Let $G$ be a connected Lie group acting on the symplectic manifold $(M,\omega)$. In the definition of a Hamiltonian action one requires that the moment map $\mu\colon M\xrightarrow{} \mathfrak{g}^\ast$ satisfies $$\omega(X_\xi,\cdot) = d\langle \mu,\xi\rangle$$ for all $\xi\in\mathfrak{g}$. In other words, we want $X_\xi$ to be the Hamiltonian vector field of the function $\langle \mu,\xi\rangle$. While I understand what the definition says, I don't really have a geometric understanding of it. The second condition in the definition is that $\mu$ should be $G$-equivariant, which is more intuitive to me. What exactly does the above condition ensure? Is there a more geometric way to think about it?

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    $\begingroup$ I refer you to Quantum Theory, Groups and Representations: An Introduction by Peter Woit. In my copy it's Chapter 15 on Hamiltonian vector fields and the moment map. It's the most lucid explanation of the moment map I've read. The author (and I) prefers to think about the moment map as a function $\mathfrak{g} \to C^\infty(M)$ which assigns to each $L \in \mathfrak{g}$ a function on the manifold $\mu_L$. Your condition then becomes $X_L = X_{\mu_L}$ where $X_L$ is the vector field defined by the action of $L$ on functions and $X_{\mu_L}$ is the Hamiltonian vector field of $\mu_L$. $\endgroup$ Jul 6, 2023 at 23:38
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    $\begingroup$ Maybe more geometrically, $\{\mu_L, \cdot \}$ and $-X_L$ act on functions identically as vector fields. To recover your definition of the momentum map, observe that $\mu_L(p) = \langle \mu(p), L \rangle$. If this has been helpful, I'll type it up into a more formal answer with more details. $\endgroup$ Jul 6, 2023 at 23:43
  • $\begingroup$ @CharlesHudgins I will take a look at it, thanks a lot! If I am understanding correctly, what you are saying is that the moment map can be reinterpreted (the comoment map is what this is called I believe) so that for each $L\in\mathfrak{g}$ it picks a function on $M$ whose Hamiltonian vector field is the fundamental vector field of $L$. $\endgroup$ Jul 6, 2023 at 23:44
  • $\begingroup$ @CharlesHudgins I would really appreciate it if you can explain the first sentence in your second comment in more detail. It seems to be what I have been looking for. $\endgroup$ Jul 6, 2023 at 23:46
  • $\begingroup$ I'll type up an answer. $\endgroup$ Jul 7, 2023 at 0:03

2 Answers 2


Per my comment, this answer is largely cribbed from an excellent text by Peter Woit called Quantum Theory, Groups and Representations: An Introduction. In my edition this discussion occurs in Chapter 15.

The Setup

Suppose we have a symplectic manifold $(M, \omega)$. As usual, the Hamiltonian vector field associated to a function $f : M \to \mathbb{R}$ (called the Hamiltonian) is by definition a vector field $X_f$ which satisfies $$ df = i_{X_f} \omega $$ (NB: by non-degeneracy of $\omega$, this $X_f$ is in fact unique)

We'll also define a Poisson bracket $\{\cdot , \cdot \}$ by $$ \{f, h\} = \omega(X_f, X_h) $$

Finally, we'll consider the action of a Lie group $G$ on $M$. We won't make any assumptions about how $G$ acts on $M$, though we'll see that if we want a momentum map to actually exist, certain (very subtle) restrictions have to be put in place.

The Vector Field $X_L$

As we said above, $\mathfrak{g}$ is the set of left-invariant vector fields on $G$. We want to study the action of $G$ on $M$, so a reasonable idea is to see if we can naturally turn each vector field $L \in \mathfrak{g}$ into a vector field $X_L$ on $M$.

This is done easily enough. Recall that $\exp(tL)$ is the integral curve of $L$ (almost by definition). Fix an $x \in M$ to create a curve $\gamma(t) = \exp(tL) \cdot x$ in $M$. We'll define $X_L$ to be the vector field $$ X_L(x) = \left.\frac{d}{dt}\right|_{t = 0} \exp(tL) \cdot x $$

What is this $X_L$? Based on how we've defined things, $X_L$ is just the generator of the flow on $M$ induced by $L$.

