\circ \Psi_{E_b, E_c}) \circ \Psi_{E_a, E_b}$$ the cocycle identity $$\zeta ( E_b, E_c, E_d ) \zeta ( E_a, E_b, E_d ) = \zeta ( E_a, E_c,
E_d) \zeta ( E_a, E_b, E_c ) .$$ The square root satisfying the same identity, the product ${\circ_{{{\mathcal{D}}}}}$ is associative. Since $\Psi_{E,E}$ is the identity map, the identity of $\delta$ is a half-form morphism. Furthermore, $$\Psi_{E_b, E_b } \circ \Psi_{E_a, E_b} = \Psi_{E_a, E_b} , \qquad
\Psi_{E_a, E_b } \circ \Psi_{E_a, E_a} = \Psi_{E_a, E_b}$$ imply that $$\zeta( E_a, E_b, E_b ) = \zeta ( E_a, E_a, E_b) = 1 .$$ So the square root of $\zeta$ satisfies the same equation, consequently the identity of $\delta$ is a unit of $(E,  \delta, \varphi)$.

Since $\overline{E}_{a}$ and $E_{b}$ are Lagrangian spaces, one has $${\omega}( x - \pi_{E_a, E_b} x , \overline{ \pi_{ E_b, E_a} y} ) = 0 , \qquad  {\omega}(
\pi_{E_a, E_b} x , \overline{y} - \overline{\pi_{E_b, E_a} y} ) = 0$$ for any $x \in E_{a}$ and $y \in E_{b}$. Consequently $${\omega}( x , \overline{ \pi_{ E_b, E_a} y} ) = {\omega}( \pi_{E_a, E_b} x
,\overline{y}  )$$ which proves that $\pi_{E_b, E_a}$ is the adjoint of $\pi_{E_a, E_b}$. So $\Psi_{E_b, E_a}$ is the 