Is a symplectic submanifold of a Kähler manifold Kähler?

That is, if $X$ is a Kähler manifold with symplectic form $\omega$ and $i:Y\hookrightarrow X$ is an embedded submanifold such that $i^*\omega$ is symplectic, is $Y$ a Kähler submanifold of $X$?

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    $\begingroup$ According to this question mathoverflow.net/questions/59733/…, the Kodaira-Thurston manifold is a symplectic submanifold of $\mathbb{P}^5$. Not only is this not a Kahler submanifold, it is not even Kahler (for any metric). $\endgroup$ – Oliver Nash Jan 3 '18 at 18:04
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    $\begingroup$ Another comment. The fact that Y is symplectic is an open condition on the embedding of Y in X, it will be preserved by perturbation... so a deformation of a Kähler submanifold (among embeddings which are not necessarily holomorphic) will remain symplectic if the deformation is small enough. $\endgroup$ – Pierre Apr 12 '18 at 19:47

No. In $\mathbf C^2$ with standard 2-form and complex structure, the real span of $U=\left(\begin{smallmatrix}1\\0\end{smallmatrix}\right)$ and $V=\left(\begin{smallmatrix}i\\1\end{smallmatrix}\right)$ is a symplectic but not complex (hence not Kähler) subspace, since $$ \omega(U,V)=1, \qquad JU=\left(\begin{smallmatrix}i\\0\end{smallmatrix}\right)\notin\operatorname{span}_{\mathbf R}(U,V). $$

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  • $\begingroup$ Also, Kostant and Sternberg have a nice paper (1982) on homogeneous symplectic, non-Kähler submanifolds of $\mathbf{CP}^n$. $\endgroup$ – Francois Ziegler Jan 3 '18 at 18:48

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