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

In particular, if and are surfaces inside such that for some , then (provided we orient things correctly),

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

This defines a map
where is defined by
We say that is "nondegenerate" if the map is an isomorphism for all .


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.

  1. 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.
  2. 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

for all vector fields on .


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

For any vector field
So taking and , we get
So above is a Hamiltonian vector field.

 


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

where is the vector field defined by and is the 2-form given by . (The left hand side of this equation is called the "Lie derivative" of in the direction .) In particular, since is symplectic, is if and only if is a symplectic vector field. In particular, specialising to the case where is constant in , we see that is a symplectic vector field if and only if flowing along preserves the symplectic form .

 


Example 6.4

Consider with coordinates and symplectic form . Consider the Hamiltonian

where is some smooth real-valued function. The associated Hamiltonian vector field is
The equations for flow along are therefore
Interpreting as position, as momentum, as mass and as potential enegery, these are precisely the equations of motion in classical mechanics.

 


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 .

 
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