How to visualize symplectic transformations?

Question

This is a follow up question to this question.

Let $\omega$ be an skew-symmetric bilinear form on $\mathbb{R}^{2n}$, which is unique up to change of basis. It is given by the formula $$\omega(\mathbf{x},\mathbf{y}) = \sum_{i=1}^n{x_iy_{i+n}-y_ix_{i+n}}$$

We can then write out the definition of $\mathrm{Sp}(n,\mathbb{R})$ as the group of linear operators $A: \mathbb{R}^{2n} \to \mathbb{R}^{2n}$ such that they satisfy the condition:

$$\omega(A\mathbf{x},A\mathbf{y}) = \omega(\mathbf{x},\mathbf{y})$$ for all $\mathbf{x},\mathbf{y} \in \mathbb{R}^{2n}$.

Elements of $\mathrm{Sp}(n,\mathbb{R})$ are called symplectic transformations.

This is an analog of orthogonal transformations. If $\omega$ were chosen to be an inner product $\langle \cdot , \cdot \rangle$ instead, then the condition $$\langle A \mathbf{x} , A \mathbf{y} \rangle = \langle \mathbf{x,y} \rangle$$ is equivalent to preserving distances between points in $\mathbb{R}^n$, so a more down-to-earth point of view.

So my question is, is there a geometric meaning to the definition of symplectic transformations, like "distance-preserving transformation" in the orthogonal case? If not, is there maybe some other way to think about them?

Answer 1

Write $x=\left(\begin{array}{c} a\\ b \end{array}\right),\,y=\left(\begin{array}{c} c\\ d \end{array}\right)$ with $a,\,b,\,c,\,d\in\mathbb{R}^{n}$ so $\omega\left(x,\,y\right)=a\cdot d-b\cdot c$. In the case $n=1$, this is a determinant, and $\left(\begin{array}{c} a\\ b\\ 0 \end{array}\right)\times\left(\begin{array}{c} c\\ d\\ 0 \end{array}\right)=\omega\left(x,\,y\right)\hat{k}$. More generally, $\omega\left(x,\,y\right)$ is a sum of $n$ such expressions.

Symplectic transformations are of especial interest in studying systems obeying Hamilton's equations. Any function of the phase space coordinates varies over time as documented here. One of the conservation laws that results is that Poisson brackets are preserved. It leads to an elegant phase space flow $q(t)=\exp(-t\{H,\,\cdot\})q(0),\,p(t)=\exp(-t\{H,\,\cdot\})p(0)$. Such a flow is probably the best way one can "visualize" the consequences.

Answer 2

Here is a try: ${\Bbb R}^{2n}$ is the direct sum of the subspaces $E_i={\rm span}\{e_i,e_{i+n}\}$ for $i=1,...,n$. These subspaces come an area form and with projections which (deceptively) appears to be orthogonal projections in these coordinates. A symplectic transformation should preserve the sum of areas in these subspaces.

I say deceptively because it is simply our choice of coordinates which seem to imply some orthogonallity between the $E_i$'s. In fact, you could take any $n$ two-dimensional subspaces $E_1,...,E_n$ whose direct sum yields ${\Bbb R}^{2n}$. Define area forms $\omega_i$ in each $E_i$, evaluated on projections along the other spaces and $\sum_i \omega_i$ gives you a symplectic form. A symplectic transformation is then a transformation that preserves the sum or these 2-area forms. Any symplectic form can be transformed to the above by a suitable change of coordinates.

But this, of course, does not give you a good global geometric picture of what the symplectic transformation does. Also this picture is not independent of the choices made.