Let $X$ be a compact Kahler manifold, let $D$ be a smooth divisor in $X$, and let $U$ be a tubular neighbourhood of $D$ in $X$. Suppose that $D$ is Fano. Is it possible to extend every closed (1, 1)-form on $D$ to a closed (1, 1)-form on $U$?
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Yes. For the proof, see e.g. http://arxiv.org/abs/math/0609617 Theorem 4.1. Here a stronger result is actually proven:
Theorem: Let $(M, \omega)$ be a compact Kahler manifold, and $Z\subset M$ a closed complex submanifold. Denote by $[\omega]\in H^2(M)$ the Kahler class of $M$. Consider a Kahler form $\omega_0$ on $Z$ such that its Kahler class coinsides with the restriction $[\omega]| Z$. Then there exists a Kahler form $\omega'$ on $M$ in the same Kahler class as $\omega$, such that $\omega| Z=\omega_0$.
An additional cohomological assumption is needed, because we build a global extension, and for an extension to a local neighbourhood you don't need it; the positivity (needed in assumption) is achieved by adding a big multiple of a Kaehler form.
answered Mar 15, 2011 at 5:30
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$\begingroup$ Let \alpha be a closed (1, 1)-form on Z. Add a big multiple of a Kahler form \omega' on Z to \alpha, so that k\omega'+\alpha is a positive (1, 1)-form on Z for some large positive real number k. How do I know that there exists a Kahler class [\omega] on M that restricts to the Kahler class [k\omega'+\alpha] on Z? This may be obvious... $\endgroup$– user3566Commented Mar 16, 2011 at 13:24
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$\begingroup$ No, it's not obvious, in fact, it can be false - the existence of a Kaehler class which restricts to $[\omega_0]$ is not guaranteed. However, the proof of the above theorem works to extend the form to a tubular neighbourhood. $\endgroup$ Commented Mar 19, 2011 at 22:13
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$\begingroup$ Do you mean $\omega'|Z = \omega_0$? $\endgroup$ Commented Oct 20, 2021 at 7:51
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