Let $X$ be a Kähler manifold and $E\to X$ a holomorphic vector bundle. Is there a Kähler structure on $E$ compatible with its complex structure?
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5$\begingroup$ I recall that the result is true, and the proof uses: (1) pulling back the Kaehler form from the base and then (2) adding $\partial\bar\partial u$ for a suitable function $u$. Locally, such a function $u$ exists, obviously. Globally, you will need to use a partition of unity. But I can't remember how you ensure that the form remains positive globally. $\endgroup$– Ben McKayCommented Mar 13, 2018 at 7:47
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$\begingroup$ @BenMcKay: when you glue locally defined Riemannian metrics via a partition of unity the resulting tensor is positive definite because being positive definite is a convex property and so it's preserved by convex linear combinations. Can the same thing be used here here, or I misunderstood the construction? $\endgroup$– QfwfqCommented Mar 13, 2018 at 11:33
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1$\begingroup$ @Qfwfq: The difficulty is the Kaehler condition. Partitions of unity gluing together a Riemannian or Hermitian metric will not ensure Kaehlerity, because that involves first derivatives of the functions in the partition of unity. $\endgroup$– Ben McKayCommented Mar 13, 2018 at 16:08
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1$\begingroup$ You can take euclidean norm $|| \cdot ||^2$ as fibervise Kahler potential; then $\epsilon \partial \bar \partial ||v||^2 + \pi^* \omega$ is Kahler form for small $\epsilon$ — derivatives of unity partition are irrelevant because you can take $\epsilon$ arbitrarily small and positivity is open condition. For affine bundles situation is analogous, but they may fail to be Stein. $\endgroup$– Denis TCommented Mar 13, 2018 at 19:23
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1$\begingroup$ You are right, I was assuming that base is compact. (If base is noncompact, but bundle admits connection of bounded curvature, it still works, but otherwise, of course, not). $\endgroup$– Denis TCommented Mar 14, 2018 at 11:30
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Proposition 3.18 of Voisin's Hodge Theory and Complex Algebraic Geometry I says that, if $X$ is compact Kahler and $E$ is a holomorphic vector bundle over $X$, then $\mathbb{P}(E)$ is Kahler. Since $E$ embeds as an open submanifold of $\mathbb{P}(E \oplus \mathbb{C})$, this establishes your result for $X$ compact, and I think her proof could be simplified if you just want the vector bundle version and not the projective bundle version. But it looks to me like she actually is using compactness in a nontrivial way.
answered Mar 13, 2018 at 18:22
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$\begingroup$ Perhaps "the vector bundle version"? $\endgroup$ Commented Mar 14, 2018 at 1:35
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$\begingroup$ Fixed, thanks! (Unless you meant something more significant than the typo "tthe" for "the".) $\endgroup$ Commented Mar 14, 2018 at 1:58