I'm trying to learn about moment maps in symplectic topology (suppose our Lie group is $G$ with Lie algebra $\mathfrak g$, acting on the symplectic manifold $(M,\omega)$ by symplectomorphisms). I'm having a hard time, and I've realized this is because I don't have a good conceptual understanding of the Lie bracket, either on the Lie algebra $\mathfrak g$, or on the group of symplectomorphisms of $(M,\omega)$, or on the space of functions $\mathcal C^\infty(M,\mathbb R)$. Therefore I can't "visualize" the Hamiltonian condition, which requires that the linear map $\mathfrak g \rightarrow \mathcal C^\infty(M,\mathbb R)$, which exists when the action by $G$ is "exact," be a Lie algebra homomorphism.

Please tell me how you personally understand/intuit/conceptualize this situation, both the Lie bracket stuff and moment maps more generally! Any help is greatly appreciated.

EDIT: I didn't realize how non-standard some of this terminology is, so my question might be confusing. I call the action $\rho: G \rightarrow {\rm Symp}(M,\omega)$ "exact" if the image of the induced map $\rho: {\rm Lie}(G) \rightarrow {\rm Lie}({\rm Symp}(M,\omega))$ is contained in the sub-lie-algebra of Hamiltonian vector fields. The condition that was confusing me, I now realize, is just a technical point: that we choose a set of representative Hamiltonian functions for the image $\rho({\rm Lie}(G))$ which is a sub-Lie-algebra of $\mathcal C^\infty(M,\mathbb R)$ with its Poisson bracket. Thanks to all the helpful answers I think I understand this much better now.

In particular, if we present ${\rm Lie}(G)$ (assumed finite dimensional, semi-simple, etc) by Lie algebra generators (with some relations), then we can probably just choose appropriate elements in $\mathcal C^\infty(M,\mathbb R)$ for these generators, and then the rest of the map from ${\rm Lie}(G)$ to $\mathcal C^\infty(M,\mathbb R)$ is just forced on us, and this gives a Hamiltonian action? Is that right?

  • $\begingroup$ I am choosing to accept Ben's answer because it was the one that made me realize that this Hamiltonian condition is just about futzing around with the constants, and in particular less deep (and less confusing) than I thought it was. But Ilya, your response was also very helpful in clarifying my fuzzy understanding. $\endgroup$ Feb 7, 2010 at 0:15
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    $\begingroup$ I just found a pretty decent reference that discusses these exact questions. Perhaps it'll be useful to you. books.google.com/… $\endgroup$ Feb 8, 2010 at 2:15

5 Answers 5


This question is (at least as I read it) about the Poisson bracket; the Poisson bracket is a Lie bracket structure on the functions on a symplectic manifold.

So how should one think about Poisson bracket? Well, remember that for every function on a symplectic manifold, one has a Hamiltonian vector field $X_f$. One way to think about this is that if your symplectic manifold is the phase space of a physical system (the space of possible positions and momenta), and $f$ is the energy function, then the resulting vector field is the derivative of the time evolution of the system.

The Poisson bracket $\{f,g\}$ is defined to be $X_f(g)$, the derivative of $g$ along the vector field $X_f$. That is the Poisson bracket of $f$ and $g$ is the time derivative of $g$ if you use $f$ as the energy function. Remarkably, this operation is anti-symmetric, and defines a Lie algebra structure.

As you mentioned, a moment map is equivalent to a Lie algebra homomorphism from $\mathfrak{g}$ to the space of functions on your manifold. I'm not sure how you're exactly supposed to visualize that, but let me explain how I think about it.

So, imagine you have your favorite G-action on a symplectic manifold, preserving the symplectic structure. Taking derivative, you get a map of Lie algebras from $\mathfrak{g}$ to vector fields on your manifold. It sounds from your question like how to think about such Lie algebra homomorphisms is actually what is confusing you. This just says that if you were to integrate the vector fields coming from $\mathfrak{g}$, you would get the group G (or maybe a finite cover).

Now, each of these vector fields corresponds under the symplectic form to a 1-form. If your manifold has no $H^1$, then you can integrate these to functions, but of course, you can't do this uniquely; it's only unique up to a constant. So taking all of these lifts, you get a vector space of functions which is $\dim \mathfrak{g}+1$ dimensional (assuming $G$ acted faithfully). This is closed under Lie bracket, so it's a finite dimensional Lie algebra $\tilde {\mathfrak{g}}$ with a map $\tilde {\mathfrak{g}}\to {\mathfrak{g}}$.

It might be that $\tilde {\mathfrak{g}}\cong {\mathfrak{g}}\times \mathbb{R}$ as a Lie algebra, in which case you can get a moment map by picking a splitting of the map above, or it might not, in which case you don't have a moment map. If $\mathfrak{g}$ is semi-simple, then the latter case is impossible, so you always have a moment map.


I believe the following way (Kostant's, 1970) to be the best way to think about the Hamiltonian condition.

