When does a symplectic bundle split? Let $(E,\omega)$ be a symplectic bundle over $X$. I know that $E$ splits as a vector bundle $E = F \oplus G$. 
I want wo check, if this is a splitting of symplectic bundles. What do I need to check?
Is it enough to find $\omega_F, \omega_G$ such that $\omega = \omega_F + \omega_G$ and  such that $\omega_F|_F$ is symplectic on $F$ and $\omega_G|_G$ is symplectic on $G$?
Do I also need to check some condition on $\omega_F|_G$?
 A: Given a morphism of vector bundles $\psi: F \oplus G \rightarrow E$, you want to check that there are symplectic forms $\omega_F$ on $F$ and $\omega_G$ on $G$ such that $\psi$ is a symplectomorphism with respect to the symplectic forms $\omega_F\oplus \omega_G$ on $F\oplus G$ and $\omega$ on $E$. This is equivalent to the following two lines being equal for all $f_1$, $f_2\in F$, $g_1$, $g_2\in G$:
\begin{align}
(\omega_F\oplus\omega_G)(f_1+g_1,f_2+g_2) &= \omega_F(f_1,f_2) + \omega_G(g_1,g_2), \\ \omega(\psi(f_1+g_1),\psi(f_2 + g_2)) &= \omega(\psi(f_1),\psi(f_2)) + \omega(\psi(g_1),\psi(g_2)) + \omega(\psi(f_1),\psi(g_2)) + \omega(\psi(g_1),\psi(f_2)).
\end{align}
The first line is a definition, and in the second we used the linearity of $\psi$ and the bilinearity of $\omega$. By setting some of $f_1$, $f_2$, $g_1$, $g_2$ zero and using the antisymmetry of $\omega$, this requirement is equivalent to
\begin{align}
\omega_F(f_1,f_2) &= \omega(\psi(f_1),\psi(f_2)),\\
\omega_G(g_1,g_2) & =\omega(\psi(g_1),\psi(g_2)),\\
\omega(\psi(f),\psi(g)) & = 0,
\end{align}
for all $f_1$, $f_2$, $f\in F$, $g_1$, $g_2$, $g\in G$. These are therefore the conditions you have to check.
