$\det ( g - {\operatorname{id}})$. So ${\operatorname{Mp}}_*(S,E)$ has at most 4 components. To conclude it suffices to prove that $m$ is onto. This follows from equation (\[eq:square\]), because the right hand side can be positive and negative. Actually, it is also proved in [@CoZe] lemma 1.7 that any loop in ${\operatorname{Sp}}_* (S)$ is contractible in ${\operatorname{Sp}}(S)$, which gives another proof that ${\operatorname{Mp}}_* (S,E)$ has 4 components.

A characterization of the index {#sec:char-index}
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Assume $S$ is two-dimensional. By choosing a basis $(e,f)$ of $S$ such that the symplectic product of $e$ with $f$ is equal to 1, we identify $S$ with ${{\mathbb{R}}}^2$ and ${\operatorname{Sp}}(S)$ with ${\operatorname{Sl}}(
2, {{\mathbb{R}}})$. For any $g \in {\operatorname{Sp}}(S)$, $$\det ( {\operatorname{id}}- g ) = 2 - \operatorname{tr}( g ) .$$ So the set of $g$ with $\det ( {\operatorname{id}}- g ) <0$ consists of the hyperbolic elements with negative trace, whereas the set of $g$ with $ \det (
{\operatorname{id}}- g ) >0$ consists of the elliptic elements together with the hyperbolic ones with a positive trace.

Recall that ${\operatorname{Sl}}( 2, {{\mathbb{R}}})$ is diffeomorphic to the product of the circle and the unit disc. Such a diffeomorphism is the map sending ${\theta}\in S^1$ and $(u,v) 