Hamiltonian group actions and moment maps

Definition 6.3

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

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

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

 

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

$$\mathrm {Lie} (\mathrm {Symp} (X,\omega ))=\mathrm {symp} (X,\omega ),$$

$$\mathrm {Lie} (\mathrm {Symp} (X,\omega ))=\mathrm {symp} (X,\omega ),$$

the space of symplectic vector fields on $X$$X$. (This sits inside the space of all smooth vector fields, which is the Lie algebra of the group of diffeomorphisms of $X$$X$.)

Exercise 6.5

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

$$\{F,G\}=\omega (v_{F},v_{G}).$$

$$\{F,G\}=\omega (v_{F},v_{G}).$$

Show that $\{\,,\,\}$$\{\,,\,\}$ is a Lie bracket on $C^{\infty }(X,\mathbb {R} )$$C^{\infty }(X,\mathbb {R} )$, and that the map

$${\begin{aligned}C^{\infty }(X,\mathbb {R} )&\to \mathrm {symp} (X)\\F&\mapsto v_{F}\end{aligned}}$$

$${\begin{aligned}C^{\infty }(X,\mathbb {R} )&\to \mathrm {symp} (X)\\F&\mapsto v_{F}\end{aligned}}$$

is a Lie algebra homomorphism with respect to the usual Lie bracket of vector fields. Deduce that the space of Hamiltonian vector fields $\mathrm {ham} (X)\subseteq \mathrm {symp} (X)$$\mathrm {ham} (X)\subseteq \mathrm {symp} (X)$ is a Lie subalgebra.

 

Definition 6.4

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

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

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

 

Definition 6.5

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

$$G\to \mathrm {Ham} (X,\omega )\to \mathrm {Diff} (X).$$

$$G\to \mathrm {Ham} (X,\omega )\to \mathrm {Diff} (X).$$

Equivalently, the action is Hamiltonian if for every $\xi \in \mathrm {Lie} (G)$$\xi \in \mathrm {Lie} (G)$, associated vector field $v_{\xi }$$v_{\xi }$ given by

$$(v_{\xi })_{x}={\frac {d}{dt}}\exp(t\xi )\cdot x|_{t=0}$$

$$(v_{\xi })_{x}={\frac {d}{dt}}\exp(t\xi )\cdot x|_{t=0}$$

is Hamiltonian.

 

Example 6.5

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

$$v_{\{H_{i},H_{j}\}}=[v_{H_{i}},v_{H_{j}}]=0$$

$$v_{\{H_{i},H_{j}\}}=[v_{H_{i}},v_{H_{j}}]=0$$

and hence $\{H_{i},H_{j}\}$$\{H_{i},H_{j}\}$ is constant for all $i,j$$i,j$. With a little more work, we can actually show that

$$\{H_{i},H_{j}\}=0\quad {\text{for all }}i,j.$$

$$\{H_{i},H_{j}\}=0\quad {\text{for all }}i,j.$$

 

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

Definition 6.6

A submanifold $L\subseteq X$$L\subseteq X$ is called isotropic if $\omega |_{L}=0$$\omega |_{L}=0$, and Lagrangian if in addition $\dim L={\frac {1}{2}}\dim X$$\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 $T^{n}$$T^{n}$ acting on $\mathbb {C} ^{n}$$\mathbb {C} ^{n}$ via

$$(e^{i\theta _{1}},e^{i\theta _{2}},\ldots ,e^{i\theta _{n}})\cdot (z_{1},z_{2},\ldots ,z_{n})=(e^{i\theta _{1}}z_{1},e^{i\theta _{2}}z_{2},\ldots ,e^{i\theta _{n}}z_{n}).$$

$$(e^{i\theta _{1}},e^{i\theta _{2}},\ldots ,e^{i\theta _{n}})\cdot (z_{1},z_{2},\ldots ,z_{n})=(e^{i\theta _{1}}z_{1},e^{i\theta _{2}}z_{2},\ldots ,e^{i\theta _{n}}z_{n}).$$

The corresponding Hamiltonians are $H_{i}={\frac {1}{2}}|z_{i}|^{2}$$H_{i}={\frac {1}{2}}|z_{i}|^{2}$ generating the $\theta _{i}$$\theta _{i}$ rotation.

