Hamiltonian flows, cotangent lifts, and momentum maps

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.

2 Symplectic gradient

If $(M,\omega )$ is a symplectic manifold and $H\in C^{\mathrm{\infty }}(M)$, using the nondegeneracy of the symplectic form $\omega$ one can prove that there is a unique vector field $X_H\in {\mathrm{\Gamma }}^{\mathrm{\infty }}(M)$ 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

$$(i_X\omega )(Y)=(X\mathrm{⌟}\omega )(Y)=\omega (X,Y).$$

We call $X_H$ the symplectic gradient of $H$. If $X\in {\mathrm{\Gamma }}^{\mathrm{\infty }}(M)$ and $X=X_H$ for some $H\in C^{\mathrm{\infty }}(M)$, we say that $X$ is a Hamiltonian 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 $d{q}_i\frac{\partial }{\partial q_j}={\delta }_{ij}$, $d{p}_i\frac{\partial }{\partial p_j}={\delta }_{ij}$, $d{q}_i\frac{\partial }{\partial p_j}=0$ and $d{p}_i\frac{\partial }{\partial q_j}=0$, and because $d{q}_j\wedge d{p}_j=-d{p}_j\wedge d{q}_j$,

$i_{X_H}\omega$ $=$ ${\displaystyle \sum _{i=1}^n}d{q}_i\wedge d{p}_i{\displaystyle \sum _{j=1}^n}\left({\displaystyle \frac{\partial H}{\partial p_j}}{\displaystyle \frac{\partial }{\partial q_j}}-{\displaystyle \frac{\partial H}{\partial q_j}}{\displaystyle \frac{\partial }{\partial p_j}}\right)$

$=$ ${\displaystyle \sum _{i=1}^n}d{q}_i\wedge d{p}_i\left({\displaystyle \frac{\partial H}{\partial p_i}}{\displaystyle \frac{\partial }{\partial q_i}}-{\displaystyle \frac{\partial H}{\partial q_i}}{\displaystyle \frac{\partial }{\partial p_i}}\right)$

$=$ ${\displaystyle \sum _{i=1}^n}{\displaystyle \frac{\partial H}{\partial p_i}}d{p}_i+{\displaystyle \frac{\partial H}{\partial q_i}}d{q}_i$

$=$ $dH.$

3 Flows

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

$$D^x=\{t\in ℝ:(x,t)\in D\}$$

is an open interval including $0$. A flow on $M$ is a smooth map $\varphi :D\to M$ such that if $x\in M$ then ${\varphi }_0(x)=x$ and such that if $x\in M$, $s\in D^x,t\in D^{{\varphi }_s(x)}$ and $s+t\in D^x$, then

$${\varphi }_t({\varphi }_s(x))={\varphi }_{s+t}(x).$$

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

$$V_x={\frac{d}{dt}|}_{t=0}{\varphi }^x(t).$$

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 $ℝ\times M$, 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

$${ℒ}_X\omega =0,$$

where ${ℒ}_X\omega$ is the Lie derivative of $\omega$ along the flow of $X$. A Hamiltonian flow is the flow of a Hamiltonian vector field. If $X$ is a complete symplectic vector field and $\varphi :M\times ℝ$ is the flow of $X$, then for all $t\in ℝ$, the map ${\varphi }_t:M\to M$ is a symplectomorphism.

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

${{\displaystyle \frac{d}{dt}}|}_{t=s}H({\varphi }^x(t))$ $=$ $(d_{{\varphi }^x(s)}H)({({\varphi }^x)}^{\prime }(s))$

$=$ $(d_{{\varphi }^x(s)}H)(X_H({\varphi }^x(s)))$

$=$ ${\omega }_{{\varphi }^x(s)}(X_H({\varphi }^x(s)),X_H({\varphi }^x(s)))$

$=$ $0.$

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

${\displaystyle \frac{d}{dt}}(H\circ {\varphi }_t)$ $=$ ${\displaystyle \frac{d}{dt}}({\varphi }_t^{*}H)$

