MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $E\rightarrow M$ be a holomorphic bundle over a Kähler manifold. Does its projectivisation $\mathbb{P}(E)$ always admit a Kähler metric? If yes, how to see that?

share|cite|improve this question

As pointed out by Georges Elencwajg, the answer is yes.

However, if one substitutes the assumption "holomorphic vector bundle" with the weaker "complex vector bundle", the answer is no.

In fact, there is the following result proven by C. Voisin.

Start with a complex Kähler manifold $X$ having a given class $\alpha \in H^4(X, \mathbb{Q})$ such that, for any given compatible Hodge decomposition on $H^*(X)$, $\alpha$ is not of type $(2,2)$.

Then if $E$ is any complex vector bundle on $X$ satisfying $$c_1(E)=0, \quad c_2(E)=\alpha,$$ the projective bundle $\mathbb{P}(E)$ admits no Kähler metric (even better, it is not homeomorhic to any Kähler manifold).

The simplest example of such a pair $(X, \alpha)$ is obtained by choosing for $X$ a complex torus of dimension $4$ and for $\alpha$ a class satisfying the property that the cup product map $$\alpha \cup \colon H^1(X, \mathbb{Q}) \longrightarrow H^5(X, \mathbb{Q})$$ has odd rank.

See these notes by C. Voisin for more details.

share|cite|improve this answer

Let me recall you briefly how to obtain such a Kähler metric on the total space of the projectivized bundle.

Start with any given hermitian metric $h$ on $E$ and consider on the projectivized bundle $\mathbb P(E)$ of hyperplanes of $E$ the tautological line bundle $\mathcal O_E(1)\to\mathbb P(E)$ of rank one quotients of $E$. Then, on $\mathcal O_E(1)$ you have a natural induced quotient hermitian metric, which I shall call again $h$.

If you compute the Chern curvature $\Theta(\mathcal O_E(1))$ of $\mathcal O_E(1)$ with respect to $h$, you will find a closed $(1,1)$-form on $\mathbb P(E)$ which is positive along the relative tangent bundle (after all, the restricion of $\mathcal O_E(1)$ to fibers $\simeq\mathbb P^{\operatorname{rk}E-1}$is just the usual $\mathcal O(1)$, so that its curvature restricted to fibers is just the usual Fubini-Study metric). No more can be said along "horizontal" directions.

Now, suppose that $M$ is compact Kähler, with Kähler form $\omega$ and call $\pi\colon\mathbb P(E)\to M$ the natural projection. Since $E$ is holomorphic (see Francesco's answer), $\pi$ is holomorphic as well. Thus, $\pi^*\omega$ is again a closed $(1,1)$-form on $\mathbb P(E)$ which is zero on vertical directions (as a pull-back) and strictly positive on horizontal ones.

Finally, by compactness of $M$, and so of $\mathbb P(E)$, you can find a large constant $C$ such that the closed $(1,1)$-form $$ C\,\pi^*\omega+i\,\Theta(\mathcal O_E(1)) $$ is positive definite everywhere (such a large multiple is chosen in such a way that $\omega$ compensates the possible lack of positivity of $\Theta(\mathcal O_E(1))$ along horizontal directions, and observe that $\omega$ does not interfere with the vertical ones).

This gives you the desired Kähler metric on $\mathbb P(E)$.

share|cite|improve this answer

If $M$ is compact ( the usual assumption in Kähler manifold theory) the answer is "yes". You can look it up in Claire Voisin's book Proposition 3.18, page 78.

share|cite|improve this answer
Quite amusing: Francesco posted his counter-example while I was writing up my positive answer! I can only hope for Voisin's sake that the $X$ mentioned in Francesco's post is not compact... – Georges Elencwajg Jun 26 '11 at 9:51
Dear Geoges, your answer is correct. I think the point is that in Voisin's examples the vector bundle $E$ is complex but not holomorphic – Francesco Polizzi Jun 26 '11 at 9:53
Dear Francesco, thanks for the quick clarification. – Georges Elencwajg Jun 26 '11 at 11:32

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.