# decomposition of general symplectic matrix

Suppse $S$ be a $2n \times 2n$ symplectic matrix $$S = \left( \begin{array}{cc} A & B\\ C & D \end{array} \right) .$$

By definition, $S^tJS$=J, where

$$J = \left( \begin{array}{cc} 0 & I\\ -I & 0 \end{array} \right) .$$

Now there is a theorem states that $S=S_1S_2$ where $S_1$ and $S_2$ are free symplectic matrices. This means,

$$S_i = \left( \begin{array}{cc} A_i & B_i\\ C_i & D_i \end{array} \right), det(B_i)\neq 0.$$

Now suppose $S$ be symplectic matrix with respect to different skew bilinear form i.e $S^t \theta S=\theta$, where

$$\theta = \left( \begin{array}{cc} 0 & \theta _1\\ -\theta _1^t & 0 \end{array} \right) .$$ Is the decomposition of of $S$ into two free symplectic matrices still possible?

• Have you tried generalizing the steps of the proof of the theorem? Do you have a reference for that theorem which has a proof? – Omnomnomnom Jun 15 '16 at 16:20
• books.google.de/… – Alisha Jun 15 '16 at 18:24