# Basic Symplectic Geometry

**Definition 6.1**

Let be a smooth manifold. A 2-form is called a "symplectic form" if and is nondegenerate. In this case, we call the pair a "symplectic manifold".

One can think of the closed condition as follows. By Stokes' Theorem, for any 3-dimensional submanifold with boundary , we have

What about non-degeneracy? Since is a 2-form, for each , we get an alternating map

**Exercise 6.1**

Let be a vector space with . Show that satisfies if and only if the induced map as described above is an isomorphism.

Suppose we also have an inner product , and so an isomorphism . Show that, if the composition is orthogonal with respect to , then , so defines a complex structure on .

Conversely, suppose we have a complex structure and inner product on , which is compatible with in the sense that . Prove that is a skew non-degenerate 2-form.

**Example 6.1**

The following are examples of symplectic manifolds.

- Kähler manifolds. If is an almost Hermitian manifold with Hermitian form , then is a nondegenerate 2-form by Exercise on local Kahler. So is a symplectic form if and only if , i.e., if and only if is an almost Kähler manifold.
- Cotangent bundles. Consider . This has a canonical 1-form on it, defined by for and . Then (claim) the 2-form is actually a symplectic 2-form. So is naturally a symplectic manifold (and is a good model for the phase space of classical physics---think about the case )

**Exercise 6.2**

Take local coordinates on . Let be coordinates on the fibres of .

Check that in these coordinates and and that this is indeed non-degenerate.

The construction of the isomorphism above globalises to give an isomorphism between vector fields and 1-forms: given a vector field on a symplectic manifold , we get a 1-form satisfying

**Definition 6.2**

We say that a vector field on is "symplectic" if the 1-form is closed, i.e., . We say that is "Hamiltonian" if is exact, i.e., for some smooth function on . In this case, we call the function a "Hamiltonian" for the vector field . We write and respectively for the spaces of symplectic and Hamiltonian vector fields on .

**Remark 6.1**

Note that the Hamiltonian is only defined up to constant translation. Indeed given by (where is the unique vector field satisfying ) has kernel precisely the constants .

**Example 6.2**

Consider with coordinates , and the symplectic form . Consider the Hamiltonian . We have

**Remark 6.2**

On , every symplectic vector field is Hamiltonian, since every closed -form is exact.

**Example 6.3**

Consider the the torus . If are coordinates on , then the symplectic form descends to a symplectic form on . The vector field has , which is closed but not exact (as is not a function on ). Hence is symplectic but not Hamiltonian.

**Remark 6.3**

Let be a symplectic manifold and assume that is a smoothly varying family of diffeomorphisms. Then we have

**Example 6.4**

Consider with coordinates and symplectic form . Consider the Hamiltonian

**Remark 6.4**

The vector field is called the "symplectic gradient of ". It is the vector field dual to under and, once we have picked a compatible almost complex structure and metric, is precisely . Since this is orthogonal to , flowing under preserves . Formally this follows from .

**Exercise 6.3**

Let act on diagonally in the usual way. Show, by differentiating, that the infinitesimal vector field defined by , is for . Putting , show that this is really where . Furthermore, verify that this is a Hamiltonian vector field with Hamiltonian for any constant .

**Exercise 6.4**

Let be a symplectic vector field on and a path in . Denote by the surface swept out after time by the flow along the vector field starting at . Then define . Prove that is Hamiltonian for every .