The Function $\mu_L$

At this point, one wonders: is $X_L$ is a Hamiltonian vector field and, if so, which Hamiltonian generates it? It turns out that $X_L$ is not in general a Hamiltonian vector field and the conditions under which it is one depend on both $G$ and $M$.

Let's suppose there is a Hamiltonian $\mu_L$ for each $L \in \mathfrak{g}$, i.e. $X_{\mu_L} = X_L$. We have $$ \{\mu_L, f\} = \omega(X_{\mu_L}, X_f) = \omega(X_L, X_f) = -\omega(X_f, X_L) = - df(X_L) = - X_L f $$ And so we see that $\{\cdot, \mu_L \} = X_L$ as vector fields. Hamilton's equations of motion tell us that for Hamiltonian $\mu_L$, we have $$ \frac{df}{dt} = \{f, \mu_L\} = X_L f $$ That is, if $\mu_L$ is our Hamiltonian, then $\frac{d}{dt} = X_L$. This, to me, is the clearest way to think about what the moment map is, even though, as we shall soon see, authors tend to define things in a way where it is not clear that this is what's going on.

The Moment Map?

You may have noticed that $L$ comes from a Lie algebra $\mathfrak{g}$ and that $\mu_L$ lives in the Lie algebra of functions on $M$. You may wonder: is $L \to \mu_L$ a Lie algebra homomorphism? The answer is, again, in general no.

Let's suppose $L \to \mu_L$ is a Lie algebra homomorphism. Let's temporarily give this map a name, $N$, so that $N(L) = \mu_L$. If we evaluate $N(L)$ at a point $x \in M$, we get $N(L)(x) = \mu_L(x)$. Observe that $N$ is linear in the first argument (it is a homomorphism), so $N(\cdot)(x)$ is a linear function $\mathfrak{g} \to \mathbb{R}$ for every $x \in M$. In other words, $N(\cdot)(x) \in \mathfrak{g}^*$.

This motivates the definition of the moment map in the OP. We say that the moment map is a function $\mu : M \to \mathfrak{g}^*$ defined by $\mu(x)(L) = N(L)(x)$.

Unwinding the OP Condition

At last we can unwind the condition in the OP and try to state things in a more intuitive way. We require $$ \omega(X_\xi , \cdot) = d\langle \mu, \xi \rangle $$ This means \begin{align} \omega(X_\xi, Y) &= d\langle \mu, \xi \rangle (Y) \\&= d(\mu(\cdot)(\xi))(Y) \\&= d(N(\xi)(\cdot))(Y) \\&= d\mu_{\xi} (Y) \\&= (i_{X_{\mu_{\xi}}} \omega)(Y) \\&= \omega(X_{\mu_{\xi}}, Y) \end{align} By non-degeneracy of $\omega$, this is the now familiar condition $X_\xi = X_{\mu_{\xi}}$, where $\mu_\xi$ is just $\langle \mu, \xi\rangle$.

Final Thoughts

Hopefully this gives you some geometric intuition for the condition in the OP. It is nothing but the statement that $X_\xi = X_{\mu_{\xi}}$ where $\mu_{\xi} = \langle \mu, \xi\rangle$. In the language of Hamiltonian mechanics, this means that if $\langle \mu, \xi \rangle$ is the Hamiltonian, then $\frac{d}{dt} = X_\xi$.

This condition is automatically satisfied as soon as we can assign a Hamiltonian vector field to each $L \in \mathfrak{g}$, no matter what that assignment looks like.

If that assignment can be chosen to be a lie algebra homomorphism, only then do we call it a moment map. The condition that the assignment be a lie algebra homomorphism is the really interesting part. Notice that the choice of $\mu_L$ for each $L$ is only unique up to a constant locally. So maybe we should pick constants so that $L \to \mu_L$ is a lie-algebra homomorphism. It turns out we can't always do this. When we can't, we say that the action by $G$ has an anamoly. This depends on very subtle properties of the action and is of course of great interest to physicists.


