Does every almost complex manifold admit an almost-Kähler structure?

Question

I couldn't find a conclusive answer to this question online. Here is my reasoning.

Let $M$ be an almost complex manifold. Then, from what I understand, we can define almost complex structure $J$ on $M$. Moreover, $M$ being second countable implies that it is paracompact. Hence we can define a Riemannian metric $h$ on $M$. Then we define $g(X,Y) = h(X,Y) + h(JX,JY)$ so that $g(X,Y) = g(JX,JY)$.

Now that we have $J$ and $g$, we can define $\omega(X,Y) := g(X,JY)$ so that $\omega$ is skew symmetric and non-degenerate (since $g$ is non-degenerate because it is a Riemannian metric). Thus $\omega$ is a symplectic form.

Then $(M,g,\omega,J)$ is an almost Kähler manifold (almost because we don't have that $J$ is integrable).

I'd like to know if all this works or if there is something I am not understanding.

Thanks a lot!

Answer 1

It is not the case that $\omega$ is a symplectic form, that requires the additional condition $d\omega = 0$. Instead, $\omega$ is called an almost symplectic form.

We say that $(M, g, J, \omega)$ is almost-Kähler if $d\omega = 0$. In particular, we do not require $J$ to be integrable.

An almost-Kähler manifold is a symplectic manifold with a choice of compatible almost complex structure. With this in mind, we see that not every complex manifold admits an almost-Kähler structure. For example, Hopf manifolds are not symplectic so they cannot be almost-Kähler.