linear isomorphism $g$ of $S$ such that ${\operatorname{id}}- g $ is invertible. For any such $g$, it is easily checked that $$A( g) :=  \tfrac{1}{2}  ( {\operatorname{id}}+ g  ) \circ ( {\operatorname{id}}- g ) ^{-1} \circ j$$ belongs to ${\operatorname{Sym}}(S, j)$. Denote by ${\operatorname{Mp}}_* (S,E)$ the subset of ${\operatorname{Mp}}(S,E) $ consisting of the pairs $( g, z)$ such that $g \in {\operatorname{Sp}}_* (S) $.

\[prop:def-index\] For any automorphism $(g, z) \in {\operatorname{Mp}}_* (S,E) $, one has $$z \operatorname{det}^{1/2} \bigl( \tfrac{1}{2} {\operatorname{id}}+  i A ( g)  \bigr) =  \frac{ i ^{m ( g, z)}
}{|\det({\operatorname{id}}- g)  |^{1/2}}$$ where $m(g, z ) \in {{\mathbb{Z}}}/  4 {{\mathbb{Z}}}$.

Since the inverse of $({\operatorname{id}}- g )^{-1} j $ is $-j ( {\operatorname{id}}-
  g)$, one has

[2]{} + i A (g) = & ( -j ( - g ) + i (+ g ) ) ( - g )\^[-1]{} j\
= & -j ( - g )(- g )\^[-1]{} j

where $\pi = \tfrac{1}{2}( {\operatorname{id}}-  i j )$ is the projector of $S
\otimes {{\mathbb{C}}}$ with image $E$ and kernel $\overline{E}$. From this one deduces that $$\begin{gathered}
 \det \bigl( \tfrac{1}{2} {\operatorname{id}}+  i A (g) \bigr) =   \frac{\det(
   \overline{\pi} - \pi g) }{\det ( {\operatorname{id}}- g )} 