A moment map is a set of rules that measure your symplectic actions.

On a symplectic manifold, you can use one Hamiltonian function $H$ to test action/energy/momentum at every point of your symplectic manifold equipped with the symmetry (precisely, a flow generated by the Hamiltonian vector field $X_H$) corresponding to the Hamiltonian. This is a version of Noether's theorem. So you can think of $H$ as A ruler of the symmetry.

Now, suppose you have more than one symmetries! Precisely, you have a Lie group $G$ of dimension greater than 1 acts on your symplectic manifold. Then for every element $\zeta\in \mathfrak{g}$ of Lie algebra, you can first assume the flow corresponding to a $\zeta$ is a Hamiltonian flow: i.e. there exists $H_\zeta$ such that $X_\zeta=X_{H_\zeta}$. Therefore, you have an assignment: for a symmetry $\zeta$, you have a Hamiltonian function $H_\zeta$ (a ruler for each $\zeta$). By definition of the Hamiltonian field, you have $dH_\zeta=\iota_{X_\zeta}\omega$. (In fact, here you just need to solve a ODE for each $\zeta$).

Now, you may expect that the assignment $\zeta \mapsto H_\zeta$ should be linear since momentum should be linear with respect to speed. In particular, the ruler $\zeta \mapsto H_\zeta(x)$ at every point $x$ should be linear with respect to every symmetry $\zeta$. Consequently, we have that $\zeta \mapsto H_\zeta(x),\, \mathfrak{g}\rightarrow \mathbb{R}$ is an element in $\mathfrak{g}^*$ for every $x$. We call this linear function $\mu(x)$. Then we have a map $\mu: M \rightarrow \mathfrak{g}^*$. And by the definition, we have $\langle \mu,\zeta\rangle(x)=H_\zeta(x)$. So we put them together to see that $$d\langle \mu,\zeta\rangle=dH_\zeta=\iota_{X_\zeta}\omega.$$

Presently, we actually didn't check that the map $\mu$ we constructed from this scheme is well-defined! Because the choice of $H_\zeta$ has freedom since we only know its differential. So, we can also choose $H_\zeta+c(\zeta)$ as the Hamiltonian of $X_\zeta$. In particular, for $\zeta, \eta\in \mathfrak{g}$, you can check that we can take both $\{H_\zeta,H_\eta\}$ and $H_{[\zeta,\eta]}$ as the Hamiltonian function of $[\zeta,\eta]$. But definitively, they can differ by a constant $C(\zeta, \eta)$. By linearity of $c,C$ with respect $\zeta, \eta\in \mathfrak{g}$, we have that $c\in \mathfrak{g}^*$ and $C\in \wedge^2\mathfrak{g}^*$.

However, you can check that to guarantee the $G$-equivarance, equivalently, we should require $\{H_\zeta,H_\eta\}=H_{[\zeta,\eta]}$, which we don't know so far. So, if you want to correct the issue by carefully choosing the constants $c,C$, we need the following coherence condition (which is not a prior known) $$C(\zeta, \eta) = c([\zeta,\eta]),\forall \zeta,\eta\in \mathfrak{g}.$$ If you can choose $c,C$ with this condition, then you will see that after the correction by $c$, the map $\mu+c$ gives you a moment map. Actually, this procedure is a typical scheme to construct your moment map!

In conclusion, a moment map is a collection of rulers that measure your group of symmetries in a consistent way (compared to a single Hamiltonian function).

Bonnes: In fact, we have that $c\in C^1(\mathfrak{g},\mathbb{R})=\mathfrak{g}^*$ and $C\in C^2(\mathfrak{g},\mathbb{R})=\wedge^2\mathfrak{g}^*$, where $\wedge^q\mathfrak{g}^*=C^q(\mathfrak{g},\mathbb{R})$ is the Lie algebra cochain with $\mathbb{R}$ coefficient. Moreover, the coherence condition means that $C=-\partial_{CE}c$, where $\partial_{CE}$ is Chevalley-Eilenberg differential of the Lie algebra cochain. So, under certain conditions of Lie algebra cohomology, you can prove existence/uniqueness of moment maps.


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