# Hamiltonian group actions and moment maps

Definition 6.3

If ${\displaystyle (X,\omega )}$ is a symplectic manifold, the group of symplectomorphisms of ${\displaystyle X}$ is the infinite-dimensional Lie group

${\displaystyle \mathrm {Symp} (X,\omega )=\{\phi :X\to X\mid \phi ^{*}\omega =\omega \}.}$

By Remark 6.3, the Lie algebra of ${\displaystyle \mathrm {Symp} (X,\omega )}$ is

${\displaystyle \mathrm {Lie} (\mathrm {Symp} (X,\omega ))=\mathrm {symp} (X,\omega ),}$
the space of symplectic vector fields on ${\displaystyle X}$. (This sits inside the space of all smooth vector fields, which is the Lie algebra of the group of diffeomorphisms of ${\displaystyle X}$.)

Exercise 6.5

If ${\displaystyle F}$ and ${\displaystyle G}$ are smooth functions on ${\displaystyle X}$, define the Poisson bracket of ${\displaystyle F}$ and ${\displaystyle G}$ by

${\displaystyle \{F,G\}=\omega (v_{F},v_{G}).}$
Show that ${\displaystyle \{\,,\,\}}$ is a Lie bracket on ${\displaystyle C^{\infty }(X,\mathbb {R} )}$, and that the map
{\displaystyle {\begin{aligned}C^{\infty }(X,\mathbb {R} )&\to \mathrm {symp} (X)\\F&\mapsto v_{F}\end{aligned}}}
is a Lie algebra homomorphism with respect to the usual Lie bracket of vector fields. Deduce that the space of Hamiltonian vector fields ${\displaystyle \mathrm {ham} (X)\subseteq \mathrm {symp} (X)}$ is a Lie subalgebra.

Definition 6.4

We denote by ${\displaystyle \mathrm {Ham} (X,\omega )}$ the unique (connected) subgroup of ${\displaystyle \mathrm {Symp} (X,\omega )}$ with

${\displaystyle \mathrm {Lie} (\mathrm {Ham} (X,\omega ))=\mathrm {ham} (X,\omega ).}$

Definition 6.5

Let ${\displaystyle G}$ be a connected Lie group acting on a symplectic manifold ${\displaystyle (X,\omega )}$. We say that the action is Hamiltonian if the action of ${\displaystyle G}$ factors through

${\displaystyle G\to \mathrm {Ham} (X,\omega )\to \mathrm {Diff} (X).}$
Equivalently, the action is Hamiltonian if for every ${\displaystyle \xi \in \mathrm {Lie} (G)}$, associated vector field ${\displaystyle v_{\xi }}$ given by
${\displaystyle (v_{\xi })_{x}={\frac {d}{dt}}\exp(t\xi )\cdot x|_{t=0}}$
is Hamiltonian.

Example 6.5

Let ${\displaystyle G=T^{n}=U(1)^{n}}$ be a torus. Then ${\displaystyle \mathrm {Lie} (G)=\mathbb {R} ^{n}}$ with Lie bracket ${\displaystyle [\,,\,]=0}$. If we have a Hamiltonian action of ${\displaystyle G}$ on ${\displaystyle (X,\omega )}$, then if ${\displaystyle \xi _{1},\ldots ,\xi _{n}}$ is a basis for ${\displaystyle \mathrm {Lie} (G)}$, we get a collection ${\displaystyle H_{1},\ldots ,H_{n}}$ of Hamiltonians on ${\displaystyle X}$. Since ${\displaystyle [\xi _{i},\xi _{j}]=0}$, passing to the corresponding vector fields ${\displaystyle v_{H_{i}}}$, we have

${\displaystyle v_{\{H_{i},H_{j}\}}=[v_{H_{i}},v_{H_{j}}]=0}$
and hence ${\displaystyle \{H_{i},H_{j}\}}$ is constant for all ${\displaystyle i,j}$. With a little more work, we can actually show that
${\displaystyle \{H_{i},H_{j}\}=0\quad {\text{for all }}i,j.}$

As a slight aside, note that since the ${\displaystyle v_{H_{i}}}$ span the tangent space to each orbit, this implies that ${\displaystyle \omega =0}$ when restricted to a ${\displaystyle T^{n}}$-orbit. Submanifolds of this type appear sufficiently often in symplectic topology that they have a special name.