 

Example 6.7

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

 

Definition 6.7

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

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

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

 

Remark 6.6

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

 

Example 6.8

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

$$\mu (z_{1},\ldots ,z_{n})=({\frac {1}{2}}|z_{1}|^{2},\ldots ,{\frac {1}{2}}|z_{n}|^{2}).$$

$$\mu (z_{1},\ldots ,z_{n})=({\frac {1}{2}}|z_{1}|^{2},\ldots ,{\frac {1}{2}}|z_{n}|^{2}).$$

The moment image is

$$\mu (\mathbb {C} ^{n})=\{(x_{1},\ldots ,x_{n})|x_{i}\geq 0\}.$$

$$\mu (\mathbb {C} ^{n})=\{(x_{1},\ldots ,x_{n})|x_{i}\geq 0\}.$$

For example, for $n=2$$n=2$, the moment image is the quadrant in $\mathbb {R} ^{2}$$\mathbb {R} ^{2}$ below.

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

$$\mu ^{-1}(c)=\{(e^{i\theta _{1}}r_{1},e^{i\theta _{2}}r_{2})\in \mathbb {C} ^{2}\}$$

$$\mu ^{-1}(c)=\{(e^{i\theta _{1}}r_{1},e^{i\theta _{2}}r_{2})\in \mathbb {C} ^{2}\}$$

is a free $T^{2}$$T^{2}$-orbit.

 

In general, we have the following cool fact.

Theorem 6.1

If $\mu :X\to \mathbb {R} ^{n}$$\mu :X\to \mathbb {R} ^{n}$ is the moment map for a Hamiltonian torus action, then $\mu (X)\subseteq \mathbb {R} ^{n}$$\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 $G$$G$ be a connected compact Lie group with Lie algebra ${\mathfrak {g}}$${\mathfrak {g}}$ and a Hamiltonian action on a symplectic manifold $(X,\omega )$$(X,\omega )$. A map $\mu :X\to {\mathfrak {g}}^{*}$$\mu :X\to {\mathfrak {g}}^{*}$ is called an equivariant moment map if

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

 

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

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

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

Exercise 6.6

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

$$\iota _{v_{\xi }}\omega =-d\langle \mu ,\xi \rangle .$$

$$\iota _{v_{\xi }}\omega =-d\langle \mu ,\xi \rangle .$$

Show that the following are equivalent.

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

$${\tilde {\mu }}:{\mathfrak {g}}\to C^{\infty }(X,\mathbb {R} )\quad {\text{given by}}\quad {\tilde {\mu }}(\xi )(x)=\langle \mu (x),\xi \rangle$$

$${\tilde {\mu }}:{\mathfrak {g}}\to C^{\infty }(X,\mathbb {R} )\quad {\text{given by}}\quad {\tilde {\mu }}(\xi )(x)=\langle \mu (x),\xi \rangle$$

is a homomorphism of (left) $G$$G$-modules, where $G$$G$ acts on ${\mathfrak {g}}$${\mathfrak {g}}$ via the adjoint action, and on $C^{\infty }(X,\mathbb {R} )$$C^{\infty }(X,\mathbb {R} )$ by $(gf)(x)=f(g^{-1}x)$$(gf)(x)=f(g^{-1}x)$.

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

 

Lemma 6.1

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

 

Proof

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

$$V=\{(F,\xi )\in C^{\infty }(X,\mathbb {R} )\oplus {\mathfrak {g}}\mid v_{F}=v_{\xi }\}.$$

$$V=\{(F,\xi )\in C^{\infty }(X,\mathbb {R} )\oplus {\mathfrak {g}}\mid v_{F}=v_{\xi }\}.$$

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