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)?


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 $$\Gamma(\mathbb P^n,\mathcal O(-1))=0$$

  • $\begingroup$ 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 '16 at 1:43
  • $\begingroup$ 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 '16 at 1:47
  • $\begingroup$ 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 '16 at 1:59
  • $\begingroup$ @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 '16 at 5:10
  • $\begingroup$ Obviously I forgot that $\mathbb{S}^n$ is not a complex manifold. Now it makes sense, thank you! $\endgroup$ – user113988 Oct 21 '16 at 5:45

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 $\mathcal O(-1)$.


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.