Standard symplectic space

Let $𝑉 = 𝕂 2 𝑛$ with basis $(𝑥 1, \dots, 𝑥 𝑛, 𝑦 1, \dots, 𝑦 𝑛)$ . The $2 𝑛$ -dimensional standard symplectic space is $𝑉$ with the symplectic form $𝜔$ defined by #m/def/geoalg

𝜔 (𝑥 𝑖, 𝑦 𝑖) = - 𝜔 (𝑦 𝑖, 𝑥 𝑖) = 𝛿 𝑖 𝑗 𝜔 (𝑥 𝑖, 𝑥 𝑗) = 𝜔 (𝑦 𝑖, 𝑦 𝑗) = 0

or in terms of matrices

𝜔 (𝑥, 𝑦) = 𝑥 𝖳 [0 1 𝑛 - 1 𝑛 0] 𝑦

Every finite-dimensional symplectic space has a Darboux basis satisfying the above relations, hence every finite-dimensional symplectic space is equivalent to $𝕂 2 𝑛$ . In the case $𝑛 = 1$ , we get the Determinant form.

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