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

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).

• The Brauer group of $P^n(\mathbb{C})$ equals the unramified Brauer group of $\mathbb{C}(t_1,\dots,t_n)$. For every field $k$, one has $Br_{nr}(k(t))=Br_{nr}(k)$, and for an algebraically closed field the unramified Brauer group vanishes (being a subgroup of the Brauer group); see Gille-Szamuely's book "Central simple algebras and Galois cohomology". Using the short exact sequence for $1\to G_m \to GL_n \to PGL_n\to 1$, and passing to etale cohomology, one deduces your statement on projective bundles. – A.Garcia Nov 30 '19 at 17:21
• See e.g. this question for a proof. mathoverflow.net/questions/75774/… I'm not sure why you want a reference for the equivalent statement, but Grothendieck's "Dix Exposes" will certainly suffice as a reference for the equivalence. – Daniel Litt Dec 1 '19 at 1:11
• Since you mention Hartshorne's book, I'll just add that this is the content of Exercise II.7.10(c), which states that projective bundles over any regular noetherian scheme arise as the projectivization of a vector bundle on the base. (There is also a hint in the exercise on how to show this.) – Devlin Mallory Dec 2 '19 at 16:04
• Dear Delvin, now I am really puzzled. I look in the paper of Beauville, arxiv.org/pdf/1507.02476.pdf and on the top of Page 20 it is written there : "Thus we associate to each $P^1$-bundle $p : P \to V$ a class $[p]$ in $H^2(V, \mathbb G_m)$, and this class is trivial if and only if $p$ is a projective bundle". I deduce from this that there exist bundles for which this class is non-trivial and so they are not projective. I guess I am missing something here... – aglearner Dec 2 '19 at 16:51
• @aglearner There is no contradiction. A projective space bundle, which is locally trivial for the Zariski topology is not the same as a bundle in the analytic or etale topologies. You and the references probably use the latter,and Hartshorne the former. – Donu Arapura Dec 2 '19 at 17:36


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.

• This only shows the resulting statement up to diffeomorphism, right? – Mike Miller Nov 30 '19 at 19:00
• @aglearner I had not seen this statement before, though this kind of argument is related to things I spend a lot of time thinking about. – Arun Debray Nov 30 '19 at 19:26
• Dear Arun, I see. I realised that I have not stated in my question that I need a reference for the fact that these two spaces are isomorphic as $\it projective\; varieties$. So, this is not quite a question in algebraic topology, it is rather a question in algebraic geometry. It is a statement that there is an invertible holomorphic map between two spaces. I'll update the question so that this is stated explicitly. – aglearner Nov 30 '19 at 20:43
• You are trying to lift a holomorphic $\mathrm{PGL}_{k+1}$-principal bundle to a $\mathrm{GL}_{k+1}$-principal bundle. Using standard methods in bundle theory, this reduces to the statement that $\mathrm H^2(\mathbb P^n, \mathcal O^*) = 0$, where $\mathcal O^*$ is the sheaf of invertible holomorphic bundle. From the exponential sequence this reduces to the statements that $\mathrm H^2(\mathbb P^n, \mathcal O) = \mathrm H^3(\mathbb P^n, \mathbb Z) = 0$, which are standard. – Angelo Dec 1 '19 at 12:47
• @aglearner: If you so desire, you can take the proof Angelo has sketched and rewrite it purely in terms of Cech cocycles. This is as elementary a proof as I can imagine... – Daniel Litt Dec 2 '19 at 18:50

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.

• @algearner Let me recapitulate Angelo's comment, which gives a very direct proof. A rank $k$ projective space bundle in the analytic topology is given by a 1-cocycle in $h_{ij}$ with values in $PGL_k$. After refining the covering, if necessary, you can lift it to a 1-cochain $g_{ij}$ with values in $GL_{k+1}$. The obstruction for this to be cocycle is $g_{ij}g_{jk}g_{ki}$, which can be viewed a $2$-cocycle with values in $O^*$. By your assumption, this is a coboundary of a $1$-cocycle $l_{ij}$ in $O^*$. – Donu Arapura Dec 9 '19 at 16:20
• ... Then $g_{ij}'=g_{ij}l_{ij}$ would give a cocycle, which gives your desired vector bundle. No need to say "Brauer group", if that's what bothers you. – Donu Arapura Dec 9 '19 at 16:20
• Dear Donu, thank you for this comment! When you write "By your assumption" you mean $H^{2,0}(Y)=0$? – aglearner Dec 9 '19 at 22:51
• Yes, that's what I meant. – Donu Arapura Dec 10 '19 at 2:33

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.

• Did you ever have a look at proposition 1.4 of numdam.org/article/SB_1964-1966__9__199_0.pdf ? – Libli Dec 11 '19 at 19:39
• Never. J'ai trop peur de Grothendieck. Thanks a lot for giving this reference. It looks very much like what one needs...? Though my rudimentary level of knowledge of algebraic geometry doesn't let me understand this fully – aglearner Dec 18 '19 at 17:27