-{\operatorname{id}}, -\epsilon)$ with $\epsilon ^2 = ( -1) ^n$. So the index $m$ takes two distinct values on $\tilde{U}_*(E)$.

Metalinear group of a Lagrangian subspace
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Let $\Lambda$ be a Lagrangian subspace of $S$. Let ${\operatorname{Sp}}( S, \Lambda,
j \Lambda)$ be the subgroup of ${\operatorname{Sp}}(S)$ consisting of the elements preserving $\Lambda$ and $j \Lambda$. Let ${\operatorname{Ml}}( \Lambda)$ be the metalinear group of $\Lambda$, i.e. the subgroup of ${\operatorname{Gl}}( \Lambda) \times {{\mathbb{C}}}^*$ consisting of the pairs $(h , z)$ such that $ z^2 = \det h $.

For any $(h,z) \in {\operatorname{Ml}}( \Lambda)$, there is a unique pair $( g, z') \in {\operatorname{Sp}}( S, \Lambda,
j \Lambda) \times {{\mathbb{C}}}^*$ such that

-   $h$ is the restriction of $g$ to $\Lambda$.

-   $z'/z \in {{\mathbb{R}}}_+$.

The map $ j_\Lambda : {\operatorname{Ml}}( \Lambda) \rightarrow {\operatorname{Mp}}(S,E)$ sending $(h,z )$ into $(g, z')$ is an injective group morphisms, with image the set of $(g,z')$ such that $g \in  {\operatorname{Sp}}( S, \Lambda,
j \Lambda)$.

Let $(e_i)$ be a basis of $\Lambda$, orthonormal for the scalar product ${\omega}( x, j y)$. Then $g \in {\operatorname{Sp}}( S, \Lambda,
j \Lambda)$ if and only if its matrix in the basis $e_1, \ldots , e_n,
je_1$, $\ldots$, $j e_n $ 