an isomorphism $\varphi : \delta
^{\otimes 2} \rightarrow  \wedge ^{{\operatorname{top}}} E^*$. We endow $\delta$ with the scalar product making $\varphi$ a unitary map.

Let us consider the category ${{\mathcal{D}}}(S)$ with objects the triples $(E, \delta,
\varphi)$ consisting of a positive polarization together with a half-form line. The morphisms from $(E_a , \delta_a, \varphi_a)$ to $(E_b,  \delta_b, \varphi_b)$ are the linear maps $\Psi : \delta_a \rightarrow
\delta_b$ satisfying $$\varphi_b \circ \Psi^{\otimes 2} = \Psi_{E_a, E_b} \circ
\varphi_a.$$ The composition of a morphism $\Psi: (E_a , \delta_a, \varphi_a)
\rightarrow (E_b,  \delta_b, \varphi_b)$ with a morphism $\Psi ' : (E_b , \delta_b, \varphi_b)
\rightarrow (E_c,  \delta_c, \varphi_c)$ is defined as $$\Psi' {\circ_{{{\mathcal{D}}}}}\Psi := \zeta ^{1/2} ( E_a, E_b, E_c) \Psi' \circ \Psi$$ where the product in the right-hand side is the usual composition of maps.

\[prop:cat\_demi\_forme\] The product ${\circ_{{{\mathcal{D}}}}}$ is well-defined and associative. For any object $(E, j , \varphi)$, the identity of $\delta$ is a unit of $( E, \delta , \varphi)$. So ${{\mathcal{D}}}(S)$ is a category. Furthermore each morphism is invertible, its inverse being its adjoint.

Because of equation (\[eq:def\_zeta\]), $\Psi' {\circ_{{{\mathcal{D}}}}}\Psi$ is a morphism. We deduce from the associativity $$\Psi_{E_c, E_d} \circ (\Psi_{E_b, E_c} \circ \Psi_{E_a, E_b}) =
(\Psi_{E_c, E_d} 