If $(M,\omega)$ is a Kähler manifold, then we have the following structures on $M:$
- A smooth manifold structure on $M$.
- An almost complex structure $J.$
- A Riemannian metric $g,$ satisfying $$g(X,Y)=g(JX,JY).$$
- A closed 2-form $\omega,$ satisfying $$\omega(X,Y)=g(JX,Y).$$
We can state all of these things without referencing the complex structure on $(M,\omega)$, so my question is the following:
If $(M,J,g,\omega)$ is a smooth manifold $M$ with the structures $J,g,\omega$ and relations as above, does this induce a complex structure on $M?$
This certainly seems plausible to me, here is my attempt at proving this:
The only step I cannot justify rigorously is that $\omega$ is parallel. $d\omega=0,$ so $\nabla \omega=0$ should be true as well since $\omega$ is defined in terms of $J$ and $g,$ which is parallel, and we also have some compatibility of $J$ and $g.$
Assuming $\nabla \omega=0,$ we get \begin{equation*} \begin{aligned} 0=&(\nabla_Z\omega)(X,Y)=\nabla_Z(\omega(X,Y))-\omega(\nabla_Z,Y)-\omega(X,\nabla_Z Y)\\ =&\nabla_Z(g(JX,Y))-\omega(\nabla_Z X,Y)-\omega(X,\nabla_Z Y)\\ =&g((\nabla_Z J)X,Y)-\omega(X,\nabla_Z Y) \end{aligned} \end{equation*} Using normal coordinates we can now eliminate the $\nabla_Z Y$ term and use nondegeneracy of $g$ to deduce $\nabla J=0.$
Now we can show that the Nijenhuis tensor vanishes and apply the Newlander-Nirenberg theorem to get the complex structure: \begin{equation*} \begin{aligned} N_J(X,Y)=&[X,Y]+J([JX,Y]+[X,JY])-[JX,JY]=\\ =&\nabla_X Y-\nabla_Y X+J(\nabla_{JX}Y-\nabla_Y (JX)+\nabla_X(JY)-\nabla_{JY}X)-\nabla_{JX}(JY)-\nabla_{JY}(JX)\\ =&\nabla_X Y-\nabla_Y X+J\nabla_{JX} Y- JJ\nabla_Y X+JJ\nabla_X Y-J\nabla_{JY} X-J\nabla_{JX}Y+J\nabla_{JY}X\\ =&0 \end{aligned} \end{equation*} I guess I am also fine with putting $\nabla\omega=0$ or $\nabla J=0$ as an assumption but I am wondering if it is really necessary.