Page 205 of the book Classical Groups and Algebraic K-Theory defines a symplectic module to be an arbitrary quadratic module $(M,h,q)$ over a form ring $(R,\Lambda)$ with $(J,\varepsilon)$ where $J=\operatorname{id}_R$, $\varepsilon=-1$ and $\Lambda=R$. In particular, this implies that $R$ is commutative.
Notably, the above differs from any instance of a trace-valued skew-Hermitian module (defined on the same page) in that the latter can't have $J=\operatorname{id}_R$ (but must still have $J^2=\operatorname{id}_R$). However, I'm thinking we can turn the former into an instance of the latter, making the former seemingly redundant!
My observation is that given the above form ring, we can define the following form ring $(R',\Lambda')$:
- $R'$ is the $2 \times 2$ matrices over $R$,
- $\Lambda' = \{\lambda I_{2\times2} \mid \lambda \in R\}$,
- $\begin{pmatrix}a & b \\ c & d \end{pmatrix}^{J'} = \begin{pmatrix}d & -b \\ -c & a \end{pmatrix}$,
- $\varepsilon' = -1$.
Any quadratic module over $(R',\Lambda')$ becomes a quadratic module over $(R,\Lambda)$ by restriction of scalars along the inclusion $(R,\Lambda) \hookrightarrow(R',\Lambda')$. Question: Can all quadratic modules over the symplectic form ring $(R,\Lambda)$ be obtained this way?
