=  ( -1)^n \frac{\det ( \pi g : E \rightarrow
E )}{ \det ( {\operatorname{id}}- g ) }   \end{gathered}$$ The inverse of $\pi_{E, gE}$ is the restriction of $\pi$ from $gE$ to $E$. So $$\begin{gathered}
 \label{eq:2}
 z^2 = \det ( g^{-1} \pi_{E, gE} : E \rightarrow E) = \operatorname{det}^{-1} ( \pi g : E \rightarrow
E )  .\end{gathered}$$ The two previous equations imply that $$\begin{gathered}
 \label{eq:square}
 z^ 2 \det \bigl( \tfrac{1}{2} {\operatorname{id}}+  i A (g) \bigr) =  \frac{(-1)^n}{  \det ( {\operatorname{id}}- g )
},\end{gathered}$$ which concludes the proof.

We call $m(g,z)$ the index of $(g,z)$. This defines a locally constant map $m$ from ${\operatorname{Mp}}_* (S,E)$ to $ {{\mathbb{Z}}}/ 4 {{\mathbb{Z}}}$ which distinguishes the components of ${\operatorname{Mp}}_* (S,E)$, as proves the following proposition.

${\operatorname{Mp}}_* (S,E) $ has four components: $m^{-1} (0)$, $m^{-1} (1)$, $m^{-1} (2)$ and $m^{-1} (3)$.

Observe also that $m( -{\operatorname{id}}, e^{i  \frac{\pi}{2} p } )= p $ for $p
 =n$ mod 2. When $S$ is two-dimensional one can can easily characterize the index of $(g,z)$ in term of the trace of $g$ and the argument of $z$, cf. equation (\[eq:ind2d\]).

It is proved in [@CoZe] lemma 1.7, that ${\operatorname{Sp}}_* (S)$ has two components, distinguished by the sign of 