First, "why" is there a central extension $H^0(M; {\mathbb R}) \to C^\infty (M) \to symp(M)$ of Lie algebras? Of what is $C^\infty (M)$ supposed to be the Lie group? For $symp(M)$, the Lie algebra of vector fields annihilating the symplectic form $\omega$, it's clear it should the Lie algebra of the group $Symp(M)$ of symplectomorphisms.

Assume now that $[\omega]$ is integral. Then it is $c_1$ of some "prequantization" line bundle $\mathcal L$, and $\omega$ is the curvature of some Hermitian connection $\alpha$ on that line bundle. Let $Aut(M,{\mathcal L},\alpha)$ denote the group of Hermitian bundle automorphisms (moving the base around) of $\mathcal L$ preserving $\alpha$. This group obviously maps to $Diff(M)$, forgetting the action on the fibers, but because it preserves $\alpha$ on $\mathcal L$ it preserves $\omega$ on $M$, so the image lies inside $Symp(M)$. The kernel consists of bundle automorphisms that only act fiberwise, and for them to preserve the flat connection they must, on each component, rotate all the fibers by the same element of $U(1)$.

The Hamiltonian condition, then, is about whether one can lift the action of $G$ on $M$ to an action on the line bundle over $M$. It's very easy, given such a lift, to write down a moment map. (Basically, now that you're dealing with a $1$-form $\alpha$ instead of a $2$-form $\omega$, you can pair vector fields from $\mathfrak g$ with it.)

One example I find instructive is ${\mathbb R}^{2n}$ acting on itself by translation, with the space given the usual symplectic structure. That's acting as symplectomorphisms, and the space is simply connected, so there's no $H^1$ obstruction (as when $T^1$ acts on $T^2$). But one can't lift the action to preserve the (non-flat) connection on the (trivial) line bundle; it only lifts to an action of the Heisenberg group.

Another subtle example is $SO(3)$ acting on $S^2$ with the area $1$ symplectic structure. On the Lie algebra level, yes, the action is Hamiltonian. But actually $SO(3)$ doesn't act on the line bundle; only its double cover $SU(2)$ does.

Finally, think about the case that $G$ acts algebraically on $X \subseteq {\mathbb P}V$. I like to say that $X$ is "equivariantly projective" if $G$ acts on ${\mathbb P}V$ preserving $X$, and this is pretty nearly an algebro-geometric replacement for the Hamiltonian condition. (Non-example: $X$ is a nodal cubic curve, whose smooth locus is ${\mathbb C}^\times$, acted on by ${\mathbb C}^\times$.)

  • $\begingroup$ Thanks, this is very interesting. I was about to ask what the condition was for. $\endgroup$ Feb 8, 2010 at 18:52

I'm not an expert, but I don't find the situation very hard to imagine(see footnote below). It doesn't use anything beyond the fact that "the Lie algebra is the tangent space of the Lie group at the identity". Moreover, for this purpose it is enough to imagine the tangent space as arrows at the identity of the Lie group, pointing in the directions you can move.

Oh, and a side remark that might be helpful to some. Having a map $M \to \mathfrak g^*$ that I usually thought of as a "moment map" is exactly the same as having a map $\mathfrak g \to C^\infty(M,\mathbb R)$ the questioner is talking about, and the latter is easier to visualize.

The simplest case is when $G=\mathbb R$. This is the standard case of the (time-independent) Hamiltonian flow. The Lie-algebra $\mathfrak g$ is one-dimensional, so a linear moment map $f:\mathfrak g \to C^\infty (M,\mathbb R)$ is determined by its value on the unit vector $\vec e \in \mathfrak g$. If $H=f(\vec e)$, this is precisely the Hamiltonian on your manifold. The flow with respect to the moment map is precisely the same as the flow of this Hamiltonian.

The next simplest case is when $G=\mathbb R^n$. The only difference here is that there are several directions in which you could go. Let ${\vec e_1 ,\ldots \vec e_n} \in \mathfrak g$ be the vectors at the identity of $G$ that represent the possible directions. Now, a moment map $f:\mathfrak g \to C^\infty (M,\mathbb R)$ is determined by n Hamiltonians; for each $1\leq i \leq n$ we have $H_i = f(\vec e_i)$.

Now, any path in $G=\mathbb R^n$ will correspond to some flow on the manifold $M$. In particular, if you always go "right" (in the direction of $\vec e_1$), the flow will be precisely that of the Hamiltonian $H_1$; the existence of other directions won't matter. If you only go in the direction of $\vec e_2$, the flow will be that of $H_2$. For any path, you can approximate it by a piecewise-linear path that's always parallel to a coordinate axes; the flow will be that of $H_i$ whenever you go parallel to the axes of $\vec e_i$.

Finally, if G is any Lie group, it's a manifold, so it locally looks like $\mathbb R^n$. Everything I said above about this case still holds, with one exception: the coordinate directions may no longer be "independent". (So, going right, then up, then left, then down, might not bring you exactly to the same point you started) Unfortunately, at this point my expertness runs out, so see the other answers for a detailed explanation. However, algebraically, the condition that you need to add is precisely that the moment map f is a Lie-algebra homomorphism; think of this as an error term you need to add to account for the lack of independence when you change the direction of your piecewise-linear path.

