Hamiltonian action

Definitions

Weakly Hamiltonian action

A symplectic action of on is called weakly Hamiltonian if every fundamental vector field is a Hamiltonian vector field.

Hamiltonian action

A symplectic action of on is called Hamiltonian if it is weakly Hamiltonian, and the diagram below commutes:

i.e. an action is called Hamiltonian if there exists an equivariant moment map.

Motivation

The fundamental vector fields of a symplectic action are symplectic. The goal is to identify an analogous definition for Hamiltonian vector fields where the fundamental vector fields to the action are related to some function such that

Note that this is already a stronger condition than a symplectic action since each Hamiltonian vector field must be symplectic. However, this definition has issues.

Most importantly, the function isn’t unique. In fact, there can be infinitely many of them, since can differ by a constant. Therefore, we aim to pick the function for each such that each of the fundamental vector fields were Hamiltonian?

So, we are looking for some map

such that

We also want this map to be a Lie algebra homomorphism so that it preserves the Lie bracket on and the Poisson bracket on seen as a Lie bracket.

Measuring obstructions to being Hamiltonian

Cochain condition

We can study how far away a weakly Hamiltonian -action is from being Hamiltonian. (Most of this is followed from the discussion in @mcduff2017 chapter 5).

Let there be a weakly Hamiltonian -action on , and choose a linear map such that and where denotes the Hamiltonian vector field of . Then there is a (unique) map such that, for all , This function satisfies

Proof

First note that since is a Lie algebra homomorphism. Note that both and are Hamiltonian (and thus symplectic) vector fields, so is a Hamiltonian vector field with Hamiltonian function thus,

This implies that the function is always constant. Therefore,

Hence, since is linear, and (and thus ) follows the Jacobi identity, then it also follows it as seen in the equation in the statement above.

Why is this interesting?

Given a Lie algebra , we can build the Lie algebra cohomology of by taking the cochains to be then building the cohomology groups from there.

Now consider a cocycle, i.e. a cochain such that . For completeness, here is the explicit formula for the differential operator

So, looking specifically at a 2-cocycle we have

Since is a trivial -representation, this simplifies to

This is exactly the condition given in the statement of the proposition above. Hence, we see that from the proposition is a cocycle in the Lie algebra cohomology.

Since the action in proposition is weakly Hamiltonian, there may exist other functions , these will all differ by a coboundary of the form

Therefore, we can see that given a weakly Hamiltonian action of on , then we can find the associated equivalence class

This equivalence class measures “how far away” the action is to being Hamiltonian.

NOTE

Some authors call a measurement of obstructions to ALL weakly Hamiltonian actions being Hamiltonian, which is very intuitive.

For example, if (or vanishes) then and the comoment map is a Lie algebra homomorphism, so the action is Hamiltonian.