Fundamental form of an almost complex manifold Let $(M^{2m},h,J)$ be an almost Hermitian manifold with fundamental form $\Omega$.
I want a local expression for this $\Omega (X,Y)=h(JX,Y)$ in this almost-complex manifold, but during my computations I struggle with the almost complex structure $J$. The only thing I know is that $J^2=-Id$.
So my questions so far are:

*

*What does the (1,1)-tensor $J$ do with an arbitrary vector field X (local/global)?


*Would it be better to do the computations in a local orthonormal frame?


*Does a Riemannian metric in a local orthonormal frame behave like the standard inner product?
 A: Your comments have now led me to a proper understanding of the question, and I can now write an answer.
We have the following set up. $M$ is a $2m$-manifold, with almost complex structure $J$ and a Riemannian metric $h$ that preserves the almost complex structure. I.e., so that $h(Ju,Jv)=h(u,v)$ for all tangent vectors $u$ and $v$ at every point. 
You then define the fundamental form $\Omega(X,Y) = h(JX,Y)$, which is skew symmetric, since $$\Omega(Y,X) = h(JY,X) = h(X,JY) = h(JX,J^2Y) = h(JX,-Y) = -\Omega(X,Y).$$
I'll note at this point that your choice of letters is slightly misleading. I would expect $g$ for the Riemannian metric, and wikipedia leads me to expect $\omega$ for the fundamental form. However, now that the meanings have been clarified, it doesn't really matter what letters you picked.
Let $e_i$ be a local orthonormal frame. It's kind of poor notation to regard the vectors $e_i$ as covectors, even though they are canonically identified with such via $h$. Thus let $\theta_i$ be the dual basis, $\theta_i(x) = h(e_i,x)$. Observe that $h(Je_i,-) = -h(e_i,J-)$, so the dual of $Je_i$ is $-\theta_i J$.
Now you want to show that 
$$\Omega = -\frac{1}{2} \sum_i \theta_i \wedge \theta_i J.$$
First, since $e_i$ are orthonormal, observe that the entries of the matrix for $J$ with respect to the $e_i$ are $J_{ij} = h(e_i,Je_j)$. Also since $h(Jx,Jy)=h(x,y)$, $J$ is orthonormal, so $J^T = J^{-1}$.
To do so, observe that $Je_i$ is also an orthonormal frame, so it suffices to show that both sides agree for all pairs $(e_i,Je_j)$ that we evaluate them at.
Evaluating $\Omega$ first, we get $$\Omega(e_i,Je_j) = h(Je_i,Je_j) = \delta_{ij}.$$
Now evaluating the right hand side, we have
$$-\frac{1}{2} \sum_k \theta_k(e_i)\theta_k(J^2e_j) - \theta_k(Je_j)\theta_k(Je_i)$$
$$=-\frac{1}{2} \sum_k -\delta_{ik}\delta_{jk} - h(e_k,Je_j)h(e_k,Je_i)$$
$$ = \frac{1}{2}\sum_k \delta_{ik}\delta_{jk} + \frac{1}{2}\sum_k J_{kj}J_{ki}$$
$$ = \frac{1}{2}\delta_{ij} + \frac{1}{2} \sum_k J^T_{ik}J_{kj}$$
$$ = \frac{1}{2}\delta_{ij} + \frac{1}{2} \delta_{ij} = \delta_{ij},$$
as desired.
