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As pointed out by GeorgeGeorges 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.

Post Undeleted by Francesco Polizzi

The

As pointed out by George, the answer is yes.

However, if one substitutes the assumption "holomorphic vector bundle" with "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.

Post Deleted by Francesco Polizzi
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