# Hamiltonian flows, cotangent lifts, and momentum maps

Jordan Bell
April 3, 2014

## 1 Symplectic manifolds

Let $(M,\omega)$ and $(N,\eta)$ be symplectic manifolds. A symplectomorphism $F:M\to N$ is a diffeomorphism such that $\omega=F^{*}\eta$. Recall that for $x\in M$ and $v_{1},v_{2}\in T_{x}M$,

 $(F^{*}\eta)_{x}(v_{1},v_{2})=\eta_{F(x)}((T_{x}F)v_{1},(T_{x}F)v_{2});$

$T_{x}F:T_{x}M\to T_{F(x)}N$. (A tangent vector at $x\in M$ is pushed forward to a tangent vector at $F(x)\in N$, while a differential 2-form on $N$ is pulled back to a differential 2-form on $M$.) In these notes the only symplectomorphisms in which we are interested are those from a symplectic manifold to itself.11 1 I am interested in flows on a phase space and this phase space is a symplectic manifold. For some motivation for why we want phase space to be a symplectic manifold, read: http://research.microsoft.com/en-us/um/people/cohn/thoughts/symplectic.html

If $(M,\omega)$ is a symplectic manifold and $H\in C^{\infty}(M)$, using the nondegeneracy of the symplectic form $\omega$ one can prove that there is a unique vector field $X_{H}\in\Gamma^{\infty}(M)$ such that, for all $x\in M,v\in T_{x}M$,

 $\omega_{x}(X_{H}(x),v)=(dH)_{x}(v).$

This can also be written as

 $i_{X_{H}}\omega=dH,$

where

 $(i_{X}\omega)(Y)=(X\lrcorner\omega)(Y)=\omega(X,Y).$

We call $X_{H}$ the symplectic gradient of $H$. If $X\in\Gamma^{\infty}(M)$ and $X=X_{H}$ for some $H\in C^{\infty}(M)$, we say that $X$ is a Hamiltonian vector field.22 2 On a Riemannian manifold, a vector field that is the gradient of a smooth function is called a gradient vector field or a conservative vector field.

Let’s check that

 $X_{H}=\sum_{i=1}^{n}\frac{\partial H}{\partial p_{i}}\frac{\partial}{\partial q% _{i}}-\frac{\partial H}{\partial q_{i}}\frac{\partial}{\partial p_{i}}.$

We have, because $dq_{i}\frac{\partial}{\partial q_{j}}=\delta_{ij}$, $dp_{i}\frac{\partial}{\partial p_{j}}=\delta_{ij}$, $dq_{i}\frac{\partial}{\partial p_{j}}=0$ and $dp_{i}\frac{\partial}{\partial q_{j}}=0$, and because $dq_{j}\wedge dp_{j}=-dp_{j}\wedge dq_{j}$,

 $\displaystyle i_{X_{H}}\omega$ $\displaystyle=$ $\displaystyle\sum_{i=1}^{n}dq_{i}\wedge dp_{i}\sum_{j=1}^{n}\left(\frac{% \partial H}{\partial p_{j}}\frac{\partial}{\partial q_{j}}-\frac{\partial H}{% \partial q_{j}}\frac{\partial}{\partial p_{j}}\right)$ $\displaystyle=$ $\displaystyle\sum_{i=1}^{n}dq_{i}\wedge dp_{i}\left(\frac{\partial H}{\partial p% _{i}}\frac{\partial}{\partial q_{i}}-\frac{\partial H}{\partial q_{i}}\frac{% \partial}{\partial p_{i}}\right)$ $\displaystyle=$ $\displaystyle\sum_{i=1}^{n}\frac{\partial H}{\partial p_{i}}dp_{i}+\frac{% \partial H}{\partial q_{i}}dq_{i}$ $\displaystyle=$ $\displaystyle dH.$

## 3 Flows

Let $M$ be a smooth manifold. Let $D$ be an open subset of $M\times\mathbb{R}$, and for each $x\in M$ suppose that

 $D^{x}=\{t\in\mathbb{R}:(x,t)\in D\}$

is an open interval including $0$. A flow on $M$ is a smooth map $\phi:D\to M$ such that if $x\in M$ then $\phi_{0}(x)=x$ and such that if $x\in M$, $s\in D^{x},t\in D^{\phi_{s}(x)}$ and $s+t\in D^{x}$, then

