Find neccessary and sufficient condition on $A,B,C,D$ in order that $M^{t}JM = J$

Question

Let $n$ be a positive integer, $I$ the $n \times n$ identity matrix over $\Bbb C$, and $J$ the $2n \times 2n$ matrix given by $$J = \begin{bmatrix} 0 & I \\ -I & 0 \\ \end{bmatrix} $$

Let $M$ be a $2n \times 2n$ matrix over $\Bbb C$ of the form:

$$M = \begin{bmatrix} A & B \\ C & D \\ \end{bmatrix} $$ where $A,B,C,D$ are $n \times n$ matrices over $\Bbb C$. Find neccessary and sufficient conditions on $A,B,C,D$ in order that $M^{t}JM =J$

In the case $n=1$, we can easily prove that $M$ must have $\det M = 1$. So in this general case, I guess that the condition will be $\det(AD-BC) = 1$, but it seems more difficult. Suppose we have this bilinear form:

$$f((x_{1},x_{2},..., x_{2n}),(y_{1}, y_{2},...y_{2n})) = \sum_{i=1}^{n}x_{i}y_{n+i} - \sum_{i=1}^{n}y_{i}x_{n+i}$$ Then $f$ has matrix in standard basis is $J$. So what we must do is to find all linear operator which preserves $f$. That means find all $T$ which satisfies $f(T\alpha, T\beta) = f(\alpha,\beta)$. What I guess is $f(T\alpha, T\beta) = \det (T)f(\alpha,\beta)$, but it's quite difficult for me. Anyone here can help me solve this problem using this approach. Thanks so much

Answer 1

$$\begin{align}M^tJM &= \left[\begin{array}{cc}A^t & C^t\\B^t & D^t\end{array}\right]\left[\begin{array}{cc}0 & I\\-I & 0\end{array}\right]\left[\begin{array}{cc}A & B\\C & D\end{array}\right]\\ &= \left[\begin{array}{cc}-C^t & A^t\\-D^t & B^t\end{array}\right]\left[\begin{array}{cc}A & B\\C & D\end{array}\right]\\ &= \left[\begin{array}{cc}A^tC-C^tA & A^tD-C^tB\\B^tC-D^tA & B^tD-D^tB\end{array}\right]\\ &= \left[\begin{array}{cc}A^tC-(A^tC)^t & A^tD-C^tB\\-(A^tD-C^tB)^t & B^tD-(B^tD)^t\end{array}\right]\end{align}$$

Hence, it is necessary and sufficient that $A^tC$ and $B^tD$ are symmetric and that $A^tD-C^tB=I$. (Note that that last condition is equivalent to $\det M=1$ in the $n=1$ case, and that scalars--that is, $1\times 1$ matrices--are always symmetric.)

Answer 2

Note, if the matrix is defined over $\Bbb C$ you may (or may not) want the conjugate transpose..

Cameron has answered your question but it might be worth adding that as you have defined it $M$ is in the symplectic group $Sp(2n, \Bbb C)$ and it always satisfies $\det M=1$. To limit the possibilities to $\det M=\pm 1$ is easy enough:
$\det M^{t}JM=\det J$
$\det M^{t}\det J\det M= \left( \det M \right)^{2} \det J=\det J$
$\det M=\pm 1$
To prove $\det M \ne-1$ is a little more tricky but possible, see this review paper "On the determinant of symplectic matrices" by Mackey & Mackey http://homepages.wmich.edu/~mackey/detsymp.pdf

The symplectic group is the name for the group which preserves the nondegenerate, skew symmetric bilinear form; which is the bilinear form you have defined with $J$ in the standard basis.

To see why $M$ satisfies the symplectic matrix condition note that what you have written is equivalent to the standard way to write a skew-symmetric nondegenerate (i.e. symplectic) form, only you have expanded the matrix in components:
$f(x,y)=\sum_{i = 1}^{n}x_{i}y_{n+i} - \sum_{i=1}^{n}y_{i}x_{n+i}=x^{t}Jy$.
Now take an arbitrary transformation acting on this form, i.e. a matrix $M$, and you have:
$f(Mx,My)=(Mx)^{t}J(My)=x^{t}M^{t}JMy$.
Now demand $M$ preserves the symplectic form, i.e we want $f(Mx,My)=f(x,y)$, this implies:
$x^{t}M^{t}JMy=x^{t}Jy$
which is true for all $x,y$ iff:
$M^{t}JM=J$