by $$\left( \begin{matrix} D   & 0  \\ 0 &  \overline{D}  \end{matrix} \right)$$ with $D$ the diagonal matrix with entries $ d_i = \frac{i}{2} (1+u_i)
( 1 - u_i)^{-1} 
$. Here the $u_i$’s are the eigenvalues of $h$, $h(e_i) = u _i e_i$. A straightforward computation using that $u_i$ is a complex number with modulus 1 gives $$( \tfrac{1}{2} + i d_i ) ( \tfrac{1}{2} +i \overline{d_i}) = \frac{1}{ (1- \overline{u}_i)^2}$$ so that $$\operatorname{det} \bigl( \tfrac{1}{2} {\operatorname{id}}+  i A ( j(h))  \bigr) =  \frac{1}{
  \operatorname{det}^2 ( {\operatorname{id}}- h^{-1})  }$$ This proves the result up to a plus or minus sign. $U_*(E)$ being connected, it suffices now to check the result for one element $h$. It is obvious for $h = -{\operatorname{id}}$, because $A(\iota (h) ) =0$.

Let $\tilde{U} (E)$ be the subgroup of $U(E) \times {{\mathbb{C}}}$ $$\tilde{U} (E) = \{ ( h, z) / \; z^2 = \det h \}$$ It is isomorphic to the twofold cover of $U(E)$. By equation (\[eq:2\]), $\iota$ lifts to the embedding from $\tilde{U}(E)$ to ${\operatorname{Mp}}(S,E)$ sending $(h, z)$ to $(\iota (h) , z^{-1})$. This map is a group morphism. Finally observe that $\tilde{U}_*(E)$ has 2 components, one containing $( -{\operatorname{id}}, \epsilon)$ and the other $( 