Linear Symplectic Geometry

Symplectic Vector Space

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.

Symplectic Complement

Let be a symplectic vector space. The symplectic complement of a subspace is the subspace

Classification of Subspaces

Let be a symplectic vector space. A subspace is called

• isotropic if , i.e. ,
• coisotropic if ,
• symplectic if , i.e. is non-degenerate.
• Lagrangian if .
  

Lemma

For any subspace , there holds

Proof Let be the symplectic duality map. i.e., it satisfies . It identifies with the annihilator of in .

Corollary

From the above lemma, we immediately deduce that:

• is symplectic if and only if is symplectic.
• is isotropic if and only if is coisotropic.
• is Lagrangian if and only if it is isotropic and has half the dimension of .
  

Symplectic Basis

Let be a symplectic vector space of dimension . Then a basis such that is called a symplectic basis.

e.g.

• Suppose is the standard symplectic form on . Then is a symplectic basis of .
• is a symplectic form on , then , , , and forms a symplectic basis.

Proposition

Every symplectic vector space has a symplectic basis. Moreover, there exists a vector space isomorphism such that , for which is the standard symplectic form on . i.e. all symplectic vector spaces of the same dimension are linearly symplectomorphic.

Proof We prove by induction on the dimension. Since is nondegenerate, there exists vectors such that . Hence the subspace spanned by and is symplectic. Let be its symplectic complement, then is symplectic as well and . By the induction hypothesis, there exists a symplectic basis of . Then is a symplectic basis of . The linear map is defined by which satisfies .

Corollary

Let be a -dimensional real vector space and let be a skew-symmetric bilinear form on . Then is nondegenerate if and only if its -fold exterior power is nonzero, i.e. .

Lemma

Every isotropic subspace of is contained in a Lagrangian subspace. Moreover, every basis of a Lagrangian subspace can be extended to a symplectic basis of .

Proof Let be an isotropic subspace of , that is, . Observe that if we add any vector to , then the new subspace is still isotropic, since for all and . Thus we can keep adding vectors from to until , which means . Hence we get a Lagrangian subspace which is generated by and vectors from .
To prove the second part, we have to utilize the compatible almost complex structure on . Given a Lagrangian subspace , the subspace is also Lagrangian, and can be identified with the dual space . In fact, we observe the following diagram commutes:

Because is non-degenerate, the map defined by is injective. Moreover, , so it is an isomorphism. Thus we can identify with , and hence we can extend the basis of to a basis for which . As both and are Lagrangian, we have . Finally, we have Thus is a symplectic basis of .

e.g. Consider the direct sum of a vector space and its dual space . The canonical symplectic form on is given by where and . Then it has standard symplectic basis given by , where is a basis of with dual basis .

Almost Complex Vector Spaces

Almost Complex Structures on Vector Spaces

Let be a vector space, a complex structure on is a linear map with . The pair is called an alomost complex vector space.

-Compatible Complex Structures

Let be a symplectic vector space. A complex structure on is said to be compatible (with , or -compatible) if and for all non-zero .

e.g. On , the multiplication by is a complex structure. This complex structure is compatible with the standard symplectic form : and Hence .

Proposition

Let be a symplectic vector space with an almost complex structure . Then is -compatible if and only if it induces a real inner product on defined by

Proposition

Let be a symplectic vector space. Then there is a compatible complex structure on .