$=$ ${\varphi }_t^{*}({ℒ}_{X_H}H)$

$=$ ${\varphi }_t^{*}((i_{X_H}\omega )(X_H))$

$=$ ${\varphi }_t^{*}(\omega (X_H,X_H))$

$=$ ${\varphi }_t^{*}(0)$

$=$ $0.$

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

5 Poisson bracket

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

$\mathrm{\{}f,g\}$ $=$ $\omega (X_f,X_g)$

$=$ ${\displaystyle \sum _{i=1}^n}(d{q}_i\wedge d{p}_i){\displaystyle \sum _{j=1}^n}\left({\displaystyle \frac{\partial f}{\partial p_j}}{\displaystyle \frac{\partial }{\partial q_j}}-{\displaystyle \frac{\partial f}{\partial q_j}}{\displaystyle \frac{\partial }{\partial p_j}}\right){\displaystyle \sum _{k=1}^n}\left({\displaystyle \frac{\partial g}{\partial p_k}}{\displaystyle \frac{\partial }{\partial q_k}}-{\displaystyle \frac{\partial g}{\partial q_k}}{\displaystyle \frac{\partial }{\partial p_k}}\right)$

$=$ ${\displaystyle \sum _{i=1}^n}\left({\displaystyle \frac{\partial f}{\partial p_i}}d{p}_i+{\displaystyle \frac{\partial f}{\partial q_i}}d{q}_i\right){\displaystyle \sum _{k=1}^n}\left({\displaystyle \frac{\partial g}{\partial p_k}}{\displaystyle \frac{\partial }{\partial q_k}}-{\displaystyle \frac{\partial g}{\partial q_k}}{\displaystyle \frac{\partial }{\partial p_k}}\right)$

$=$ ${\displaystyle \sum _{i=1}^n}-{\displaystyle \frac{\partial f}{\partial p_i}}{\displaystyle \frac{\partial g}{\partial q_i}}+{\displaystyle \frac{\partial f}{\partial q_i}}{\displaystyle \frac{\partial g}{\partial p_i}}.$

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 $f\in C^{\mathrm{\infty }}(M)$ 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 $Xf\in C^{\mathrm{\infty }}(M)$, 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 $\varphi$ is the flow of $X$, then

$$({ℒ}_X\tau )(x)={\frac{d}{dt}|}_{t=0}({\varphi }_t^{*}\tau )(x),$$

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

$$({ℒ}_{X}f)(x)={\frac{d}{dt}|}_{t=0}({\varphi }_t^{*}f)(x)={\frac{d}{dt}|}_{t=0}f({\varphi }_t(x))=X_{x}f=(Xf)(x).$$

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

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

$X_{\{f,g\}}\mathrm{⌟}\omega$ $=$ $d\{f,g\}$

$=$ $d(X_{g}f)$

$=$ $d({ℒ}_{X_g}f)$

$=$ ${ℒ}_{X_g}(df)$

$=$ ${ℒ}_{X_g}(X_f\mathrm{⌟}\omega )$

$=$ $({ℒ}_{X_g}{X}_f)\mathrm{⌟}\omega +X_f\mathrm{⌟}{ℒ}_{X_g}\omega$

$=$ $[X_g,X_f]\mathrm{⌟}\omega +X_f\mathrm{⌟}0$

$=$ $[X_g,X_f]\mathrm{⌟}\omega$

$=$ $-[X_f,X_g]\mathrm{⌟}\omega .$

Since the symplectic form $\omega$ is nondegenerate, if $X\mathrm{⌟}\omega =Y\mathrm{⌟}\omega$ then $X=Y$, so

