Given a symplectic manifold $(X, \omega)$ and a group $G$ acting on $X$ preserving the symplectic form, we define the moment map $\mu : X \to \mathfrak{g}^*$ so that $$ \langle d\mu(v), \xi\rangle = \omega\big(\xi^*(x), v\big) $$ where $\xi \in \mathfrak{g}$, $\xi^*$ is the vector field generated by $\xi$, and $v \in T_x(X)$.

Why can we do this? Why is $\mu$ defined?

I read the above definition as follows. Define the $\mathfrak{g}^*$-valued 1-form $\theta$ via $$ \theta(v) = \omega(\xi^*,v). $$ Then this form is exact, and so we define $\mu$ so that $d\mu = \theta$.

Why can we do this? Having done some simple computations, the fact that $G$ preserves the symplectic form seems to be imply that exactness holds (and from the above definition it is necessary), but I don't know why this should be the case.

My question is:

Why does the fact that $G$ preserves the symplectic form imply that the 1-form $\theta$ defined above is exact?


Is the preservation of the symplectic form equivalent to the exactness of $\theta$?


Both answers are "No."

There are well-known obstructions to the existence of an equivariant momentum mapping arising from the action by symplectomorphisms of a group $G$ on a symplectic manifold. They can be phrased in many ways, but if $G$ is connected and its Lie algebra is semisimple, for example, the obstructions vanish.

A nice treatment can be found in the classic paper by Atiyah and Bott: "The moment map and equivariant cohomology" in Topology 1984.

After-dinner update:

Let me try to add some more details, since it seems I was a little too quick both reading the question and also the comments! (In my defense, I was getting really hungry and wanted to get home!)

Let $(M,\omega)$ be a symplectic manifold and let $G$ be a connected Lie group acting on $M$ via symplectomorphisms. Let $\mathfrak{g}$ be the Lie algebra of $G$ and for every $X \in \mathfrak{g}$ let $\xi_X$ denote the corresponding vector field on $M$. Since $G$ acts symplectomorphically, $\xi_X$ is a symplectic vector field; that is, $\mathcal{L}_{\xi_X} \omega = 0$, which, as Ben points out, implies that $d i_{\xi_X}\omega = 0$. Let $\mathrm{Sym}(M)$ denote the Lie algebra of symplectic vector fields on $M$. We have a Lie algebra homomorphism $\mathfrak{g} \to \mathrm{Sym}(M)$.

A symplectic vector field $\xi$ is said to be hamiltonian if $i_\xi\omega$ is not merely closed, but also exact. The $G$-action is hamiltonian if the $\xi_X$ are hamiltonian for all $X \in \mathfrak{g}$. Let $\mathrm{Ham}(M)$ denote the Lie algebra of hamiltonian vector fields. It is not hard to show that the Lie bracket of two symplectic vector fields is hamiltonian, so $\mathrm{Ham}(M)$ is an ideal in $\mathrm{Sym}(M)$ with abelian quotient. This gives rise to a short exact sequence of Lie algebras $$ 0 \longrightarrow \mathrm{Ham}(M) \longrightarrow \mathrm{Sym}(M) \longrightarrow H^1(M) \longrightarrow 0 $$ where $H^1(M)$ is the first de Rham cohomology group thought of as an abelian Lie algebra.

If $\xi \in \mathrm{Ham}(M)$, then $i_\xi \omega = df$ for some $f \in C^\infty(M)$. This defines a map $C^\infty(M) \to \mathrm{Ham}(M)$ which is also a Lie algebra homomorphism if we make $C^\infty(M)$ into a Lie algebra via the Poisson bracket. The kernel of this map consists of the locally constant functions, whence we have another short exact sequence of Lie algebras $$ 0 \longrightarrow H^0(M) \longrightarrow C^\infty(M) \longrightarrow \mathrm{Ham}(M) \longrightarrow 0 $$

Putting these two sequences together we have a four-term exact sequence starting and ending at the first two de Rham cohomology groups: $$ 0 \longrightarrow H^0(M) \longrightarrow C^\infty(M) \longrightarrow \mathrm{Sym}(M) \longrightarrow H^1(M) \longrightarrow 0 $$

Now the symplectic $G$-action defines a Lie algebra homomorphism $\mathfrak{g} \to \mathrm{Sym}(M)$ and for the existence of a momentum map, we want this to lift to a Lie algebra morphism $\mathfrak{g} \to C^\infty(M)$.

