Real Symplectic Group

The symplectic group, denoted as $ Sp(2n, R) $, consists of all 2n x 2n matrices with real entries that satisfy a unique geometric condition: when any matrix `A` from this group is transposed, multiplied by a specific matrix J, and then by `A` again, the outcome is the original matrix J. This relationship is captured by the formula $$ Sp(2n, R) = \{ A \in M_{2n}(R) \ : \ A^TJA = J \} $$.

The matrices within the symplectic group $ Sp(2n,\mathbb{R}) $ are characterized by their square shape and their dimensions, $ 2n \times 2n $, filled with real numbers. These matrices, referred to as $ A $, play a crucial role in preserving a certain geometric structure encapsulated by the equation $ A^T J A = J $, where $ J $ signifies a predetermined antisymmetric matrix, and $ A^T $ represents the transpose of $ A $.

This critical relationship, $ A^T J A = J $, ensures that the transformations depicted by $ A $ maintain specific geometric and physical relationships, pivotal in the study of classical and quantum mechanics.

The matrix `J` described here takes the form of a block matrix, with the top right quadrant being an identity matrix of size `n` and its bottom left quadrant the negative of an identity matrix of the same dimension. The remaining quadrants, top left and bottom right, are filled with zeroes. The configuration of $ J $ is thus given by: $$ J = \begin{pmatrix} 0 & I_n \\ -I_n & 0 \end{pmatrix} $$.

This collection of matrices, apart from their geometric significance, ensures the conservation of a bilinear form related to matrix J. In other words, symplectic transformations keep the scalar product, as defined by J, unchanged, which has profound implications, particularly in Hamiltonian mechanics where the conservation of motion equations is essential, achieved through symplectic transformations.

An Illustrative Example

To grasp a simple instance of a matrix within the symplectic group $ Sp(2n, \mathbb{R}) $, consider $ n = 1 $. At this level, we encounter a symplectic matrix with dimensions $ 2 \times 2 $.

For $ n = 1 $, we're given the matrix $ J $ as follows:

$$ J = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} $$

For a matrix $ A $ to be part of $ Sp(2, \mathbb{R}) $, it must fulfill the condition $ A^T J A = J $.

A straightforward illustration of such a matrix $ A $ is given by the rotation matrix:

$$ A = \begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{pmatrix} $$

Here, $ \theta $ represents any real angle.

To confirm that $ A $ indeed belongs to $ Sp(2, \mathbb{R}) $, let's perform the operation $ A^T J A $ and demonstrate that the outcome precisely matches $ J $:

$$ A^T = \begin{pmatrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{pmatrix} $$

$$ A^T J = \begin{pmatrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{pmatrix} \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} = \begin{pmatrix} -\sin(\theta) & \cos(\theta) \\ -\cos(\theta) & -\sin(\theta) \end{pmatrix} $$

$$ A^T J A = \begin{pmatrix} -\sin(\theta) & \cos(\theta) \\ -\cos(\theta) & -\sin(\theta) \end{pmatrix} \begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} = J $$

This demonstrates that, irrespective of the angle $ \theta $, the rotation matrix $ A $ constitutes an element of the symplectic group $ Sp(2, \mathbb{R}) $.

Example 2

Here's a practical example of a symplectic matrix when $ n=2 $.

\[
A = \begin{pmatrix}
0 & 1 & 0 & 0 \\
-1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & -1 & 0
\end{pmatrix}
\]

In this instance, the matrix J under consideration is:

\[
J = \begin{pmatrix}
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
-1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0
\end{pmatrix}
\]

This J matrix is structured into two $ 2 \times 2 $ blocks: the top-right block is an identity matrix $ I_2 $, and the bottom-left block is the negative of an identity matrix $ -I_2 $, with the principal diagonal blocks being zeros.

The matrix $ A $ exhibits the symplectic property that $ A^T J A = J $, with $ J $ being the standard symplectic matrix for $ n=2 $:

$$
A^T = \begin{pmatrix}
0 & -1 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 0 & -1 \\
0 & 0 & 1 & 0
\end{pmatrix}
$$

$$
A^T J = \begin{pmatrix}
0 & -1 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 0 & -1 \\
0 & 0 & 1 & 0
\end{pmatrix} \begin{pmatrix}
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
-1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0
\end{pmatrix} =
\begin{pmatrix}
0 & 0 & 0 & -1 \\
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 \\
-1 & 0 & 0 & 0
\end{pmatrix}
$$

$$ A^TJA = \begin{pmatrix}
0 & 0 & 0 & -1 \\
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 \\
-1 & 0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
0 & 1 & 0 & 0 \\
-1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & -1 & 0
\end{pmatrix} =
\begin{pmatrix}
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
-1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0
\end{pmatrix}
$$

$$ A^T J A = J $$

This calculation solidly confirms that $ A $ qualifies as a symplectic matrix.