Determinant of symplectic matrix

Great timing! Check out this recent preprint: Donsub Rim, "An Elementary Proof That Symplectic Matrices Have Unit Determinant", http://arxiv.org/abs/1505.04240. It proceeds by decomposing the inequality $\det(S^TS+I)\geq1$.

If $S$ is a symplectic matrix, it preserves the standard symplectic form $\omega=\sum_{i=1}^n dx_i\wedge dy_i$, i.e. $S^*\omega=\omega$. Note that $\displaystyle\frac{\omega^n}{n!}$ is the standard volume form of $\mathbb{R}^{2n}$. Now we have \begin{eqnarray} \det(S)\frac{\omega^n}{n!}=S^*\left(\frac{\omega^n}{n!}\right)=\frac{\omega^n}{n!} \end{eqnarray} So $\det(S)=1$.