 $\phi_{t}(\phi_{s}(x))=\phi_{s+t}(x).$

For $x\in M$, define $\phi^{x}:D^{x}\to M$ by $\phi^{x}(t)=\phi_{t}(x)$. The infinitesimal generator of a flow $\phi$ is the vector field $V$ on $M$ defined for $x\in M$ by

 $V_{x}=\frac{d}{dt}\Big{|}_{t=0}\phi^{x}(t).$

It is a fact that every vector field on $M$ is the infinitesimal generator of a flow on $M$, and furthermore that there is a unique flow whose domain is maximal that has that vector field as its infinitesimal generator, and we thus speak of the flow of a vector field.

We say that a vector field is complete if it is the infinitesimal generator of a flow whose domain is $\mathbb{R}\times M$, in other words if it is the infinitesimal generator of a global flow. It is a fact that if $V$ is a vector field on a compact smooth manifold then $V$ is complete.

## 4 Hamiltonian flows

Let $(M,\omega)$ be a symplectic manifold. We say that a vector field $X$ on $M$ is symplectic if

 $\mathcal{L}_{X}\omega=0,$

where $\mathcal{L}_{X}\omega$ is the Lie derivative of $\omega$ along the flow of $X$. A Hamiltonian flow is the flow of a Hamiltonian vector field.33 3 cf. gradient flow. If $X$ is a complete symplectic vector field and $\phi:M\times\mathbb{R}$ is the flow of $X$, then for all $t\in\mathbb{R}$, the map $\phi_{t}:M\to M$ is a symplectomorphism.

Let $H\in C^{\infty}(M)$, and let $\phi$ be the flow of the vector field $X_{H}$. If $(x,s)$ is in the domain of the flow $\phi$, we have

 $\displaystyle\frac{d}{dt}\Big{|}_{t=s}H(\phi^{x}(t))$ $\displaystyle=$ $\displaystyle(d_{\phi^{x}(s)}H)((\phi^{x})^{\prime}(s))$ $\displaystyle=$ $\displaystyle(d_{\phi^{x}(s)}H)(X_{H}(\phi^{x}(s)))$ $\displaystyle=$ $\displaystyle\omega_{\phi^{x}(s)}(X_{H}(\phi^{x}(s)),X_{H}(\phi^{x}(s)))$ $\displaystyle=$ $\displaystyle 0.$

Thus a Hamiltonian vector field is symplectic: $H$ does not change along the flow of $X_{H}$. We can also write this as

 $\displaystyle\frac{d}{dt}(H\circ\phi_{t})$ $\displaystyle=$ $\displaystyle\frac{d}{dt}(\phi_{t}^{*}H)$ $\displaystyle=$ $\displaystyle\phi_{t}^{*}(\mathcal{L}_{X_{H}}H)$ $\displaystyle=$ $\displaystyle\phi_{t}^{*}((i_{X_{H}}\omega)(X_{H}))$ $\displaystyle=$ $\displaystyle\phi_{t}^{*}(\omega(X_{H},X_{H}))$ $\displaystyle=$ $\displaystyle\phi_{t}^{*}(0)$ $\displaystyle=$ $\displaystyle 0.$

It is a fact that if $H_{\textrm{dR}}^{1}(M)=\{0\}$ (i.e. if $\alpha$ is a 1-form on $M$ and $d\alpha=0$ then there is some $f\in C^{\infty}(M)$ such that $\alpha=df$) then every symplectic vector field on $M$ is Hamiltonian. In particular, if $M$ is simply connected then $H_{\textrm{dR}}^{1}(M)=\{0\}$, and hence if $M$ is simply connected then every symplectic vector field on $M$ is Hamiltonian.

## 5 Poisson bracket

For $f,g\in C^{\infty}(M)$, we define $\{f,g\}\in C^{\infty}(M)$ for $x\in M$ by

 $\{f,g\}(x)=\omega_{x}(X_{f}(x),X_{g}(x)).$

This is called the Poisson bracket of $f$ and $g$. We write

 $\{f,g\}=\omega(X_{f},X_{g}).$

We have

 $\{f,g\}=X_{f}g=(df)X_{g}.$

We say that $f$ and $g$ Poisson commute if $\{f,g\}=0$. The Poisson bracket of $f$ and $g$ tells us how $f$ changes along the Hamiltonian flow of $g$. If $f$ and $g$ Poisson commute then $f$ does not change along the flow of $X_{g}$.