Definition 6.6

A submanifold ${\displaystyle L\subseteq X}$ is called isotropic if ${\displaystyle \omega |_{L}=0}$, and Lagrangian if in addition ${\displaystyle \dim L={\frac {1}{2}}\dim X}$.

Remark 6.5

Lagrangian submanifolds appear in quantum mechanics as the smallest submanifolds on which a which a wavefunction can be localised. (cf. Heisenberg uncertainty principle)

Example 6.6

Consider ${\displaystyle T^{n}}$ acting on ${\displaystyle \mathbb {C} ^{n}}$ via

${\displaystyle (e^{i\theta _{1}},e^{i\theta _{2}},\ldots ,e^{i\theta _{n}})\cdot (z_{1},z_{2},\ldots ,z_{n})=(e^{i\theta _{1}}z_{1},e^{i\theta _{2}}z_{2},\ldots ,e^{i\theta _{n}}z_{n}).}$
The corresponding Hamiltonians are ${\displaystyle H_{i}={\frac {1}{2}}|z_{i}|^{2}}$ generating the ${\displaystyle \theta _{i}}$ rotation.

Example 6.7

Consider ${\displaystyle S^{2}\subseteq \mathbb {R} ^{3}}$ with symplectic form given by the area form. (If we identify ${\displaystyle S^{2}}$ with ${\displaystyle \mathbb {CP} ^{1}}$, then this is also the natural symplectic structure coming from the Fubini-Study form.) Then the ${\displaystyle U(1)}$ action on ${\displaystyle S^{2}}$ given by rotation about the ${\displaystyle z}$-axis is Hamiltonian, and the Hamiltonian ${\displaystyle H}$ is just the height function (projection to ${\displaystyle z}$-coordinate).

Definition 6.7

Let ${\displaystyle X}$ be a symplectic manifold with a Hamiltonian action of ${\displaystyle T^{n}}$ and associated Hamiltonians ${\displaystyle H_{1},\ldots ,H_{n}}$. The moment map of the action is ${\displaystyle \mu :X\to \mathbb {R} ^{n}}$ given by

${\displaystyle \mu (x)=(H_{1}(x),\ldots ,H_{n}(x)).}$

Remark 6.6

The moment map for a torus action is well-defined up to translation in ${\displaystyle \mathbb {R} ^{n}}$.

Example 6.8

For ${\displaystyle T^{n}}$ acting on ${\displaystyle \mathbb {C} ^{n}}$ as in Example 6.6, have

${\displaystyle \mu (z_{1},\ldots ,z_{n})=({\frac {1}{2}}|z_{1}|^{2},\ldots ,{\frac {1}{2}}|z_{n}|^{2}).}$
The moment image is
${\displaystyle \mu (\mathbb {C} ^{n})=\{(x_{1},\ldots ,x_{n})|x_{i}\geq 0\}.}$
For example, for ${\displaystyle n=2}$, the moment image is the quadrant in ${\displaystyle \mathbb {R} ^{2}}$ below.

For every point ${\displaystyle c=({\frac {1}{2}}r_{1}^{2},{\frac {1}{2}}r_{2}^{2})}$ on the interior, we have

${\displaystyle \mu ^{-1}(c)=\{(e^{i\theta _{1}}r_{1},e^{i\theta _{2}}r_{2})\in \mathbb {C} ^{2}\}}$
is a free ${\displaystyle T^{2}}$-orbit.

In general, we have the following cool fact.

Theorem 6.1

If ${\displaystyle \mu :X\to \mathbb {R} ^{n}}$ is the moment map for a Hamiltonian torus action, then ${\displaystyle \mu (X)\subseteq \mathbb {R} ^{n}}$ is a convex polytope.

We can define moment maps for more general group actions as follows.

