Symplectic vector space

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.

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 linear map $A:(V,{\omega }_V)\to (W,{\omega }_W)$ is symplectic when it pulls back the target symplectic form to the source form. Nondegenerate bilinear form makes such a map injective. In equal finite dimensions it is an isomorphism. The automorphisms form the symplectic group and preserve the symplectic orientation on the ambient space and on every symplectic subspace.

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.

Let $V=C\oplus D$ be a symplectic orthogonal complement splitting into two symplectic subspaces of dimension two. In symplectic bases, the graph of a function $A:D\to C$ pulls back the symplectic form to ${\omega }_D+A^{\ast }{\omega }_C=(1+\det A){\omega }_D$. It is a symplectic subspace exactly when $\det A\ne -1$. Its symplectic orientation agrees with the orientation transported from $D$ exactly when $\det A>-1$. When $\det A<-1$, the transverse oriented pair $(C,{\Gamma }_A)$ has negative intersection sign, whereas two distinct complex lines in a compatible complex two-space have positive intersection sign.

If $T$ preserves a nondegenerate bilinear form $\omega$ that is alternating and satisfies $T^2=I$ on a finite-dimensional vector space over a field of characteristic different from two, thenThe two eigenspaces are orthogonal for $\omega$, since $\omega (v_{+},v_{-})=\omega (T{v}_{+},T{v}_{-})=-\omega (v_{+},v_{-})$. Each restricted form is nondegenerate: a vector annihilating its own eigenspace also annihilates the other and hence all of $V$. Both dimensions are even, say $2a$ and $2b$. Consequently $\mathrm{tr}T=2a-2b=\dim V-4b$ as an integer obtained from the eigenspace dimensions. Over the real numbers this gives the congruence $\mathrm{tr}T\equiv \dim V(\mathrm{mod}4)$.

The symplectic orthogonal complement of a subspace $W$ is $W^{\omega }=\{v:\omega (v,w)=0\text{ for every }w\in W\}$. The nondegenerate bilinear form gives $\dim W^{\omega }=\dim V-\dim W$. An isotropic subspace of a symplectic vector space satisfies $W\subset W^{\omega }$; a Lagrangian subspace satisfies equality.

A subspace $W$ of a symplectic vector space $(V,\Omega )$ is isotropic when $\Omega {|}_{W\times W}=0$. It has at most half the dimension of $V$; equality makes it a Lagrangian subspace. This use of isotropic concerns an alternating bilinear form, rather than an isotropic quadratic form.

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]

Write vectors in a symplectic basis as $v={\sum }_i(q_i{e}_i+p_i{f}_i)$ and $w={\sum }_i(Q_i{e}_i+P_i{f}_i)$. Their coordinate vectors and the matrix of a bilinear form areHere $I_n$ is the identity matrix, and $T$ denotes transpose. The ordering matters: interleaving the basis as $(e_1,f_1,\dots ,e_n,f_n)$ instead gives a block diagonal matrix with $2\times 2$ blocks $\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$. Some conventions reverse every $f_i$, giving $-J$ instead. Always check the displayed pairing convention.^[ref]

Over the real numbers, each summand is the signed area of a parallelogram in the $(q_i,p_i)$ coordinate plane. For $a=(2,1,1,1{)}^T$ and $b=(1,1,0,2{)}^T$, the two signed areas are $-1$ and $1$, so $\omega (a,b)=0$ although the vectors are linearly independent. The cancellation concerns the symplectic form; it does not assert that their dot product vanishes. Over a general field, the same determinant formula is algebraic, without a signed geometric interpretation.

On $K^4$, use the standard basis $\mathcal{U}=(u_1,u_2,u_3,u_4)$ and the alternating bilinear form with matrix $J$ above for $n=2$. A second symplectic basis isFor instance, $\omega (f_1,f_2)=1-1=0$, while $\omega (e_1,f_1)=\omega (e_2,f_2)=1$ and the cross-pairings vanish. The change of basis matrix, whose columns are these new basis vectors in $\mathcal{U}$, isThe symmetric matrix $C$ makes $S^{T}JS=J$, so $S$ is a symplectic matrix. Explicitly, multiplying the blocks gives a lower-right block $C^T-C=0$; the other blocks agree with $J$. For a fixed vector, change of basis means $[v{]}_{\mathcal{U}}=S[v{]}_{\mathcal{B}}$, henceTheir symplectic form gives $3\cdot 0+2\cdot 1-1\cdot 1-1\cdot 0=1$ in $\mathcal{U}$, and $1\cdot 0+1\cdot 1-1\cdot 0-1\cdot 0=1$ in $\mathcal{B}$. The separate coordinate-plane contributions change, but their sum does not.

The same matrix also represents an active symplectic linear map: keeping one basis fixed, it sends $(q,p)$ to $(q+Cp,p)$. The following diagram shows this active interpretation over the real numbers. Its preserved total pairing is precisely the test used for linear canonical transformations.

Construct a symplectic basis two vectors at a time. Choose $e\ne 0$; the nondegenerate bilinear form supplies $g$ with $\omega (e,g)\ne 0$, and set $f=g/\omega (e,g)$. For any vector $x$, remove its paired components byThen $\omega (x^{\prime },e)=\omega (x^{\prime },f)=0$. Thus, for $U=\mathrm{span}\{e,f\}$, the symplectic orthogonal complement gives a direct sum $V=U\oplus U^{\omega }$; its restricted bilinear form remains nondegenerate. Repeat there until no vectors remain, then put all $e_i$ before all $f_i$. This proves even dimension and existence together. Only the nonzero pairing is divided out, so the construction also works in characteristic two. The pair-splitting argument is described in Bilinear Forms by Keith Conrad (Keith Conrad, Bilinear Forms, §5, proof of Theorem 5.1 and Theorem 5.4).