Existence of Symplectic Basis

TheoremSymplectic Basis Theorem

Every symplectic vector space $(V, \omega)$ of dimension $2n$ admits a symplectic basis, i.e., a basis $(e_1, \ldots, e_n, f_1, \ldots, f_n)$ such that:

$\omega(e_i, e_j) = 0, \quad \omega(f_i, f_j) = 0, \quad \omega(e_i, f_j) = \delta_{ij}$

Consequently, all symplectic vector spaces of the same dimension are symplectomorphic to $(\mathbb{R}^{2n}, \omega_0)$ with the standard symplectic form.

This fundamental theorem reveals the remarkable rigidity of symplectic linear algebra: unlike Riemannian geometry where inner products vary continuously, symplectic forms of the same dimension are all equivalent up to isomorphism.

Proof Strategy: The proof proceeds by induction on $n = \frac{1}{2}\dim V$ . The key observation is that symplectic complements of symplectic subspaces are again symplectic, allowing us to decompose the space into orthogonal symplectic planes.

Remark

This theorem is the linear analogue of Darboux's theorem for symplectic manifolds. While Darboux's theorem is a local result requiring differential geometry, the linear version is purely algebraic and global.

DefinitionCanonical Coordinates

Given a symplectic basis $(e_i, f_i)$ , we define canonical coordinates $(q_1, \ldots, q_n, p_1, \ldots, p_n)$ by:

$v = \sum_{i=1}^n q_i e_i + p_i f_i$

In these coordinates, the symplectic form takes the canonical form:

$\omega = \sum_{i=1}^n dq_i \wedge dp_i$

The canonical coordinates $(q, p)$ are the position and momentum coordinates familiar from classical mechanics. The symplectic basis theorem guarantees that such coordinates always exist locally, providing the mathematical foundation for Hamiltonian mechanics.

ExampleStandard Symplectic Structure

Consider $\mathbb{R}^{2n}$ with the standard basis $\{e_1, \ldots, e_n, f_1, \ldots, f_n\}$ where:

$e_i = (0, \ldots, 0, 1, 0, \ldots, 0), \quad f_i = (0, \ldots, 0, 1, 0, \ldots, 0)$

with 1 in the $i$ -th and $(n+i)$ -th positions respectively. The standard symplectic form is:

$\omega_0\left(\sum x_i e_i + y_i f_i, \sum x_i' e_i + y_i' f_i\right) = \sum_{i=1}^n (x_i y_i' - y_i x_i')$

Remark

The dimension constraint $\dim V = 2n$ is necessary: no odd-dimensional space admits a non-degenerate antisymmetric bilinear form. This is because the determinant of an odd-dimensional antisymmetric matrix vanishes.

The symplectic basis theorem also implies that the symplectic group $\text{Sp}(2n, \mathbb{R})$ acts transitively on the space of symplectic bases. This transitivity is crucial for understanding the geometry of the Lagrangian Grassmannian and other symplectic quotient spaces.

DefinitionSymplectic Linear Isomorphism

The theorem establishes a canonical isomorphism:

$\phi: (V, \omega) \to (\mathbb{R}^{2n}, \omega_0)$

given by $\phi(e_i) = (0, \ldots, 0, 1, 0, \ldots, 0)$ and $\phi(f_i) = (0, \ldots, 0, 1, 0, \ldots, 0)$ where the 1 appears in positions $i$ and $n+i$ respectively.