Four-dimensional symplectic basis change

On $K^4$, use the standard basis $\mathcal{U}=(u_1,u_2,u_3,u_4)$ and the alternating bilinear form with matrix $J$ above for $n=2$. A second symplectic basis isFor instance, $\omega (f_1,f_2)=1-1=0$, while $\omega (e_1,f_1)=\omega (e_2,f_2)=1$ and the cross-pairings vanish. The change of basis matrix, whose columns are these new basis vectors in $\mathcal{U}$, isThe symmetric matrix $C$ makes $S^{T}JS=J$, so $S$ is a symplectic matrix. Explicitly, multiplying the blocks gives a lower-right block $C^T-C=0$; the other blocks agree with $J$. For a fixed vector, change of basis means $[v{]}_{\mathcal{U}}=S[v{]}_{\mathcal{B}}$, henceTheir symplectic form gives $3\cdot 0+2\cdot 1-1\cdot 1-1\cdot 0=1$ in $\mathcal{U}$, and $1\cdot 0+1\cdot 1-1\cdot 0-1\cdot 0=1$ in $\mathcal{B}$. The separate coordinate-plane contributions change, but their sum does not.

The same matrix also represents an active symplectic linear map: keeping one basis fixed, it sends $(q,p)$ to $(q+Cp,p)$. The following diagram shows this active interpretation over the real numbers. Its preserved total pairing is precisely the test used for linear canonical transformations.