Given the 4-dimensional standard symplectic space $\mathbb{R}^4$ with Darboux-Coordinates $(z_1,z_2,z_3,z_4)=(x_1,x_2,y_1,y_2)$ and the symplectic structure $\omega=\sum dx_j \wedge dy_j$, i am trying to find a symplectic transformation which takes the Ellipsoid \begin{align} E: \frac{z_1^2}{a^2}+\frac{z_2^2}{b^2}+\frac{z_3^2}{c^2}+\frac{z_4^2}{d^2}=1 \end{align} to the 4-dimensional Ball with radius $r$, that is \begin{align} S^3_r(z)=\{z \in \mathbb{R}^4: z_1^2+z_2^2+z_3^2+z_4^2=r^2\}. \end{align} I know already that this is possible in $\mathbb{R}^2$ and that in general ($\mathbb{R}^{2n}$) we can write the ellipsoid $E$ in the form \begin{align} E: \sum a_i(x_i^2+y_i^2)=1, \end{align} using a linear symplectic transformation (see for example: Tabachnikov (source: https://www.math.psu.edu/tabachni/prints/Uspekhi.pdf )).

My questions are now:

1) How does this linear symplectic transformation explicitly look like, for example in $\mathbb{R}^4$?

2) Is it even possible to find a symplectic transformation $f$ (so that $\omega(fu,fv)=\omega(u,v)$ ) which takes the ellipsoid the a 4-Ball and if yes, how can one show this?


A general ellipsoid cannot be brought into a ball using only symplectic transformations.

Notice that the ellipsoid $E = \{ (z_1, z_2, z_3, z_4) | z_1^2/a^2 + z_2^2/b^2 + z_3^2/c^2 + z_4^2/d^2 \le 1\}$ is the unit ball for $\mathbb{R}^4$ equipped with the positive-definite scalar product $(v, w) := v^T \, \mathrm{diag}(a^{-2}, b^{-2}, c^{-2}, d^{-2}) \, w$. In particular, the standard unit ball is the unit ball of the standard scalar product with $a=b=c=d=1$. Hence, your question (2) amounts to the following: denoting $M$ the symmetric matrix $\mathrm{diag}(a^{-2}, b^{-2}, c^{-2}, d^{-2})$, is there a symplectic matrix $S$ such that $r^{-2} Id = S^TMS$ for some (radius) $r>0$?

Williamson's theorem is the key to answering this question: Given any $2n \times 2n$ positive-definite symmetric real matrix $M$, there exists a symplectic matrix $S \in Sp(2n)$ such that $S^T M S = \left( \begin{array}{cc} \Lambda & 0 \\ 0 & \Lambda \end{array} \right)$, where $\Lambda = \mathrm{diag}(\lambda_1, \dots, \lambda_n)$. Moreover, the $\lambda_j$'s are unique up to permutation; they form the symplectic spectrum of $M$.

Of course, any set of positive numbers $\lambda_j$'s can be obtained in this way, as one can start with $M = \left( \begin{array}{cc} \Lambda & 0 \\ 0 & \Lambda \end{array} \right)$. Hence, by the uniqueness part of Williamson's theorem, one sees that it is not always possible to have $S^TMS = r^{-2} Id$ for some $r$. In your notations, one has $a_j = 1/\lambda_j$; we deduce that there are infinitely many symplectically different ellipsoids, each determined by its symplectic spectrum.

This answers your question (2). As for your question (1), the proof of Williamson's theorem could be relevant.


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