Geometric Intuition for Hamiltonian Actions

Question

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?

Answer 1

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.

Answer 2

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.