We have

 $\displaystyle\{f,g\}$ $\displaystyle=$ $\displaystyle\omega(X_{f},X_{g})$ $\displaystyle=$ $\displaystyle\sum_{i=1}^{n}(dq_{i}\wedge dp_{i})\sum_{j=1}^{n}\left(\frac{% \partial f}{\partial p_{j}}\frac{\partial}{\partial q_{j}}-\frac{\partial f}{% \partial q_{j}}\frac{\partial}{\partial p_{j}}\right)\sum_{k=1}^{n}\left(\frac% {\partial g}{\partial p_{k}}\frac{\partial}{\partial q_{k}}-\frac{\partial g}{% \partial q_{k}}\frac{\partial}{\partial p_{k}}\right)$ $\displaystyle=$ $\displaystyle\sum_{i=1}^{n}\left(\frac{\partial f}{\partial p_{i}}dp_{i}+\frac% {\partial f}{\partial q_{i}}dq_{i}\right)\sum_{k=1}^{n}\left(\frac{\partial g}% {\partial p_{k}}\frac{\partial}{\partial q_{k}}-\frac{\partial g}{\partial q_{% k}}\frac{\partial}{\partial p_{k}}\right)$ $\displaystyle=$ $\displaystyle\sum_{i=1}^{n}-\frac{\partial f}{\partial p_{i}}\frac{\partial g}% {\partial q_{i}}+\frac{\partial f}{\partial q_{i}}\frac{\partial g}{\partial p% _{i}}.$

If $x\in M$ and $v\in T_{x}M$, then $vf$ is the directional derivative in the direction $v$. If $v=\sum_{i=1}^{n}a_{i}\frac{\partial}{\partial q_{i}}+b_{i}\frac{\partial}{% \partial p_{i}}$ and $f\in C^{\infty}(M)$ then

 $vf=\sum_{i=1}^{n}a_{i}\frac{\partial f}{\partial q_{i}}+b_{i}\frac{\partial f}% {\partial p_{i}}.$

If $X$ is a vector field on $M$ then $Xf\in C^{\infty}(M)$, defined for $x\in M$ by

 $(Xf)(x)=X_{x}f.$

If $\tau$ is a covariant tensor field and $X$ is a vector field, the Lie derivative of $\tau$ along the flow of $X$ is defined as follows: if $\phi$ is the flow of $X$, then

 $(\mathcal{L}_{X}\tau)(x)=\frac{d}{dt}\Big{|}_{t=0}(\phi_{t}^{*}\tau)(x),$

and so if $\tau$ is a function $f\in C^{\infty}(M)$, then

 $(\mathcal{L}_{X}f)(x)=\frac{d}{dt}\Big{|}_{t=0}(\phi_{t}^{*}f)(x)=\frac{d}{dt}% \Big{|}_{t=0}f(\phi_{t}(x))=X_{x}f=(Xf)(x).$

Thus if $X$ is a vector field and $f\in C^{\infty}(M)$, then $\mathcal{L}_{X}f=Xf$.

For $f,g\in C^{\infty}(M)$,

 $\displaystyle X_{\{f,g\}}\lrcorner\omega$ $\displaystyle=$ $\displaystyle d\{f,g\}$ $\displaystyle=$ $\displaystyle d(X_{g}f)$ $\displaystyle=$ $\displaystyle d(\mathcal{L}_{X_{g}}f)$ $\displaystyle=$ $\displaystyle\mathcal{L}_{X_{g}}(df)$ $\displaystyle=$ $\displaystyle\mathcal{L}_{X_{g}}(X_{f}\lrcorner\omega)$ $\displaystyle=$ $\displaystyle(\mathcal{L}_{X_{g}}X_{f})\lrcorner\omega+X_{f}\lrcorner\mathcal{% L}_{X_{g}}\omega$ $\displaystyle=$ $\displaystyle[X_{g},X_{f}]\lrcorner\omega+X_{f}\lrcorner 0$ $\displaystyle=$ $\displaystyle[X_{g},X_{f}]\lrcorner\omega$ $\displaystyle=$ $\displaystyle-[X_{f},X_{g}]\lrcorner\omega.$

