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Why is it true, that complex submanifolds of Kähler manifolds are Kähler?

A Kähler manifold is $(M,J, \omega)$, where $(M, \omega)$ is symplectic, $(M,J)$ is a complex manifold.

Now let $W \subset M$ be a complex submanifold, so $(W, J|_{W})$ is a complex manifold. Then I need to show, that $\omega|_{W \times W}$ is non-degenerate.

Now assume that for $v \in W$, $\omega(v,u)=0$ for all $u \in W$.

I need now to show that this is somehow a contradiction to the complex structure of $W$.

Can somebody give me a hint on how to do that?

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Hint: Don't think about the restriction of the Kähler form, but about the restriction of the metric.

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  • $\begingroup$ So I assume $g(u,Jv)=0$ for all $v \in W$, correct? Also, I know that $d \varphi J = J_{st} d \varphi$. But how can I use it here? $\endgroup$
    – User1
    Feb 9, 2020 at 19:47
  • $\begingroup$ Since $J^2=id$, $J$ is bijective, and $g$ is non-degenerate. $\endgroup$ Feb 10, 2020 at 7:38
  • $\begingroup$ Ok, thank you very much, got it now! I am too slow sometimes $\endgroup$
    – User1
    Feb 10, 2020 at 9:52

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