-zero global section. $\endgroup$
    – user113988
    Oct 21, 2016 at 1:47
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    $\begingroup$ Actually $\sigma([l]) = s(l/|l|)l$ is probably more correct. Then $s:\mathbb{S}^n \rightarrow \mathbb{C}$ is constant for $\mathbb{S}^n$ is compact. For the global section to be well-defined we need $\sigma([l]) = \sigma([-l])$ so $s(-l/|l|) = -s(l/|l|)$. Therefore $s = 0$ identically since it is constant. Please correct me if I misunderstood something. Thank you :) $\endgroup$
    – user113988
    Oct 21, 2016 at 1:59
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    $\begingroup$ @user113988. What you write is incorrect because $S^n$ is not an algebraic manifold nor even a complex manifold. Your argument that $s$ is constant because $S^n$ is compact is false since it only applies to complex manifolds, and indeed there are many non constant smooth functions $S^n\to \mathbb C$. $\endgroup$
    – Georges Elencwajg
    Oct 21, 2016 at 5:10
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    $\begingroup$ Obviously I forgot that $\mathbb{S}^n$ is not a complex manifold. Now it makes sense, thank you! $\endgroup$
    – user113988
    Oct 21, 2016 at 5:45
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The line bundle $\mathcal O(1)$ has as global sections the functions $x_0,\ldots,x_n$ which give coordinates on $\mathbb C^{n+1}$; these are not vectors in $\mathbb C^{n+1}$ but in its dual.

The tautological bundle is as you described, and the elements of its fibres are vectors in $\mathbb C^{n+1}$. Thus its sheaf of sections is dual to $\mathcal O(1)$, and so equals $\mathcal O(-1)$.

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