Since the symplectic form $\omega$ is nondegenerate, if $X\lrcorner\omega=Y\lrcorner\omega$ then $X=Y$, so

 $X_{\{f,g\}}=-[X_{f},X_{g}].$

It follows that $C^{\infty}(M)$ is a Lie algebra using the Poisson bracket as the Lie bracket.

The set $\Gamma^{\infty}(M)$ of vector fields on $M$ are a Lie algebra using the vector field commutator $[\cdot,\cdot]$. The symplectic vector fields are a Lie subalgebra: it is clear that they are a linear subspace of the Lie algebra of vector fields, and one shows that the commutator of two symplectic vector fields is itself a symplectic vector field. One can further show that the set of Hamiltonian vector fields is a Lie subalgebra of the Lie algebra of symplectic vector fields. It is a fact that the vector space quotient of the vector space of symplectic vector fields modulo the vector space of Hamiltonian vector fields is isomorphic to the vector space $H_{\textrm{dR}}^{1}(M)$; this is why if $H_{\textrm{dR}}^{1}(M)=\{0\}$ (in particular if $M$ is simply connected) then any symplectic vector field on $M$ is Hamiltonian.

## 6 Tautological 1-form

Let $Q$ be a smooth manifold and let $\pi:T^{*}Q\to Q$, $\pi(q,p)=q$. For $x=(q,p)\in T^{*}Q$, we have

 $d_{x}\pi:T_{x}T^{*}Q\to T_{q}Q.$

Let

 $\theta_{x}=(d_{x}\pi)^{*}(p)=p\circ d_{x}\pi:T_{x}T^{*}Q\to\mathbb{R}.$

Thus $\theta:T^{*}Q\to T^{*}T^{*}Q$. $\theta$ is called the tautological 1-form on $T^{*}Q$.

If $(Q_{1},\ldots,Q_{n})$ are coordinates on an open subset $U$ of $Q$, $Q_{i}:U\to\mathbb{R}$, then for each $q\in U$ we have that $d_{q}Q_{i}\in T_{q}^{*}U=T_{q}^{*}Q$, $1\leq i\leq n$, are a basis for $T_{q}^{*}Q$ and $\frac{\partial}{\partial Q_{i}}\big{|}_{q}$, $1\leq i\leq n$, are a basis for $T_{q}Q$. For each $p\in T_{q}^{*}Q$,

 $p=\sum_{i=1}^{n}p\left(\frac{\partial}{\partial Q_{i}}\Big{|}_{q}\right)d_{q}Q% _{i}.$

On $T^{*}U$, define coordinates $(q_{1},\ldots,q_{n},p_{1},\ldots,p_{n})$ by

 $q_{i}(q,p)=Q_{i}(q),$

and

 $p_{i}(q,p)=p\left(\frac{\partial}{\partial Q_{i}}\Big{|}_{q}\right).$

On $T^{*}U$ we can write $\theta$ using these coordinates: for $x=(q,p)\in T^{*}Q$,

 $\theta_{x}=p\circ d_{x}\pi=\sum_{i=1}^{n}p_{i}(x)d_{x}q_{i}.$

Thus, on $T^{*}U$,

 $\theta=\sum_{i=1}^{n}p_{i}dq_{i}.$

Let $\omega=-d\theta$. We have, on $T^{*}U$,

 $\displaystyle\omega$ $\displaystyle=$ $\displaystyle-d\sum_{i=1}^{n}p_{i}dq_{i}$ $\displaystyle=$ $\displaystyle-\sum_{i=1}^{n}\left(dp_{i}\wedge dq_{i}+p_{i}d(dq_{i})\right)$ $\displaystyle=$ $\displaystyle-\sum_{i=1}^{n}dp_{i}\wedge dq_{i}$ $\displaystyle=$ $\displaystyle\sum_{i=1}^{n}dq_{i}\wedge dp_{i}.$

$T^{*}Q$ is a symplectic manifold with the symplectic form $\omega$.

