I found something strange when I was working on some other problems.

I thought the triple intersection description of the unitary group said that any two of $(g, \omega, J)$ determines the third pointwisely. Then I found I was very wrong: when one tries to find a $J$ from some $(g,\omega)$ , one need not have $g(Jx, Jy)=g(x, y)$ for the resulting $J$, and hence may not have $J^2=−1$... very strange.

Consider the usual inner product on $\mathfrak{gl}(n)$:

$g(X,Y) = tr(XY)$

and this "symplectic structure"

$\omega(X,Y)=tr(A X A^{-1} Y - A Y A^{-1} X)$

where $A$ is a fixed constant element in $GL(n)$.

Now the $J$ corresponding to it seems to be

$J(X) = AXA^{-1} - A^{-1}XA$

but then $J^2 \neq -1$.

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Now the above symplectic form is degenerate, but I thought the real cause shall be something different.

Define $\omega$ on $\mathfrak{g} \times \mathfrak{g}$ as follows:

$\omega((x_1, x_2),(y_1, y_2)) = tr(x_1 y_2 - y_1 x_2 + Ad_g x_1 \cdot y_1 - Ad_g y_1 \cdot x_1 + Ad_h x_2 \cdot y_2 - Ad_h y_2 \cdot x_2)$

where $(g,h) \in G \times G$ is fixed, $(x_1,x_2) \in \mathfrak{g} \times \mathfrak{g}$, $(y_1,y_2) \in \mathfrak{g} \times \mathfrak{g}$.

One still have the $J$ issue.