Questions about vector spaces equipped with a symplectic form, a non-degenerate, skew-symmetric bilinear form.
A symplectic vector space is a vector space $V$ over a field $F$ (for example the real numbers $\Bbb{R}$) equipped with a symplectic bilinear form.
A symplectic bilinear form is a mapping $ω : V \times V → F$ that is
- bilinear: linear in each argument separately,
- alternating: $ω(v, v) = 0\; $ holds for all $v ∈ V$, and
- non-degenerate: $ω(u, v) = 0\; $ for all $v ∈ V$ implies that $u$ is zero.
If the underlying field has characteristic not $2$, alternation is equivalent to skew-symmetry. If the characteristic is $2$, skew-symmetry is implied by, but does not imply alternation. In this case every symplectic form is a symmetric form, but not vice versa. Working in a fixed basis, $ω$ can be represented by a matrix. The conditions above say that this matrix must be skew-symmetric, non-singular, and hollow. This is not the same thing as a symplectic matrix, which represents a symplectic transformation of the space. If $V$ is finite-dimensional, then its dimension must necessarily be even because every skew-symmetric, hollow matrix of odd size has determinant zero. Notice the condition that the matrix be hollow is not redundant if the characteristic of the field is $2$. A symplectic form behaves quite differently from a symmetric form; for example, the scalar product on Euclidean vector spaces.
