Linear Symplectic Geometry

Symplectic Vector Space

Skew-Symmetric Bilinear Maps

Let $V$ be an $m$-dimensional vector space over $\mathbb{R}$, and let $\Omega :V\times V\to \mathbb{R}$ be a bilinear map. The map $\Omega$ is skew-symmetric if $\Omega (u,v)=-\Omega (v,u)$, for all $u,v\in V$.

Standard Form for Skew-symmetric Bilinear Maps

Let $\Omega$ be a skew-symmetric bilinear map on $V$. Then there is a basis $u_1,\dots ,u_k,e_1,\dots ,e_n,f_1,\dots ,f_n\text{ of }V\text{ such that}$ The following conditions hold:

\Omega(u_i, v) &= 0, & \text{for all } i \text{ and all } v \in V \ \Omega(e_i, e_j) &= 0 = \Omega(f_i, f_j), & \text{for all } i, j, \text{ and} \ \Omega(e_i, f_j) &= \delta_{ij}, & \text{for all } i, j. \end{align*}$$

Moreover, $V=U\oplus W$, with $U=\text{span}\{u_1,\dots ,u_k\}$, $W=\text{span}\{e_1,\dots ,e_n,f_1,\dots ,f_n\}$

Symplectic Vector Space

A symplectic vector space is a finite dimensional real vector space furnished with a symplectic form, that is, a closed non-degenerate 2-form. An equivalent definition for Symplectic vector space can be considered as follow: The map $\tilde{\Omega }:V\to V^{\ast }$ is the linear map defined by $\tilde{\Omega }(v)(u)=\Omega (v,u).$ The kernel of $\tilde{\Omega }$ is the subspace $U$ in the previous section. A skew-symmetric bilinear map $\Omega$ is symplectic (or nondegenerate) if $\tilde{\Omega }$ is bijective, i.e., $U=\{0\}$. The map $\Omega$ is then called a linear symplectic structure on $V$, and $(V,\Omega )$ is called a symplectic vector space.

Symplectic Complement

Let $(V,\omega )$ be a symplectic vector space. The symplectic complement of a subspace $W\subset V$ is the subspace $W^{\omega }=\{v\in V\mid \omega (v,w)=0,\text{ for all }w\in W\}.$

Classification of Subspaces

Let $(V,\omega )$ be a symplectic vector space. A subspace $W\subset V$ is called

• isotropic if $W\subset W^{\omega }$, i.e. $\omega {|}_W=0$,
• coisotropic if $W^{\omega }\subset W$,
• symplectic if $W\cap W^{\omega }=\{0\}$, i.e. $\omega {|}_W$ is non-degenerate.
• Lagrangian if $W=W^{\omega }$.

Lemma

For any subspace $W\subset (V,\omega )$, there holds $\dim W+\dim W^{\omega }=\dim V,W^{\omega \omega }=W.$

Proof Let $\iota :V\to V^{\ast }$ be the symplectic duality map. i.e., it satisfies $\iota (v)(w)=\omega (v,w)$. It identifies $W^{\omega }$ with the annihilator $W^{\perp }$ of $W$ in $V^{\ast }$.

Corollary

From the above lemma, we immediately deduce that:

• $W$ is symplectic if and only if $W^{\omega }$ is symplectic.
• $W$ is isotropic if and only if $W^{\omega }$ is coisotropic.
• $W$ is Lagrangian if and only if it is isotropic and has half the dimension of $V$.

Symplectic Basis

Let $(V,\omega )$ be a symplectic vector space of dimension $2n$. Then a basis $u_1,\dots ,u_n,v_1,\dots ,v_n$ such that $\omega (u_j,u_k)=\omega (v_j,v_k)=0,\omega (u_j,v_k)={\delta }_{jk}$is called a symplectic basis.

e.g.

• Suppose $\text{d}{x}^1\wedge \text{d}{y}^1+\text{d}{x}^2\wedge \text{d}{y}^2$ is the standard symplectic form on ${\mathbb{R}}^4$. Then $\left\{{\partial }_{x^1},{\partial }_{x^2},{\partial }_{y^1},{\partial }_{y^2}\right\}$ is a symplectic basis of ${\mathbb{R}}^4$.
• $\omega =2\text{d}{x}^1\wedge \text{d}{x}^3+2\text{d}{x}^2\wedge \text{d}{x}^4+\text{d}{x}^1\wedge \text{d}{x}^4$ is a symplectic form on ${\mathbb{R}}^4$, then $u_1=\frac{1}{2}{\partial }_{x^1}$, $u_2={\partial }_{x^2}$, $v_1={\partial }_{x^3}$, and $v_2=\frac{1}{4}(2{\partial }_{x^4}-{\partial }_{x^3})$ forms a symplectic basis.

Proposition

Every symplectic vector space has a symplectic basis. Moreover, there exists a vector space isomorphism $\Psi :{\mathbb{R}}^{2n}\to V$ such that ${\Psi }^{\ast }\omega ={\omega }_0$, for which ${\omega }_0$ is the standard symplectic form on ${\mathbb{R}}^{2n}$. i.e. all symplectic vector spaces of the same dimension are linearly symplectomorphic.

