# Projectivity of projective bundles over algebraic varieties.

In Kodaira's famous 1954 paper 《On Kähler Varieties of Restricted Type 》 https://www.jstor.org/stable/pdf/1969701.pdf?refreqid=excelsior%3A3c883bd969da3e5252b449ceb0e723b1, the author proved that "the projective bundles over a projective manifold is a projective manifold", the full statement is:

Let $$B$$ be an analytic fibre bundle over a complex analytic variety $$V$$ whose fibre is a complex projective space and whose structure group is the group of projective transformations. If the base space $$V$$ is a non-singular algebraic variety imbedded in a projective space, then the bundle $$B$$ is also a non-singular algebraic variety imbedded in a projective space.

and my question is: what if we remove the condition "whose structure group is the group of projective transformations"? or if we only require the fiber is projective space, is the statement still remained true?

Here, for the fiber space $$\pi:B\to V$$, if we only require that all the fibers are projective spaces, not requiring anything about the structure group or transition functions, can we say that this fiber space is a projective bundle?

And I think this is a similar question, for complex manifolds $$E,X$$, and the holomorphic map $$\pi:E\to X$$, if we only ruquire that all the fibers of $$\pi^{-1}(t)$$ are $$\mathbb C^n$$, not requiring there exist local trivalisation $$\pi^{-1}(U_i)\cong U_i\times \mathbb C^n$$, can we still conclude that $$E$$ is a vector bundle?

• Hint: Every biholomorphic automorphism of $CP^n$ is a projective transformation. Mar 3, 2021 at 10:13

Question: "..if we only ruquire that all the fibers of $$π^{−1}(t)$$ are $$\mathbb C^n$$, not requiring there exist local trivalisation $$π^{−1}(U_i)≅U_i×\mathbb C^n$$, can we still conclude that $$E$$ is a vector bundle?"

Answer: I believe the answer is no. A vector bundle has transition maps linear maps. There are locally trivial fibrations with fiber $$\mathbb{C}^n$$ where the transition maps are not linear (affine bundles): For any quasi projective variety $$X \subseteq \mathbb{P}^n_{\mathbb{C}}$$ there is a "torsor" $$\pi: Y \rightarrow X$$. The map $$\pi$$ is surjective and locally trivial with fiber $$\mathbb{C}^n$$ but the transition maps are affine maps. The variety $$Y$$ is an affine variety. The construction is explicit, and you should try yourself to construct non-trivial examples. I would guess you get non-trivial examples for $$\mathbb{P}^1$$ or $$\mathbb{P}^n$$:

https://en.wikipedia.org/wiki/Jouanolou%27s_trick

Question: "Here, for the fiber space π:B→V, if we only require that all the fibers are projective spaces, not requiring anything about the structure group or transition functions, can we say that this fiber space is a projective bundle?"

Answer: If $$\pi: B \rightarrow X$$ is a surjective map with fibers $$\pi^{-1}(x)\cong \mathbb{P}^n_{\kappa(x)}$$ for all $$x\in X$$, it is not immediate that this implies that $$\pi$$ is locally trivial with transition maps being linear. For real smooth manifolds there is the "Ehresmann fibration lemma" and for complex manifolds there is the "period mapping":

https://en.wikipedia.org/wiki/Ehresmann%27s_lemma

https://en.wikipedia.org/wiki/Period_mapping

• Can you give a more concrete example of which all the fiber is $\mathbb C^n$ but there exist no local trivalisation $\pi^{-1}(U_i)\cong U_i \times \mathbb C^n$?
– Tom
Mar 2, 2021 at 9:58
• It seems the link you provide for "Ehresmann fibration lemma" ensure that the fibration is locally trivial, although say nothing about whether the transition maps are linear, so it is not a counter-example to my question: if we only requiring fiber being projective space can we say this fiber space is a projective bundle?
– Tom
Mar 3, 2021 at 7:54
• You must look for a "holomorphic Ehresmann lemma". It is unlikely that having projective spaces as fibers imply that fibration is a projective bundle, but I do not have an example. Mar 3, 2021 at 8:42