TITLE: Kähler metric on projectivised bundle
QUESTION [7 upvotes]: 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?

REPLY [9 votes]: 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 $\mathbb P(E)$.