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Let $X$ be a compact Kähler manifold and $S \subset X$ a closed complex submanifold. Given a closed $(1, 1)$-form $\alpha$ on $S$, is there always a closed $(1, 1)$-form $\beta$ on a neighborhood of $S$ in $X$ such that $[\beta|_S] = [\alpha]$ in $H^{1, 1}(S, \mathbb{C})$?

Note. This mathoverflow answer seems related, but I don't think it answers the above question. The proof of Theorem 4.1 in arXiv:math/0609617 begins by assuming that we have such an extension and pursue by proving positivity results. I fail to see how any of the ideas in the proof would help with the question above. But I might be wrong, so any help in that direction would be great too.

Addendum (In reply to Donu Arapura's answer). There is related work by Griffiths (The extension problem in complex analysis. II. Embeddings with positive normal bundle. Amer. J. Math. 88 (1966)), showing the obstruction to extending certain analytic objects from $S$ to a neighborhood of $S$ in $X$. For instance, it is proved that if $F \to S$ is a vector bundle which is the restriction of a vector bundle $E \to U$ on a neighborhood $U$ of $S$ in $X$, and $\alpha \in H^q(S, F)$, then the first order obstruction to extending $\alpha$ to an element of $H^q(U, E)$ (possibly after shrinking $U$) lies in $H^{q+1}(S, F \otimes N_S^*)$, where $N_S^*$ is the conormal bundle of $S$ in $X$, and it does not vanish in general. In our case, we can interpret $\alpha$ as an element of $H^1(S, T^*S)$. But this does not immediately fit into Griffiths work, since $T^*S$ does not generally extend to a vector bundle on a neighborhood of $S$ in $X$.

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  • $\begingroup$ Wouldn't be enough to let $\beta=\pi^*\alpha$, where $\pi:T\to S$ is a tubular neighbourhood? $\endgroup$ Commented Oct 13, 2021 at 13:43
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    $\begingroup$ @DonuArapura We would need a holomorphic tubular neighbourhood, which fails to exist in general. $\endgroup$ Commented Oct 13, 2021 at 13:43
  • $\begingroup$ @DonuArapura See, e.g., mathoverflow.net/questions/114414/… $\endgroup$ Commented Oct 13, 2021 at 13:44
  • $\begingroup$ I got downvoted for this. I'm new here, and I'm not sure I see what's wrong with it. Can someone explain? $\endgroup$ Commented Oct 15, 2021 at 14:01

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Contrary to my earlier comments, I now believe (but I haven't completely checked) that this is false in general. (I prefer not to comment on the paper you linked.) Suppose that $X$ is a Kähler manifold. The form $\alpha$ defines a class in $H^1(S,\Omega_S^1)$. The existence of $\beta$ should imply that $\alpha$ lifts to a class $\alpha'\in H^1(S, \Omega_{S'}^1)$, where $S'$ is the first order neighbourhood of $S$. But there are obstructions, which I believe are nontrivial.

When $\alpha=c_1(L)$, with $L$ a line bundle, then I believe the existence of $\alpha'$ is close to demanding that $L$ lifts to $S'$. But here the obstruction is well known to lie in $H^2(S,\mathcal{O})$, and is generally nontrivial.

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  • $\begingroup$ Thanks for your answer. I replied to it in the original post. $\endgroup$ Commented Oct 14, 2021 at 19:29
  • $\begingroup$ +1 by the way. I'm still not sure how to definitely find a counterexample, but this is useful nonetheless. $\endgroup$ Commented Oct 15, 2021 at 14:00

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