## 7 Cotangent lifts

Let $Q$ be a smooth manifold and let $F:Q\to Q$ be a diffeomorphism. Define

 $F^{\sharp}:T^{*}Q\to T^{*}Q$

for $x=(q,p)$ by

 $F^{\sharp}(q,p)=(F(q),(d_{F(q)}(F^{-1}))^{*}(p)).$

We call $F^{\sharp}:T^{*}Q\to T^{*}Q$ the cotangent lift of $F:Q\to Q$. It is a fact that it is a diffeomorphism. It is apparent that the following diagram commutes:

The pull-back of $\theta$ by $F^{\sharp}$ satisfies, for $x=(q,p)\in T^{*}Q$ and $(\zeta,\eta)=F^{\sharp}(q,p)\in T^{*}Q$,

 $\displaystyle((F^{\sharp})^{*}\theta)_{x}$ $\displaystyle=$ $\displaystyle(d_{x}F^{\sharp})^{*}(\theta_{F^{\sharp}(x)})$ $\displaystyle=$ $\displaystyle(d_{x}F^{\sharp})^{*}((d_{F^{\sharp}(x)}\pi)^{*}(\eta))$ $\displaystyle=$ $\displaystyle(d_{x}(\pi\circ F^{\sharp}))^{*}(\eta)$ $\displaystyle=$ $\displaystyle(d_{x}(F\circ\pi))^{*}(\eta)$ $\displaystyle=$ $\displaystyle(d_{x}\pi)^{*}((d_{\pi(x)}F)^{*}(\eta))$ $\displaystyle=$ $\displaystyle(d_{x}\pi)^{*}((d_{q}F)^{*}(\eta))$ $\displaystyle=$ $\displaystyle(d_{x}\pi)^{*}(p)$ $\displaystyle=$ $\displaystyle\theta_{x}.$

Thus $(F\sharp)^{*}\theta=\theta$, i.e. $F^{\sharp}$ pulls back $\theta$ to $\theta$. The “naturality of the exterior derivative”44 4 For each $k$, $\Omega^{k}$ is a contravariant functor, and if $f:M\to N$, then the functor $\Omega^{k}$ sends $f$ to $f^{*}:\Omega^{k}(N)\to\Omega^{k}(M)$. $d$ is a natural transformation from the contravariant functor $\Omega^{k}$ to the contravariant functor $\Omega^{k+1}$. is the statement that if $G$ is a smooth map and $\eta$ is a differential form then $G^{*}(d\eta)=d(G^{*}\eta)$. Hence, with $\omega=d\theta$,

 $(F^{\sharp})^{*}\omega=(F^{\sharp})^{*}(d\theta)=d((F^{\sharp})^{*}\theta)=d(% \theta)=\omega,$

so $F^{\sharp}$ pulls back the symplectic form $\omega$ to itself. Thus $F^{\sharp}:T^{*}M\to T^{*}M$ is a symplectomorphism.

Let $\textrm{Diff}(Q)$ be the set of diffeomorphisms $Q\to Q$. $\textrm{Diff}(Q)$ is a group. Let $G$ be a group and let $\tau:G\to\textrm{Diff}(Q)$ be a homomorphism. Define $\tau^{\sharp}:G\to\textrm{Diff}(T^{*}Q)$ by $(\tau^{\sharp})_{g}=(\tau_{g})^{\sharp}:T^{*}Q\to T^{*}Q$. $\tau^{\sharp}:G\to\textrm{Diff}(T^{*}Q)$ is a homomorphism, and for each $g\in G$, $(\tau^{\sharp})_{g}:T^{*}Q\to T^{*}Q$ is a symplectomorphism. In words, if a group acts by diffeomorphisms on a smooth manifold, then the cotangent lift of the action is an action by symplectomorphisms on the cotangent bundle.

## 8 Lie groups

Recall that if $F:M\to N$ then $TF:TM\to TN$ satisfies, for $X\in\Gamma^{\infty}(M)$ and $f\in C^{\infty}(N)$,55 5 In words: $TF$ pushes forward a vector field on $M$ to a vector field on $N$.

