$\mathbb CP^k$ bundles over $\mathbb CP^n$ are projectivisations of vector bundles. Any reference?

Question

Statement. Let $X$ be a smooth complex projective variety that is a $\mathbb CP^k$ bundle over $\mathbb CP^n$ in analtytic topology. It is well known that there exists a rank $k+1$ complex vector bundle $V$ over $\mathbb CP^n$ such that $X$ is isomorphic as a projective variety to the projectivisation $\mathbb PV$.

Question. I would like to find a precise reference to this statement, is there such a reference?

(I know that one is supposed to say that the Brauer group of $\mathbb CP^n$ is trivial, and this is why the statement holds. But I can't find any place in the literature where this statement about projective bundles is stated and I need to find such a reference :( )

PS. I realised, that Section 6.2 of the beautiful paper by Arnaud Beauville https://arxiv.org/abs/1507.02476 would do as a reference to the statement. Yet this section speaks of something far more general. I still hope for something more direct, maybe along the lines of what Angelo suggested in his comment (or at least explicitly stated).

Answer 1

$\newcommand{\C}{\mathbb C} \newcommand{\CP}{\mathbb{CP}} \newcommand{\GL}{\mathrm{GL}} \newcommand{\Z}{\mathbb Z} \newcommand{\PGL}{\mathrm{PGL}}$ Here's a topological proof.

An algebraic $\CP^k$-bundle $X\to\CP^n$ is a fiber bundle with structure group $\PGL_{k+1}(\C)$. It therefore is equivalent data to a principal $\PGL_{k+1}(\C)$-bundle $P\to\CP^n$: given $X$, the fiber of $P$ at $y\in\CP^n$ is the $\PGL_{k+1}(\C)$-torsor of isomorphisms $\CP^k\overset\cong\to X_y$. Conversely, we can recover $X$ from $P$ as the associated bundle $$X = P\times_{\PGL_{k+1}(\C)} \CP^k := P\times\CP^k /\{(p\cdot g, q)\sim (p, g\cdot q)\mid g\in\PGL_{k+1}(\C)\},$$ with the map to the base induced from that of $P$.

$X$ extends to a complex vector bundle iff we can lift $P\to\CP^n$ to a principal $\GL_{k+1}(\C)$-bundle $P'\to\CP^n$; the complex vector bundle in question is the associated bundle $$P'\times_{\GL_{k+1}(\C)}\C^{k+1}\to\CP^n.$$ The isomorphism class of $P$ is equivalent data to a homotopy class of maps $f_P\colon \CP^n\to B\PGL_{k+1}(\C)$, where $BG$ denotes the classifying space of $G$; lifting $P$ to a principal $\GL_{k+1}(\C)$-bundle is equivalent to finding a map $f_{P'}\colon\CP^n\to B\GL_{k+1}(\C)$ whose composition with the map $\phi\colon B\GL_{k+1}(\C)\to B\PGL_{k+1}(\C)$ recovers $f_P$ up to homotopy.

We can use obstruction theory to prove that a lift exists, since all of these spaces are CW complexes. Suppose we have a lift on the $m$-skeleton of $\CP^n$; then the obstruction to it extending to a lift on the $(m+1)$-skeleton is an element of $H^{m+1}(\CP^n;\pi_m(F))$, where $F$ is the fiber of $\phi$, which is $B\C^\times\simeq K(\Z, 2)$. It therefore suffices to show that a lift exists on the $2$-skeleton, as $\pi_m(K(\Z, 2)) = 0$ for $m\ne 2$.

In the standard CW structure on $\CP^n$, the $2$-skeleton is $\CP^1\cong S^2$, so we want to lift from $[S^2, B\PGL_{k+1}(\C)] = \pi_2(B\PGL_{k+1}(\C))$ to $[S^2, B\GL_{k+1}(\C)] = \pi_2(B\GL_{k+1}(\C))$. Associated to the fibration $F\to B\GL_{k+1}(\C)\to B\PGL_{k+1}(\C)$ we have a long exact sequence of homotopy groups $$\dots\to\pi_2(B\GL_{k+1}(\C))\overset{\phi_*}{\to}\pi_2(B\PGL_{k+1}(\C))\to \pi_1(F)\to\dots$$ but $\pi_1(F) = \pi_1(K(\Z, 2)) = 0$, so $\phi_*$ is surjective, and a lift exists.

Answer 2

I want to explain why the above statement reduces to topology. In fact, want to explain a slightly stronger statement.

Statement. Let $X$ be a smooth complex projective variety that is a locally trivial in analytic topology $\mathbb CP^k$-bundle over a smooth projective variety $Y$ . Suppose that $H^{2,0}(Y)=0$. Then there exists a rank $k+1$ complex vector bundle $V$ over $Y$ such that $X$ is isomorphic as a projective variety to the projectivisation $\mathbb PV$ if and only if there is a class $h\in H^2(X,\mathbb Z)$ that restricts to the generator of $H^2(\mathbb CP^k,\mathbb Z)$ on each fiber.

I believe that this condition about the existence of $h$ is equivalent to saying that $X$ can be represented as a projectivisation topologically.

Sketch proof. The only if direction is obvious, just take $\mathcal O(1)$ associated to $V$ and set $h$ to be its $c_1$.

Let's see the if direction. Suppose we find such a class $h$. Then we can construct a topological line bundle with $c_1=h$. Then, since $H^{2,0}(X)=H^{2,0}(Y)=0$, the bundle $L$ has a holomorphic structure. I.e. we have a holomorphic line bundle $L$ that restricts to each $\mathbb CP^k$ as $\mathcal{O}(1)$. Now it suffices to apply the answer to the following question: Constructing a very ample line bundle on a projective bundle

Note. I guess, that in order to see whether such a class $h$ exists one needs to do something of the type that Arun did. I decided to write this answer to demystify a bit this Brauer group.

Answer 3

I would like to give a final answer to this question (it took me some time to get it, thanks to Angelo and Donu) which is, in fact, the following statement.

Statement'. Let $Y$ be a projective variety with $H^2(Y,\mathbb {\cal O}^*)=0$. Then any holomorphic $\mathbb P^m$-bundle over it is a projectivisation of a holomoprhic vector bundle over it.

Proof. Please look into the comment of Angelo to the answer of Arun and into the comments of Donu to the earlier answer that I gave.

Coming back to the original question, we just notice (as Angelo did), that $H^2(\mathbb CP^n, {\mathcal O}^*)=0$.

I will accept this answer, to finish the story. I still wonder if the Statement' was never written in any paper or book.