There is an immediate obstruction for the map to lift at the level of vector spaces, namely the map $\mathfrak{g} \to \mathrm{Sym}(M) \to H^1(M)$ is a Lie algebra cocycle with values in the trivial module $H^1(M)$, and so defines a class in $H^1(\mathfrak{g};H^1(M))$. If this class vanishes, we do get a map $\mathfrak{g} \to C^\infty(M)$ which may only be a homomorphism modulo $H^0(M)$. In other words, it defines a Lie algebra homomorphism to a central extension of $C^\infty(M)$ defined by a 2-cocycle with values in the trivial module $H^0(M)$. The moment map will exist if the class of this cocycle in $H^2(\mathfrak{g}; H^0(M))$ vanishes.

For example, if $\mathfrak{g}$ is semisimple, then both $H^1$ and $H^2$ vanish and the momentum map exists.

The beautiful observation of Atiyah and Bott is that the obstruction can be reinterpreted in terms of the Cartan model for the equivariant de Rham cohomology of $M$, where it becomes simply the obstruction to extending $\omega$ to an equivariant cocycle.

  • 3
    $\begingroup$ How can "No" answer a question that begines with "Why..."? :) $\endgroup$ – Mariano Suárez-Álvarez Feb 14 '11 at 21:02
  • 1
    $\begingroup$ Good point! The answer is "No" if you omit the "Why" :) Otherwise the question is ill-posed. $\endgroup$ – José Figueroa-O'Farrill Feb 14 '11 at 21:05
  • $\begingroup$ Jose, I'm really confused by your answer. At least what I learned in my symplectic geometry class is that the answer is "yes." $\endgroup$ – Ben Webster Feb 14 '11 at 21:45
  • $\begingroup$ To be fair, my question conflates the two, so your answer isn't actually that far off. $\endgroup$ – Simon Rose Feb 14 '11 at 22:23
  • $\begingroup$ Well, as comments show, I wasn't thinking with the highest level of clarity myself. That's the beauty of MO, you get to iteratively converge on the right answer. $\endgroup$ – Ben Webster Feb 14 '11 at 22:44

One thing that seems to be confusing you is that there is no "the moment map." There often several different moment maps for the same action, and as Jose says, sometimes none.

On the other hand, your questions seem to really be

If X is a vector field, are the conditions

  • $X$ preserves $\omega$ and
  • $i_X\omega=\omega(X,-)$ is an exact 1-form


The answer is "no", but it's "yes" if you replace "exact" by "closed". The preservation of $\omega$ by $X$ is the same as saying that the Lie derivative $\mathcal{L}_X\omega=0$. By Cartan's magic formula,

$$\mathcal{L}_X\omega=d(i_X\omega)+i_X(d\omega)=d(i_X\omega)$$ since the second term is zero by the closedness of $\omega$.

This distinction is important, since lots of symplectic actions don't have moment maps because of the existence of $H^1$. Think about, for example, the 2-torus $T^2$ acting on itself; any invariant volume form can also be thought of as an invariant symplectic form, bu the map $\mathfrak t^2\to H^1(T^2)$ given by $X\mapsto [i_X\omega]$ is an isomorphism. No non-trivial element of the Lie algebra acts by a Hamiltonian vector field, even though they all act by symplectic ones.