 $((TF)X)(f)=X(f\circ F),$

i.e. for $x\in M$ and $v\in T_{x}M$,

 $((T_{x}F)v)(f)=v(f\circ F),$

the directional derivative of $f\circ F\in C^{\infty}(M)$ in the direction of the tangent vector $v$.

Let $G$ be a Lie group and for $g\in G$ define $L_{g}:G\to G$ by $L_{g}h=gh$. If $X$ is a vector field on $G$, we say that $X$ is left-invariant if

 $(T_{h}L_{g})(X_{h})=X_{gh}$

for all $g,h\in G$. That is, $X$ is left-invariant if

 $(TL_{g})(X)=X$

for all $g\in G$.

If $X$ and $Y$ are left-invariant vector fields on $G$ then so is $[X,Y]$. This is because, for $F:G\to G$,

 $(TF)[X,Y]=[(TF)X,(TF)Y].$

Thus the set of left-invariant vector fields on $G$ is a Lie subalgebra of the Lie algebra of vector fields on $G$.

Define $\epsilon:\textrm{Lie}(G)\to T_{e}G$ by $\epsilon(X)=X_{e}$, where $e\in G$ is the identity element. It can be shown that this is a linear isomorphism. Hence, if $v\in T_{e}G$ then there is a unique left-invariant vector field $X$ on $G$ such that, for all $g\in G$,

 $V_{g}=(T_{e}L_{g})(v).$

It is a fact that every left-invariant vector field on a Lie group $G$ is complete, i.e. that its flow has domain $G\times\mathbb{R}$. For $X\in\textrm{Lie}(G)$, we call the unique integral curve of $X$ that passes through $e$ the one-parameter subgroup generated by $X$. Thus, for any $v\in T_{e}G$ there is a unique one-parameter subgroup $\gamma:\mathbb{R}\to G$ such that

 $\gamma(0)=e,\qquad\gamma^{\prime}(0)=v.$

We define $\exp:\textrm{Lie}(G)\to G$ by $\exp(X)=\gamma(1)$, where $\gamma$ is the one-parameter subgroup generated by $X$. This is called the exponential map. Thus $t\mapsto\exp(tX)$ is the one-parameter subgroup generated by $X$.

Fact: If $(TF)X=Y$ and $X$ has flow $\phi$ and $Y$ has flow $\eta$, then

 $\eta_{t}\circ F=F\circ\phi_{t}$

for all $t$ in the domain of $\phi$. Hence

 $L_{g}\circ\phi_{t}=\phi_{t}\circ L_{g}.$

Hence the flow $\phi$ of a left-invariant vector field $X$ satisfies

 $\displaystyle g\exp(tX)$ $\displaystyle=$ $\displaystyle L_{g}\exp(tX)$ $\displaystyle=$ $\displaystyle L_{g}(\phi_{t}e)$ $\displaystyle=$ $\displaystyle\phi_{t}(L_{g}e)$ $\displaystyle=$ $\displaystyle\phi_{t}(g).$

First we’ll define the adjoint action of $G$ on $\mathfrak{g}=T_{\textrm{id}_{G}}G$. For $g\in G$, define $\Psi_{g}:G\to G$ by $\Psi_{g}(h)=ghg^{-1}$; $\Psi_{g}$ is an automorphism of Lie groups. Define

 $\textrm{Ad}_{g}:\mathfrak{g}\to\mathfrak{g}$

by

 $\textrm{Ad}_{g}=T_{\textrm{id}_{G}}\Psi_{g};$

since $\Psi_{g}$ is an automorphism of Lie groups, it follows that $\textrm{Ad}_{g}$ is an automorphism of Lie algebras. We can also write $\textrm{Ad}_{g}$ as

 $\textrm{Ad}_{g}(\xi)=\frac{d}{dt}\Big{|}_{t=0}(g\exp(t\xi)g^{-1}).$

The adjoint action of $G$ on $\mathfrak{g}$ is

 $g\cdot\xi=\textrm{Ad}_{g}(\xi).$

For each $g\in G$, one proves that there is a unique map $\textrm{Ad}_{g}^{*}:\mathfrak{g}^{*}\to\mathfrak{g}^{*}$ such that for all $l\in\mathfrak{g}^{*},\xi\in\mathfrak{g}$,

 $(\textrm{Ad}_{g}^{*}l)(\xi)=l(\textrm{Ad}_{g}(\xi)).$

The coadjoint action of $G$ on $\mathfrak{g}^{*}$ is

 $g\cdot l=\textrm{Ad}_{g^{-1}}^{*}(l).$

## 10 Momentum map

Let $(M,\omega)$ be a symplectic manifold, let $G$ be a Lie group, and let $\sigma:G\to\textrm{Diff}(M)$ be a homomorphism such that for each $g$ in $G$, $\sigma_{g}$ is a symplectomorphism.

