Symplectic Matrices have determinant=1
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Let $\Omega=\begin{pmatrix} 0_n & I_n \\ -I_n & 0_n \end{pmatrix}$ be a $2n\times 2n$ matrix. A symplectic matrix $S$ is one for which

(1)
\begin{align} S \cdot \Omega\cdot S^{\textrm{T}}= \Omega\ . \end{align}

Writing $S$ as a block matrix of $n\times n$ blocks $S=\begin{pmatrix} A & B \\ C & D \end{pmatrix}$, the above equation is equivalent to the following three conditions

(2)
\begin{eqnarray} -B A^{\textrm{T}}+ A B^{\textrm{T}}&=&0_n \\ -B C^{\textrm{T}}+A D^{\textrm{T}}&=&I_n \\ -D C^{\textrm{T}} + C D^{\textrm{T}}&=&0_n \end{eqnarray}

It is easy to see that $\det(S)=\pm1$. However, one can show that $\det(S)=+1$ as we will now show using elementary methods.

• Assuming $\det(A)\neq0$ when $S$ is written as a block matrix, one has $\det(S)= \det(AD-ACA^{-1}B) =\det(A) \det(D- CA^{-1}B)$. Thus, we need to show that $\det(A^{-1}) =\det(D- CA^{-1}B)$.
• The second condition (of the three conditions given above) implies that $\det(A^{-1})=\det(D^{\textrm{T}}-A^{-1} B C^{\textrm{T}})=\det(D- C B^{\textrm{T}}(A^{-1})^{\textrm{T}})$.
• Now using the first condition, we see that $B^{\textrm{T}} (A^{-1})^{\textrm{T}}= A^{-1} B$. Hence $\det(D- C B^{\textrm{T}}(A^{-1})^{\textrm{T}}) = \det(D- CA^{-1}B)=\det(A^{-1})$.

This completes the proof that $\det(S)=+1$ under the assumption that $\det(A)\neq0$. The other cases can be dealt with in a similar fashion and we won't pursue them here. A more sophisticated proof makes use of the Pfaffian (of an antisymmetric matrix).

(3)
\begin{align} \textrm{Pf}(\Omega) = \textrm{Pf}(S \cdot \Omega\cdot S^{\textrm{T}}) = \det(S)\ \textrm{Pf}(\Omega) \implies \det(S)=+1\ , \end{align}

where we use a particular property of the Pfaffian.