Definition 6.8

Let ${\displaystyle G}$ be a connected compact Lie group with Lie algebra ${\displaystyle {\mathfrak {g}}}$ and a Hamiltonian action on a symplectic manifold ${\displaystyle (X,\omega )}$. A map ${\displaystyle \mu :X\to {\mathfrak {g}}^{*}}$ is called an equivariant moment map if

• ${\displaystyle \mu (gx)=\mathrm {Ad} (g)^{*}\mu (x)}$ for all ${\displaystyle g\in G}$ and ${\displaystyle x\in X}$, and
• ${\displaystyle \iota _{v_{\xi }}\omega =-d\langle \mu ,\xi \rangle }$ for ${\displaystyle \xi \in {\mathfrak {g}}}$.

Here ${\displaystyle \mathrm {Ad} (g)^{*}:{\mathfrak {g}}^{*}\to {\mathfrak {g}}^{*}}$ is (the inverse of) the dual to the adjoint action ${\displaystyle \mathrm {Ad} (g):{\mathfrak {g}}\to {\mathfrak {g}}}$, which is the map induced on the tangent space at the identity of ${\displaystyle G}$ by

{\displaystyle {\begin{aligned}G&\longrightarrow G\\h&\longmapsto ghg^{-1}.\end{aligned}}}

Exercise 6.6

In the setup of Definition of the moment map, assume that ${\displaystyle \mu :X\to {\mathfrak {g}}^{*}}$ satisfies

${\displaystyle \iota _{v_{\xi }}\omega =-d\langle \mu ,\xi \rangle .}$
Show that the following are equivalent.

1. ${\displaystyle \mu (gx)=\mathrm {Ad} (g)^{*}\mu (x)}$ for all ${\displaystyle g\in G}$ and ${\displaystyle x\in X}$.
2. The map

${\displaystyle {\tilde {\mu }}:{\mathfrak {g}}\to C^{\infty }(X,\mathbb {R} )\quad {\text{given by}}\quad {\tilde {\mu }}(\xi )(x)=\langle \mu (x),\xi \rangle }$
is a homomorphism of (left) ${\displaystyle G}$-modules, where ${\displaystyle G}$ acts on ${\displaystyle {\mathfrak {g}}}$ via the adjoint action, and on ${\displaystyle C^{\infty }(X,\mathbb {R} )}$ by ${\displaystyle (gf)(x)=f(g^{-1}x)}$.

1. The map ${\displaystyle {\tilde {\mu }}:{\mathfrak {g}}\to C^{\infty }(X,\mathbb {R} )}$ is a homomorphism of ${\displaystyle {\mathfrak {g}}}$-modules.
2. The map ${\displaystyle {\tilde {\mu }}:{\mathfrak {g}}\to C^{\infty }(X,\mathbb {R} )}$ is a homomorphism of Lie algebras.

Lemma 6.1

Let ${\displaystyle G}$ be a compact Lie group with a Hamiltonian action on a symplectic manifold ${\displaystyle X}$. Then there exists an equivariant moment map ${\displaystyle \mu :X\to {\mathfrak {g}}^{*}}$.

Proof

We have a commutative diagram of ${\displaystyle G}$-modules (MISSING) with exact rows, where

${\displaystyle V=\{(F,\xi )\in C^{\infty }(X,\mathbb {R} )\oplus {\mathfrak {g}}\mid v_{F}=v_{\xi }\}.}$
Since the bottom row is an exact sequence of finite-dimensional ${\displaystyle G}$-modules with ${\displaystyle G}$ compact, it has a ${\displaystyle G}$-equivariant splitting. In particular, we can construct a map of ${\displaystyle G}$-modules ${\displaystyle {\tilde {\mu }}:{\mathfrak {g}}\to C^{\infty }(X,\mathbb {R} )}$ covering ${\displaystyle {\mathfrak {g}}\to \mathrm {ham} (X)}$ as shown, which induces an equivariant moment map ${\displaystyle \mu :X\to {\mathfrak {g}}^{*}}$.