Let $\mathfrak{g}=T_{\textrm{id}_{G}}G$, and define $\rho:\mathfrak{g}\to\Gamma^{\infty}(M)$ by

 $\rho(\xi)(x)=\frac{d}{dt}\Big{|}_{t=0}\sigma_{\exp(t\xi)}(x)\in T_{x}M,\qquad% \xi\in\mathfrak{g},x\in M;$

$t\mapsto\sigma_{\exp(t\xi)}(x)$ is $\mathbb{R}\to M$ and at $t=0$ the curve passes through $x$, so indeed $\rho(\xi)(x)\in T_{x}(M)$. $\rho$ is called the infinitesimal action of $\mathfrak{g}$ on $M$. Each element of $G$ acts on $M$ as a symplectomorphism, each element of $\mathfrak{g}$ acts on $M$ as a vector field.

A momentum map for the action of $G$ on $(M,\omega)$ is a map $\mu:M\to\mathfrak{g}^{*}$ such that, for $x\in M$, $v\in T_{x}M$ and $\xi\in\mathfrak{g}$,

 $((T_{x}\mu)v)\xi=\omega_{x}(\rho(\xi)(x),v),$ (1)

where

 $T_{x}\mu:T_{x}M\to T_{\mu(x)}\mathfrak{g}^{*}=\mathfrak{g}^{*},$

and such that if $g\in G$ and $x\in M$ then

 $\mu(\sigma_{g}(x))=g\cdot\mu(x),$ (2)

where $g\cdot\mu(x)$ is the coadjoint action of $G$ on $\mathfrak{g}^{*}$, defined in section §9; we say that $\mu$ is equivariant with respect to the coadjoint action of $G$ on $\mathfrak{g}^{*}$.

## 11 Angular momentum

Let $G=\textrm{SO}(3)=\{A\in\mathbb{R}^{3\times 3}:A^{T}A=I,\det(A)=1\}$. The Lie algebra of $\textrm{SO}(3)$ is

 $\mathfrak{g}=\mathfrak{so}(3)=\{a\in\mathbb{R}^{3\times 3}:a+a^{T}=0\}.$

Let $Q=\mathbb{R}^{3}$, and define $\tau:G\to\textrm{Diff}(Q)$ by $\tau_{g}(q)=gq$.

Let $\theta$ be the tautological 1-form on $T^{*}Q$ and let $\omega=-d\theta$. $(T^{*}Q,\omega)$ is a symplectic manifold and $\tau^{\sharp}:G\to\textrm{Diff}(T^{*}Q)$ is a homomorphism such that for each $g\in G$, $(\tau^{\sharp})_{g}$ is a symplectomorphism. For $g\in G$, $(q,p)\in T^{*}Q$,

 $\displaystyle(\tau^{\sharp})_{g}(q,p)$ $\displaystyle=$ $\displaystyle(\tau_{g})^{\sharp}(q,p)$ $\displaystyle=$ $\displaystyle(\tau_{g}q,(d_{\tau_{g}q}(\tau_{g}^{-1}))^{*}p)$ $\displaystyle=$ $\displaystyle(\tau_{g}q,(d_{\tau_{g}q}(\tau_{g^{-1}}))^{*}p)$ $\displaystyle=$ $\displaystyle(\tau_{g}q,p\circ(d_{\tau_{g}q}\tau_{g^{-1}}))$ $\displaystyle=$ $\displaystyle(\tau_{g}q,p\circ\tau_{g^{-1}})$ $\displaystyle=$ $\displaystyle(gq,pg^{-1})$ $\displaystyle=$ $\displaystyle(gq,pg^{T}).$