Of course, in practice, in calculating flows you don't need to approximate anything with piecewise-linear paths, as there are algebraic equations will give you the Hamiltonian that corresponds to any direction. In the case of $\mathbb R ^n$, the map f will simply be linear, in general there is likely an error term.

Oh, and finally, most algebraic equations are probably simpler if you think of your moment maps as maps $M \to \mathfrak g^*$.

Footnote: When I wrote that everything is simple, I hadn't thought of the error terms yet (see above). However, I still think that for most conceptual purposes, it's best to think of your Lie group as $\mathbb R^n$ with some error terms added. Somebody correct me if I'm wrong, but when we describe Lie groups using "structure forms", isn't it precisely a way to make this idea precise?

  • $\begingroup$ Thanks! This is very clear, and I think you are right that the situation is not very difficult. But my confusion started as soon as we go from \R^n to some more complicated group, so that the Lie bracket is non-trivial. $\endgroup$ Feb 7, 2010 at 0:16

The moment map condition is a precise form of Noether's theorem. Whenever you have an $G$-invariant Hamiltonian, its flow will preserve the value of the moment map.
That is in some sense the moment map condition:
$\mu:(M,\omega)\rightarrow \mathfrak{g}^*$ should be equivariant and satisfy
$\langle d \mu(x).v,\xi\rangle=\omega(\xi_M(x),v)$ for all $v\in T_x M$ and $\xi\in \mathfrak{g}$

Now consider the flow $\phi^t$ of the $G$-invariant Hamiltonian $H$. Then you compute
$\frac{d}{dt}|_0 \langle\mu(\phi^t(x)),\xi\rangle=\langle d\mu(x).X_H(x),\xi\rangle=\omega(\xi_M(x),X_H (x))=dH(x).\xi_M(x)=\frac{d}{dt}|_0 H(e^{t\xi}.x)=0$
since H is $G$-invariant. There is also a motivation from symplectic reduction, but I don't know whether that is really relevant at the beginning.
Please don't be discouraged by this "bracket stuff". I don't consider it particularly enlightening anyway...


What I am about to say is implicit in Ben Webster's answer, but I figured that I would make it explicit.

A Hamiltonian group action on a symplectic manifold M should be thought of as a Lie group homomorphism $G \to Ham(M)$, so it induces a Lie algebra map $Lie(G) \to Lie(Ham(M))$.

But what is Lie(Ham(M))? It is the normalized $C^\infty(M)$ (those $f$ whose integral over M is zero, if M is closed) and the Lie bracket is the Poisson bracket.

So the Hamiltonian condition for a symplectic group action is just that you want your Lie group homomorphism $G \to Symp(M)$ to factor through the Lie group homomorphism $Ham(M) \to Symp(M)$.

  • $\begingroup$ Thank you, this seems to clear things up quite a bit. However, what is your exact definition of Ham(M)? Three possible definitions come to my mind, and I'm not sure which of them are the same. 1) Symplectomorphisms that come from time-independent Hamiltonians. 2) Symplectomorphisms that come from dependent Hamiltonians. 3) Symplectomorphisms that come from moment maps on some Lie group. (The last would make talking about it quite tautological, but I'm pretty sure it's equivalent to number 2). The reason this feels important is that I'm not sure how exactly you calculate Lie(Ham(M)). $\endgroup$ Feb 6, 2010 at 22:31
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    $\begingroup$ The definition of Ham(M) is your #2, namely symplectomorphisms that come from time dependent Hamiltonians. $\endgroup$
    – user1835
    Feb 6, 2010 at 22:40
  • $\begingroup$ Right. I guess 1) wouldn't even make sense - composition of two symplectomorphisms from time-independent Hamiltonian might not come from a time-independent Hamiltonian. 3) should be equivalent to 2), I hope. And an element of Lie(Ham(M)) is just a Hamiltonian (which determines an infinitesmall Hamiltonian symplectomorphism, as they always do). So it's in $C^\infty (M)$. Since adding constants to a Hamiltonian doesn't change anything, we can assume it's normalized. $\endgroup$ Feb 6, 2010 at 23:00
  • $\begingroup$ Hmm I thought I was starting to understand things but now this comment confuses me again. Lie(Ham(M)) is a sub-lie-algebra of Lie(Symp(M)), right, because Ham(M) is just a subgroup of Symp(M)? If this is not right then I am misunderstanding your definitions. If it is right, then by "factor through the the Lie group homomorphism..." you just mean that the image lies in the subgroup Ham(M). In which case this is the "exactness" requirement that I referred to, and the "Hamiltonian" requirement is something more. $\endgroup$ Feb 6, 2010 at 23:29
  • $\begingroup$ In particular, as Ben said, there is a "S.E.S." of lie algebras 0 --> R ---> C^inf(M) --p-> Lie(Ham(M)) ---> 0 Using our rho : g ---> Lie(Ham(M)), which exists because of the exactness condition, we get a sequence 0 ---> R ---> p^-1(rho(g)) ---> rho(g) ---> 0, and the Hamiltonian condition corresponds to being able to find a lie-algebra splitting of this sequence. is that right? $\endgroup$ Feb 6, 2010 at 23:34

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