4

I'm taking a course on complex manifolds, and in class we saw this fact for complex manifolds. I have a moderate background in algebraic geometry, and was interested in the same question for $\mathbb{P}^n$ bundles over a scheme $X$, where now we interpret the transition functions on $\operatorname{Spec} A \times \mathbb{P}^n$ as $A-$linear automorphisms of the homogeneous ring $A[x_0, ..., x_n]$. Maybe this is a good formulation of the precise question: is every such bundle isomorphic to $\mathbb{P}(\mathscr{E})$ for some locally free sheaf $\mathscr{E}$?

I'm sure that there is some really high-powered proof using tools that are above my paygrade though like GAGA or something. Personally I'd just like to see to what extent this is still true in the algebraic setting, although not necessarily over the complex numbers, or if possible, even projective space over a ring would be nice to know some things about.

A. Thomas Yerger
  • 17,862
  • 4
  • 42
  • 85
  • 2
    Related : https://mathoverflow.net/questions/306852/mathbbp1-bundle-over-compact-base Also related : in the book of Beauville "complex algebraic surfaces", it is proved that any $\Bbb P^1$-vector bundle over a smooth curve comes from a vector bundle of rank $2$. – Nicolas Hemelsoet Aug 31 '18 at 17:00

1 Answers1

4

Consider the exact sequence of sheaves $$ 0\rightarrow \mathcal{O}_X^*\rightarrow \operatorname{GL}(r)\rightarrow \operatorname{PGL}(r)\rightarrow 0 $$ Taking the long exact sequence in cohomology gives $$ H^1(X,\mathcal{O}_X^*)\rightarrow H^1(X,\operatorname{GL}(r))\rightarrow H^1(X,\operatorname{PGL}(r))\rightarrow H^2(X,\mathcal{O}_X^*) $$ The set $H^1(X,\operatorname{GL}(r))$ classifies rank $r$ vector bundles on $X$. The set $H^1(X,\operatorname{PGL}(r))$ classifies rank $r-1$ projective bundles on $X$. So the obstruction to every projective bundle being a vector bundle is an element of $H^2(X,\mathcal{O}_X^*)$ in the image of $H^1(X,\operatorname{PGL}(r))\rightarrow H^2(X,\mathcal{O}_X^*)$. The image of this map is called the Brauer group and it classifies vector bundles which are not Zariski trivial but are étale trivial. There are plenty of interesting varieties with nontrivial Brauer groups, so the answer to your question is negative in general. A theorem of Tsen says that for curves over an algebraically closed field, the Brauer group is always trivial, so the answer to your question is positive in the case of curves.

Samir Canning
  • 732
  • Hmm. While continuing to search, I found that this a problem in Hartshorne (II.7.10c), which says that this is true for all regular schemes. Your answer seems to suggest that in this case, the Brauer group must be trivial. Can I bother you to say a few words about why this is the case? I'm not familiar with this notion. Thank you though! – A. Thomas Yerger Sep 01 '18 at 03:15
  • 2
    Hartshorne defines projective space bundles in this exercise as those which are Zariski locally trivial, so the question of the Brauer group doesn't really come into play here. In particular, regular schemes do not in general have trivial Brauer group! A simple example of a Brauer-Severi variety (a projective space fibration that is not Zariski locally trivial) is the universal family over the Hilbert scheme of plane conics. This Hilbert scheme is isomorphic to $\mathbb P^5$, which is about as regular as they come. – Tabes Bridges Sep 02 '18 at 18:35
  • This thread has some nice info along the lines of the comment above by @Tabes https://mathoverflow.net/questions/213231/is-a-morphism-whose-all-fibers-are-mathbfpn-a-projective-bundle – Samir Canning Sep 02 '18 at 18:39