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The mapis a nonzero linear functional: otherwise $v$ would lie in the radical of a bilinear form, contradicting nondegeneracy. Its kernel is $v^{\perp }$, so the rank-nullity theorem gives $\dim v^{\perp }=n-1$. Antisymmetry gives $⟨v,v⟩=-⟨v,v⟩=0$, hence $v\in v^{\perp }$.

Suppose $w\in W$ is orthogonal to every vector of $W$. Since $W\subseteq v^{\perp }$, it is also orthogonal to $v$, and therefore toFor a nondegenerate bilinear form, $(v^{\perp }{)}^{\perp }=\mathbb{R}v$: both sides have dimension one and the latter is contained in the former. Thus $w\in W\cap \mathbb{R}v=\{0\}$, proving that the restriction to $W$ is nondegenerate.

The space $W$ has dimension $n-2$ and again carries a nondegenerate antisymmetric form. Induction, starting from the zero-dimensional space, shows that $\dim W$ is even. Therefore $n=\dim W+2$ is even. Equivalently, every finite-dimensional symplectic vector space has even dimension.

The primitive exterior square of a symplectic vector space is the kernel of its symplectic contraction of an exterior square. In dimension four, the invariant inverse-form bivector $\Omega$ has nonzero wedge square. Choose $\mathrm{vol}=\Omega \wedge \Omega /2$; the symmetric bilinear form $\xi \wedge \eta =B(\xi ,\eta )\mathrm{vol}$ is nondegenerate on ${\Lambda }^{2}V$. The primitive subspace is ${\Omega }^{\perp }$, and therefore inherits a nondegenerate form of dimension five. The symplectic Lie algebra acts on it by infinitesimal orthogonal transformations.

For a finite-dimensional symplectic vector space, the symplectic adjoint of a linear map $T$ is the unique linear map $T^{\ast }$ satisfying the displayed identity. Existence and uniqueness follow because the nondegenerate bilinear form identifies the vector space with its dual space. If $\omega$ has matrix $J$ in a basis, then $T^{\ast }=J^{-1}{T}^{\mathsf{T}}J$. Thus $(ST{)}^{\ast }=T^{\ast }{S}^{\ast }$ and $(T^{\ast }{)}^{\ast }=T$, and membership in the symplectic Lie algebra is equivalent to $T^{\ast }=-T$. This operation is linear over the ground field; it does not involve complex conjugation. It preserves diagonalizability and the property of being a nilpotent linear map, since it is conjugate to matrix transpose.

A symplectic basis organizes a nondegenerate alternating bilinear form into independent pairs. Let $(V,\omega )$ be a finite-dimensional vector space over a field $K$, with $\omega$ a nondegenerate bilinear form satisfying $\omega (v,v)=0$ for every vector $v$. Its dimension is even, say $2n$, and a symplectic basis is an ordered basis $\mathcal{B}=(e_1,\dots ,e_n,f_1,\dots ,f_n)$ such thatHere ${\delta }_{ij}$ is the Kronecker delta. Each $(e_i,f_i)$ is a hyperbolic pair; different pairs have zero mutual pairing. These conventions apply in every characteristic. In characteristic two, $-1=1$, but the condition $\omega (v,v)=0$ is still essential: a symmetric bilinear form need not be an alternating bilinear form. The existence theorem and the characteristic caveat are given in Bilinear Forms by Keith Conrad (Keith Conrad, Bilinear Forms, §1, Theorem 1.6; §5, Theorems 5.1 and 5.4 and Definition 5.3).

The benefit is a common coordinate language for every symplectic vector space of the same finite dimension over the same field. The span of all $e_i$, and separately of all $f_i$, is a Lagrangian subspace; each paired span is a symplectic subspace. In Hamiltonian mechanics, the corresponding pairing of position and momentum coordinates describes phase space and canonical transformations. On a symplectic manifold, a symplectic basis describes one tangent space; Darboux coordinates additionally describe a whole local neighborhood.^[ref]

Let $\mathfrak{a}$ be a toral subalgebra and an abelian Lie algebra in the symplectic Lie algebra, over an algebraically closed field of characteristic zero. Its defining representation has a weight decomposition. The symplectic Lie algebra identity gives $(\lambda (a)+\mu (a))\omega (v,w)=0$ for $v\in V_{\lambda }$, $w\in V_{\mu }$, so $\omega (V_{\lambda },V_{\mu })=0$ unless $\mu =-\lambda$. Since $\omega$ is a nondegenerate bilinear form, its restrictions give a perfect pairing between $V_{\lambda }$ and $V_{-\lambda }$, and a nondegenerate bilinear form on $V_0$.

An element of the centralizer of a Lie subalgebra preserves each weight space. On $V_0$ it is any element of $\mathfrak{sp}(V_0)$. On each nonzero pair, its action $A$ on $V_{\lambda }$ is arbitrary, and the perfect pairing uniquely forces the action on $V_{-\lambda }$ to be the negative dual map $-A^{\ast }$. These independent choices preserve Lie brackets, proving the displayed direct sum of Lie algebras; the sum runs over one representative per nonzero pair.

This centralizer of a Lie subalgebra is an abelian Lie algebra precisely when $V_0=0$ and all the nonzero weight spaces are one-dimensional. Indeed, ${\mathfrak{gl}}_m$ is nonabelian for $m\ge 2$ by $[E_{11},E_{12}]=E_{12}$, while every nonzero symplectic vector space contains a two-dimensional symplectic subspace on which $[\mathrm{diag}(1,-1),E_{12}]=2E_{12}$. In the stated case the surviving summands are copies of ${\mathfrak{gl}}_1$ and commute. Repeated weights of a representation, including a zero one, are therefore exactly what obstructs this commutativity.

For a symplectic vector space $(V,\omega )$, contraction is the map $c:{\Lambda }^{2}V\to \mathbb{C}$ defined by $c(v\wedge w)=\omega (v,w)$. It is equivariant for the symplectic Lie algebra. In dimension four its five-dimensional kernel is irreducible of highest weight ${\omega }_2$, while the invariant inverse-form bivector spans a complementary trivial Lie algebra representation.

Vertices are Lagrangian subspaces of a $2m$-dimensional symplectic vector space, joined when they intersect in codimension one. A pair with intersection dimension $m-r$ admits a symplectic basis in which $U=⟨e_1,\dots ,e_m⟩$ and $W=⟨f_1,\dots ,f_r,e_{r+1},\dots ,e_m⟩$. Replacing $e_i$ successively by $f_i$ gives an isotropic path of length $r$, while one edge can increase intersection dimension with a fixed target by at most one. Equal-distance ordered pairs have the same adapted form, so the symplectic group acts distance-transitively.

A subspace of a symplectic vector space is symplectic if the restricted symplectic form is nondegenerate. It has a symplectic orthogonal complement and the ambient space is the direct sum of those two subspaces.

A symplectic vector bundle has a smooth fiberwise symplectic form, making each fiber a symplectic vector space. Its structure group reduces to the symplectic group. A symplectic normal bundle is an important example.

A symplectic vector space is a vector space equipped with a nondegenerate bilinear form $\omega$ that is an alternating bilinear form: $\omega (v,v)=0$ for every vector $v$. This implies $\omega (v,w)=-\omega (w,v)$. When the characteristic is two, that latter equation alone does not imply the alternating condition. A symplectic vector space of finite dimension has even dimension: splitting off a hyperbolic pair leaves a smaller nondegenerate bilinear form, and repeating gives a decomposition into two-dimensional planes.