Let $E\rightarrow M$ be a holomorphic bundle over a compact Kähler manifold, then as discussed in Kähler metric on projectivised bundle, $\mathbb P(E)$ admits Kähler metric. If we just consider a projective bundle over a compact Kähler manifold (may not be the projection of an vector bundle), can we have the same conclusion?
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5$\begingroup$ Yes. The pushforward of the dual of the relative canonical (invertible) sheaf is a holomorphic vector bundle on the base Kaehler manifold. The projectivization is a Kaehler manifold. There is a closed holomorphic embedding of the total space of the original bundle in this new Kaehler manifold. $\endgroup$– Jason StarrCommented Jul 7, 2022 at 10:58
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1$\begingroup$ I guess it depends what you mean precisely (i.e. what category this bundle with fibres equal to projective space is in). For example if it is just a topological bundle then there are $\mathbb{CP}^1$-bundles over $T^2$ which are not orientable. I am guess Jason Starr's answer assumes that the bundle is a complex manifold and the bundle map is holomorphic . $\endgroup$– Nick LCommented Jul 7, 2022 at 12:53
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$\begingroup$ @NickL Yes, indeed! I did interpret the OP's question to mean the case when the total space is a complex manifold and the projection to the base is a holomorphic submersion that (analytically locally on the base) is just a product of the base and a fixed projective space. $\endgroup$– Jason StarrCommented Jul 7, 2022 at 13:58
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$\begingroup$ @JasonStarr, thanks, your answer seems promising. But I am not an expert in algebraic geometry and I don't understand your construction well. Could you explain a little about what is the relative canonical sheaf and why there is a closed holomorphic embedding as you mentioned? $\endgroup$– MjrCommented Jul 8, 2022 at 3:51
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