  • $\begingroup$ I'm a bit confused here. Isn't this just saying that $i_X \omega$ is closed? $\endgroup$ – Simon Rose Feb 14 '11 at 22:25
  • $\begingroup$ Yes, my bad. Just writing a little too fast. $\endgroup$ – Ben Webster Feb 14 '11 at 22:35
  • $\begingroup$ OK, it's fixed now. As Jose said, there are obstructions to the existence for a moment map for a symplectic action, and one of them is that $i_X\omega$ must be exact, not just closed. $\endgroup$ – Ben Webster Feb 14 '11 at 22:38
  • $\begingroup$ So I probably have to look more at José's answer then. $\endgroup$ – Simon Rose Feb 14 '11 at 22:39
  • $\begingroup$ ... or what you just said. $\endgroup$ – Simon Rose Feb 14 '11 at 22:39

There is yet another interpretation of the already mentioned obstructions for the existence and uniqueness of momentum maps (moment mappings, moment maps, momentum mappings....) in terms of Poisson geometry. This even generalizes then to Poisson manifolds and makes things perhaps a little bit more transparent:

For the existence of a (non-equivariant) momentum map you ask Poisson vector fields to be Hamiltonian, the obstruction therefore lies in the first Poisson cohomology, which is precisely the quotient of Poisson vector field modulo Hamiltonian vector fields. In the symplectic case this quotient becomes canonically isomorphic to the first deRham cohomology as symplecticl vector fields correspond to closed one-forms while Hamiltonian ones correspond to exact one-forms via the musical isomorphism induced by $\omega$.

The uniqueness (and also the equivariance) is then controlled by the zeroth Poisson cohomology which are th functions with trivial Hamiltonian vector field, also called the Poisson center. In the symplectic and connected case, these are just the constant functions, but in the general Poisson case this might be a much more interesting cohomology.

The advantage of this point of view is, perhaps, the fact that now all obstructions are somehow arising from the same complex: the canonical complex associated to a Poisson structure $\pi$. As a vector space this is just the sections of the exterior algebra of the tangent bundle (i.e. the multivector fields) and the differential is the Schouten bracet with $\pi$. In the symplectic case, this boils down to the deRham complex via the musical isomorphim.

You can find a detailed discussion of this in the (by the way very: nice) book of Ana Cannas da Silva and Alan Weinstein on "Geometric Models for Noncommutative Algebras".

  • 1
    $\begingroup$ A small nitpick is that I would not say that the obstruction lies in the Poisson cohomology. This is an equivariant problem, hence the Lie algebra has to come in somewhere. The obstructions in the da Silva and Weinstein are, respectively, a Lie algebra homomorphism from $\mathfrak{g}$ to the first Poisson cohomology (which, unless I misunderstand, need not be a 1-cocycle in general) and a class in the second cohomology of $\mathfrak{g}$ with values in the zeroth Poisson cohomology. I wonder whether there is perhaps an equivariant Poisson cohomology theory where the obstruction does lie. $\endgroup$ – José Figueroa-O'Farrill Feb 16 '11 at 10:49
  • $\begingroup$ Sorry, I was a little bit sloppy. Concerning the equivariant Poisson cohomology, I think I have seen something like that somewhere. Don't remember, though... :( In any case, this should be rather straightforward to cook up a reasonable Cartan like model for it. $\endgroup$ – Stefan Waldmann Feb 16 '11 at 14:19
  • $\begingroup$ Thanks -- googling I just found this paper by Viktor Ginzburg: arxiv.org/abs/dg-ga/9611002. Having glanced at it very quickly, he seems to apply this to the case of Poisson Lie groups acting on a Poisson manifold and the resulting momentum mapping. $\endgroup$ – José Figueroa-O'Farrill Feb 16 '11 at 20:13
  • $\begingroup$ Oh, yes. That is the one I had in mind. Sorry for not googling myself :) $\endgroup$ – Stefan Waldmann Feb 17 '11 at 8:27

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