# Calculation of the standard symplectic space matrix

I am learning there is an important connection between Hamiltonian formalisms and Symplectic Geometry. It seems like the Newtonian Mechanics are described on what is called the standard symplectic space, which is given by:

$$\Omega=\bigg(\matrix{0&I_n\\-I_n&0}\bigg)$$

This is a $2n\times2n$ matrix, where the $I_n$'s are identity matrices of $n\times n$. So the dimension of this space is proportional to the number of degrees of freedom of the system (in phase space there are $2n$ d.o.f.).

My question is: how do they construct this matrix? Can there exist new machanical systems (relativistic, quantum) so that this matrix is modified? In such case, how are these constructed?

I would appreciate good sources to study this kind of things, because I only found wikipedia pages, and also the book Mathematical Methods of Classical Mechanics by V. I. Arnold, but I find it too formal, and hard to follow.

1. The geometric setting for a Hamiltonian theory is often taken to be a $2n$-dimensional real symplectic manifold $(M,\omega)$, where $\omega$ is a closed non-degenerate real 2-form.
2. In a coordinate neighborhood $U$, the 2-form $\omega$ is given as $$\left.\omega\right|_{U}~=~\frac{1}{2}\sum_{I,J=1}^{2n}\omega_{IJ}(z)~ \mathrm{d}z^I \wedge\mathrm{d}z^J,\tag{1}$$ where $\omega_{IJ}=-\omega_{JI}$ is a non-degenerate antisymmetric real matrix.
3. The inverse matrix $$(\omega^{-1})^{IJ}~=~\{z^I,z^J\}_{PB}\tag{2}$$ gives rise to a Poisson bracket.
4. The Darboux theorem states that there locally (in a sufficiently small open neighborhood $U$) exist Darboux/canonical coordinates $$z^I~=~\left(q^1,\ldots, q^n,p_1, \ldots, p_n\right),\tag{3}$$ where $\omega_{IJ}$ is on the form $$\omega ~=~\begin{bmatrix} {\bf 0}_{n \times n} & -{\bf 1}_{n \times n} \cr {\bf 1}_{n \times n} & {\bf 0}_{n \times n} \end{bmatrix}, \tag{4}$$ or equivalently, $$\left.\omega\right|_{U}~=~\sum_{i=1}^n\mathrm{d}p_i \wedge\mathrm{d}q^i.\tag{5}$$
• thank you for your answer, so $\omega$ and its inverse only depends on the coordinates? that is why the standard basis has such form, because one always considers Darboux coordinates? – Saavestro Oct 27 '16 at 15:34
• $\uparrow$ Yes & Yes. – Qmechanic Oct 27 '16 at 18:00