Hence for $\xi\in\mathfrak{g}$ and $(q,p)\in T^{*}Q$,

 $\displaystyle\rho(\xi)(q,p)$ $\displaystyle=$ $\displaystyle\frac{d}{dt}\Big{|}_{t=0}(\tau^{\sharp})_{\exp(t\xi)}(q,p)$ $\displaystyle=$ $\displaystyle\frac{d}{dt}\Big{|}_{t=0}(\exp(t\xi)q,p\exp(t\xi^{T}))$ $\displaystyle=$ $\displaystyle(\xi q,p\xi^{T})$ $\displaystyle=$ $\displaystyle(\xi q,-p\xi).$

Define $V:\mathfrak{g}\to\mathbb{R}^{3}$ by

 $V\begin{pmatrix}0&-\xi_{3}&\xi_{2}\\ \xi_{3}&0&-\xi_{1}\\ -\xi_{2}&\xi_{1}&0\end{pmatrix}=\begin{pmatrix}\xi_{1}\\ \xi_{2}\\ \xi_{3}\end{pmatrix}.$

One checks that $\xi q=V(\xi)\times q$ and $p\xi=p^{T}\times V(\xi)$.

For $(q,p)\in T^{*}Q$, $(v,w)\in T_{(p,q)}T^{*}Q$, and $\xi\in\mathfrak{g}$, we have

 $\displaystyle\omega_{(q,p)}(\rho(\xi)(q,p),(v,w))$ $\displaystyle=$ $\displaystyle\omega_{(q,p)}((\xi q,-p\xi),(v,w))$ $\displaystyle=$ $\displaystyle\sum_{j=1}^{3}dq_{j}\wedge dp_{j}((\xi q,-p\xi),(v,w))$ $\displaystyle=$ $\displaystyle\sum_{j=1}^{3}\left((\xi q)_{j}dp_{j}+(p\xi)_{j}dq_{j}\right)(v,w)$ $\displaystyle=$ $\displaystyle\sum_{j=1}^{3}w_{j}(\xi q)_{j}+v_{j}(p\xi)_{j}$ $\displaystyle=$ $\displaystyle w\cdot(V(\xi)\times q)+v\cdot(p^{T}\times V(\xi)).$

Define $\mu:T^{*}Q\to\mathfrak{g^{*}}$ by $\mu(q,p)(\xi)=(q\times p^{T})\cdot V(\xi)$. I claim that $\mu$ satisfies (1) and (2). We have just calculated the right-hand side of (1), so it remains to calculate the left-hand side. I find the left-hand side unwieldly to calculate in a clean and precise way, so I will merely claim that it is equal to the right-hand side. I have convinced myself that it is true by symbol pushing.

For $g\in G$ and $\xi\in\mathfrak{g}$, $\textrm{Ad}_{g}\xi=g\xi g^{-1}$, and hence, for $(q,p)\in T^{*}Q$,

 $\displaystyle(g\cdot\mu(q,p))\xi$ $\displaystyle=$ $\displaystyle\left(\textrm{Ad}_{g^{-1}}^{*}\mu(q,p)\right)\xi$ $\displaystyle=$ $\displaystyle\mu(q,p)(\textrm{Ad}_{g^{-1}}\xi)$ $\displaystyle=$ $\displaystyle\mu(q,p)(g^{-1}\xi g)$ $\displaystyle=$ $\displaystyle(q\times p^{T})\cdot V(g^{-1}\xi g).$

On the other hand,

 $\displaystyle\mu((\tau^{\sharp})_{g}(q,p))\xi$ $\displaystyle=$ $\displaystyle\mu(gq,pg^{T})\xi$ $\displaystyle=$ $\displaystyle((gq)\times(pg^{T})^{T})\cdot V(\xi)$ $\displaystyle=$ $\displaystyle((gq)\times(gp^{T}))\cdot V(\xi)$ $\displaystyle=$ $\displaystyle(g(q\times p^{T}))\cdot V(\xi)$ $\displaystyle=$ $\displaystyle(q\times p^{T})\cdot(g^{T}V(\xi))$ $\displaystyle=$ $\displaystyle(q\times p^{T})\cdot(g^{-1}V(\xi)).$