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Let $\mathbb{P}(E)$ be a projective bundle over some smooth projective variety $X$, defined over $\mathbb{C}$ for definiteness. Then this bundle is also a smooth projective variety.

Smoothness is clear from the trivialization, and it is also clear using the Segre embedding that every patch can be embedded in some projective space. Does is automatically follow that the entire bundle can be embedded in some projective space? One can definitely glue them to get a variety, by using the triple intersection rules, but is it necessarily projective?

EDIT: in Tyurin's Vector Bundles, one reads that a vector bundle over a complete variety is neither affine nor complete, but the projectivization is an actual projective variety. This is what I am wondering about.

Karsten
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  • $P(E)$ is by definition projective. Notice that the Hartshorne-definition "embeds into some P^n" is not the correct definition of projectiveness. The definitions are equivalent over projective bases, though. This is all spelled out in EGA II. – Martin Brandenburg Oct 15 '14 at 10:45
  • Yes I understand that, but in the setting I'm talking about it should still be provable right? – Karsten Oct 15 '14 at 10:47

1 Answers1

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A projective bundle $\mathbb{P}(\mathcal{E})$ over a smooth projective variety $X$ (over any base field $k$) is indeed a smooth projective variety.

Such a scheme $X$ is noetherian. By Exercise II.7.10. in Hartshorne, for a locally free sheaf of rank $n+1$ on $X$, its projectivization $\mathbb{P}(\mathcal{E})$ is always a $\mathbb{P}^{n}$-bundle over $X$, and conversely, since $X$ is also regular, every $\mathbb{P}^{n}$-bundle over $X$ arises in this way.

By definition (see here) the morphism $\pi \colon \mathbb{P}(\mathcal{E})\to X$ is projective in the sense of EGA. But $X$ admits an ample invertible sheaf, so in this case EGA-projective imlpies Hartshorne-projective (see the same reference a bit below). Since composition of Hartshorne-projective morphisms is Hartshorne-projective and the structure morphism $X\to \text{Spec}(k)$ is Hartshorne-projective, so is $\mathbb{P}(\mathcal{E})\to \text{Spec}(k)$ and therefore $\mathbb{P}(\mathcal{E})$ is a closed subset of some projective space over your base field $k$.

Smoothness of $\mathbb{P}(\mathcal{E})$ over $k$ follows as you say from the trivializations: smoothness is a local property and $\mathbb{P}(\mathcal{E})$ is locally a product of an open set of $X$ (smooth) and a projective $n$-space over $k$ (also smooth).

Irreducibility can be shown as follows: if $U$ is a trivializing open in $X$, by irreducibility of $X$, $U$ is also irreducible. Now $\pi^{-1}(U)$ is the product of two (irreducible) quasi-projective varieties over $k$, hence also irreducible. But in fact $\pi^{-1}(U)$ is dense in $\mathbb{P}(\mathcal{E})$, so we get that $\mathbb{P}(\mathcal{E})$ is irreducible.

To see that $\pi^{-1}(U)$ is dense in $\mathbb{P}(\mathcal{E})$ you have two arguments:

  • The (topological) map $\pi$ is open and therefore the preimage of a dense subspace is dense.

  • For any other trivialising open $V$, the intersection $U\cap V$ is dense (again by irreducilibity of $X$) and so the preimage $\pi^{-1}(U\cap V)=\pi^{-1}(U)\cap \pi^{-1}(V)$ is dense in $\pi^{-1}(V)$. But these sets cover $\mathbb{P}(\mathcal{E})$, hence $\pi^{-1}(U)$ is dense in $\mathbb{P}(\mathcal{E})$.

Pedro
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  • 'since $X$ is also regular, every $\mathbb P^n$-bundle over $X$ arises in this way', isn't the obstruction arise in the cohomology group $H^2(X,\mathcal O^*)$, so the regular condition is not enough to deduce that? – Tom Jan 18 '21 at 03:41
  • Dear @Tom, that statement is also part of the cited exercise in Hartshorne's Algebraic Geometry book, namely Exercise II.7.10.(c) – Pedro Jan 19 '21 at 08:36
  • We may have different defintions of projective bundle, from wiki https://en.wikipedia.org/wiki/Projective_bundle: Every vector bundle over a variety X gives a projective bundle by taking the projective spaces of the fibers, but not all projective bundles arise in this way: there is an obstruction in the cohomology group $H^2(X,\mathcal O^*)$ , so is your method only applied to the case arised from vector bundle? – Tom Jan 19 '21 at 08:43
  • Dear @Tom, I think the definition I was using (the one in Hartshorne's book) agrees with the one in Wikipedia. In fact, in the second paragraph of the Wikipedia page that you linked you can find the statement that we are talking about. I believe the statement you are talking about refers to the projectivization of a vector bundle (seen as a geometric object), which strictly speaking is not the same as the projectivization of a locally free sheaf. Does this clarify the issue? – Pedro Jan 19 '21 at 09:00
  • Dear@Pedro, I still have 2 questions: 1.what is exactly the difference between the projectivization of a vector bundle and the projectivization of a locally free sheaf? 2. do you have an example that a projective bundle which is not arised by taking the projective spaces of the fibers? – Tom Jan 19 '21 at 09:19
  • Dear @Tom, I think my previous comment was misleading, I apologize. After first reading the Wikipedia article I assumed naively that the problem must be that those two constructions somehow do not commute, but I believe they should commute over a smooth variety. I suspect that the real problem may lie in the distinction between étale local triviality and Zariski local triviality, so it may be a matter of definition after all (perhaps this is why it says "clarification needed" in the Wikipedia article). See also https://mathoverflow.net/questions/306852/mathbbp1-bundle-over-compact-base – Pedro Jan 19 '21 at 10:06
  • @Tom I just found an even better reference: https://math.stackexchange.com/questions/2212997/geometric-fibers-mathbb-pn-vanishing-of-brauer-group-implies-projective-bu – Pedro Jan 19 '21 at 10:11
  • Dear@Pedro, thanks for the reference provided, I find the paper by Elencwajg and Narasimhan(1983), "Projective bundles on a complex torus" may have provide more information about this issue, in this paper, they only assume the base space $X$ to be a complex manifold, not a variety as in wiki, do you think it makes any difference to this projective bundle question? – Tom Jan 19 '21 at 11:06
  • Dear @Tom, if they talk about manifolds they are probably considering the analytic topology. And yes, this topology is much finer than the Zariski topology, so it could happen that a bundle is analytic/étale locally trivial but not Zariski locally trivial, because Zariski opens are all huge and so it could be impossible to find a Zariski open subset over which your bundle is trivial, you may need to go down to small open subsets like euclidean balls to trivialize it. I think if you have further comments or questions we should move this discussion to a chat – Pedro Jan 19 '21 at 11:18