# Symplectic transformation of ellipsoid in standard symplectic space

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?

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.