Verifying compatibility of symplectic and metric structure of $\mathbb{R}^{2n}$ I was reading about on wikipedia under the Hermitian Manifold page that for a almost-complex structure on a manifold $M$ that we have the following:
$$\omega(\cdot,\cdot) = g(J\cdot,\cdot).$$
I am having trouble with the explicit calculation of this relation between the symplectic form and the inner product on $\mathbb C^n$. 
I have tried using $\mathbb{R}^{2n}$ with its usual symplectic form $\omega_0$ as $\mathbb C^n$ where $z_j = x_j + iy_j$, and I am using $i = J$. Generally I start out with:
$$\begin{align}
g(J(u),v) 
&= \sum_{j=1}^n J(u_j)\bar v_j, \\
&= \langle J(\tilde u),\tilde v\rangle + i\omega_0(J(\tilde u),\tilde v),
\end{align}$$
where that inner product is the usual one on $\mathbb R^{2n}$ and $\tilde\cdot$ indicates the change in coordinates from $z_j$ to $x_j,y_j$. But beyond there I am a bit fuzzy on what to do since this so far seems incorrect, getting $J$ in the /real/ inner product, and also in the /real/ symplectic form.
Is it possible that someone can give me a hint/nudge in the right direction on this problem (and also perhaps correct me where I have been mistaken)? Thanks!

Using $z_j = x_j + \sqrt{-1} y_j$, the standard symplectic structure on $\mathbb C^n$ is 
$$ \omega = \sum_{i=1}^n dx_i \wedge dy_i.$$
Writing $(x, y) = \sum_{i=1}^nx^i e_i + y^i f_i$, $J$ is given by 
$$ J e_i = f_i,\ \ \ Jf _j = - e_j.$$
Then by checking directly, 
$$\begin{cases} \omega(e_i, e_j) = g(Je_i, e_j) = 0, \\ 
 \omega(f_i, f_j) = g(Jf_i, f_j) = 0, \\ 
\omega (e_i, f_j) = g(Je_i, f_j) = \delta_{ij}
\end{cases}$$
So $\omega$ and $g(J\cdot, \cdot)$ agree on basis and thus are the same. 
Note that in general the compatibility is part of the assumption. 