Proof We prove by induction on the dimension. Since $\omega$ is nondegenerate, there exists vectors $u_1,v_1\in V$ such that $\omega (u_1,v_1)=1$. Hence the subspace spanned by $u_1$ and $v_1$ is symplectic. Let $W$ be its symplectic complement, then $W$ is symplectic as well and $\dim W=\dim V-2$. By the induction hypothesis, there exists a symplectic basis $u_2,\dots ,u_n,v_2,\dots ,v_n$ of $W$. Then $u_1,\dots ,u_n,v_1,\dots ,v_n$ is a symplectic basis of $V$. The linear map $\Psi :{\mathbb{R}}^{2n}\to V$ is defined by $\Psi ((x_1,\dots ,x_n,y_1,\dots ,y_n)):={\sum }_{j=1}^n{x}_j{u}_j+{\sum }_{j=1}^n{y}_j{v}_j,$which satisfies ${\Psi }^{\ast }\omega ={\omega }_0$. $\square$

Corollary

Let $V$ be a $2n$-dimensional real vector space and let $\omega$ be a skew-symmetric bilinear form on $V$. Then $\omega$ is nondegenerate if and only if its $n$-fold exterior power is nonzero, i.e. ${\omega }^n=\omega \wedge \cdots \wedge \omega \ne 0$.

Lemma

Every isotropic subspace of $V$ is contained in a Lagrangian subspace. Moreover, every basis $u_1,\dots ,u_n$ of a Lagrangian subspace $L$ can be extended to a symplectic basis of $(V,\omega )$.

Proof Let $W$ be an isotropic subspace of $V$, that is, $W\subset W^{\omega }$. Observe that if we add any vector $v\in W^{\omega }\setminus W$ to $W$, then the new subspace $W(v)$ is still isotropic, since $\omega (w,v)=0$ for all $w\in W$ and $\omega (v,v)=0$. Thus we can keep adding vectors from $W^{\omega }\setminus W$ to $W$ until $W^{\omega }\setminus W=\varnothing$, which means $W=W^{\omega }$. Hence we get a Lagrangian subspace $L$ which is generated by $W$ and vectors from $W^{\omega }\setminus W$. To prove the second part, we have to utilize the compatible almost complex structure $J$ on $V$. Given a Lagrangian subspace $L$, the subspace $L^{\prime }:=J(L)$ is also Lagrangian, and can be identified with the dual space $L^{\ast }$. In fact, we observe the following diagram commutes: Because $\omega$ is non-degenerate, the map $\iota :L^{\prime }\to L^{\ast }$ defined by $\iota (Ju)(w)=\omega (Ju,w)$ is injective. Moreover, $\dim L^{\prime }=\dim L=\dim L^{\ast }$, so it is an isomorphism. Thus we can identify $L^{\prime }$ with $L^{\ast }$, and hence we can extend the basis $u_1,\dots ,u_n$ of $L$ to a basis $u_1,\dots ,u_n,v_1,\dots ,v_n$ for which $v_j={\iota }^{-1}(u_j^{\ast })$. As both $L$ and $L^{\prime }$ are Lagrangian, we have $\omega (u_j,u_k)=\omega (v_j,v_k)=0$. Finally, we have $\omega (u_j,v_k)=\omega (u_j,{\iota }^{-1}(u_k^{\ast }))=\omega (u_j,J{u}_k)=g(u_j,u_k)={\delta }_{jk}$Thus $u_1,\dots ,u_n,v_1,\dots ,v_n$ is a symplectic basis of $(V,\omega )$. $\square$

e.g. Consider the direct sum $E\oplus E^{\ast }$ of a vector space $E$ and its dual space $E^{\ast }$. The canonical symplectic form on $E\oplus E^{\ast }$ is given by $\omega ((v,\alpha ),(u,\beta ))=\beta (v)-\alpha (u),$where $v,u\in E$ and $\alpha ,\beta \in E^{\ast }$. Then it has standard symplectic basis given by $e_1,\dots ,e_n,e_1^{\ast },\dots ,e_n^{\ast }$, where $e_1,\dots ,e_n$ is a basis of $E$ with dual basis $e_1^{\ast },\dots ,e_n^{\ast }$.

Almost Complex Vector Spaces

Almost Complex Structures on Vector Spaces

Let $V$ be a vector space, a complex structure on $V$ is a linear map $J:V\to V$ with $J^2=-\mathrm{id}$. The pair $(V,J)$ is called an alomost complex vector space.

$\omega$-Compatible Complex Structures

Let $(V,\omega )$ be a symplectic vector space. A complex structure $J$ on $V$ is said to be compatible (with $\omega$, or $\omega$-compatible) if $\omega (Ju,Jv)=\omega (u,v)$ and $\omega (u,Ju)>0$ for all non-zero $u\in V$.

e.g. On ${\text{C}}^n$, the multiplication by $i$ is a complex structure. This complex structure $J$ is compatible with the standard symplectic form $\omega ={\sum }_{k=1}^n\text{d}{x}^k\wedge \text{d}{y}^k$: $\text{d}{x}^k\wedge \text{d}{y}^k(i(z_1,\dots ,z_n),i(w_1,\dots ,w_n))=-\Im (z_k)\Re (w_k)+\Im (w_k)\Re (z_k),$and $\text{d}{x}^k\wedge \text{d}{y}^k((z_1,\dots ,z_n),(w_1,\dots ,w_n))=\Re (z_k)\Im (w_k)-\Re (w_k)\Im (z_k).$Hence $\omega (Ju,Jv)=\omega (u,v)$.

Proposition

Let $(V,\omega ,J)$ be a symplectic vector space with an almost complex structure $J$. Then $J$ is $\omega$-compatible if and only if it induces a real inner product $g$ on $V$ defined by $g(u,v):=\omega (u,Jv).$

Proposition

Let $(V,\omega )$ be a symplectic vector space. Then there is a compatible complex structure $J$ on $V$.