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

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

The set ${\mathrm{\Gamma }}^{\mathrm{\infty }}(M)$ 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 $H_{\text{dR}}^1(M)$; this is why if $H_{\text{dR}}^1(M)=\{0\}$ (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

$${\theta }_x={(d_x\pi )}^{*}(p)=p\circ d_x\pi :T_x{T}^{*}Q\to ℝ.$$

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

If $(Q_1,\mathrm{\dots },Q_n)$ are coordinates on an open subset $U$ of $Q$, $Q_i:U\to ℝ$, then for each $q\in U$ we have that $d_q{Q}_i\in T_q^{*}U=T_q^{*}Q$, $1\le i\le n$, are a basis for $T_q^{*}Q$ and ${\frac{\partial }{\partial Q_i}|}_q$, $1\le i\le n$, are a basis for $T_{q}Q$. For each $p\in T_q^{*}Q$,

$$p=\sum _{i=1}^{n}p\left({\frac{\partial }{\partial Q_i}|}_q\right)d_q{Q}_i.$$

On $T^{*}U$, define coordinates $(q_1,\mathrm{\dots },q_n,p_1,\mathrm{\dots },p_n)$ by

$$q_i(q,p)=Q_i(q),$$

and

$$p_i(q,p)=p\left({\frac{\partial }{\partial Q_i}|}_q\right).$$

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 $T^{*}U$,

$$\theta =\sum _{i=1}^n{p}_{i}d{q}_i.$$

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

$\omega$ $=$ $-d{\displaystyle \sum _{i=1}^n}{p}_{i}d{q}_i$

$=$ $-{\displaystyle \sum _{i=1}^n}\left(d{p}_i\wedge d{q}_i+p_{i}d(d{q}_i)\right)$

$=$ $-{\displaystyle \sum _{i=1}^n}d{p}_i\wedge d{q}_i$

$=$ ${\displaystyle \sum _{i=1}^n}d{q}_i\wedge d{p}_i.$

$T^{*}Q$ 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

$$F^{\mathrm{♯}}:T^{*}Q\to T^{*}Q$$

for $x=(q,p)$ by

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

We call $F^{\mathrm{♯}}:T^{*}Q\to T^{*}Q$ 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 $F^{\mathrm{♯}}$ satisfies, for $x=(q,p)\in T^{*}Q$ and $(\zeta ,\eta )=F^{\mathrm{♯}}(q,p)\in T^{*}Q$,

${({(F^{\mathrm{♯}})}^{*}\theta )}_x$ $=$ ${(d_x{F}^{\mathrm{♯}})}^{*}({\theta }_{F^{\mathrm{♯}}(x)})$

$=$ ${(d_x{F}^{\mathrm{♯}})}^{*}({(d_{F^{\mathrm{♯}}(x)}\pi )}^{*}(\eta ))$

$=$ ${(d_x(\pi \circ F^{\mathrm{♯}}))}^{*}(\eta )$

$=$ ${(d_x(F\circ \pi ))}^{*}(\eta )$

$=$ ${(d_x\pi )}^{*}({(d_{\pi (x)}F)}^{*}(\eta ))$

$=$ ${(d_x\pi )}^{*}({(d_{q}F)}^{*}(\eta ))$

$=$ ${(d_x\pi )}^{*}(p)$

$=$ ${\theta }_x.$

Thus ${(F\mathrm{♯})}^{*}\theta =\theta$, i.e. $F^{\mathrm{♯}}$ pulls back $\theta$ to $\theta$. The “naturality of the exterior derivative” 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 $\omega =d\theta$,

$${(F^{\mathrm{♯}})}^{*}\omega ={(F^{\mathrm{♯}})}^{*}(d\theta )=d({(F^{\mathrm{♯}})}^{*}\theta )=d(\theta )=\omega ,$$

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

Let $\text{Diff}(Q)$ be the set of diffeomorphisms $Q\to Q$. $\text{Diff}(Q)$ is a group. Let $G$ be a group and let $\tau :G\to \text{Diff}(Q)$ be a homomorphism. Define ${\tau }^{\mathrm{♯}}:G\to \text{Diff}(T^{*}Q)$ by ${({\tau }^{\mathrm{♯}})}_g={({\tau }_g)}^{\mathrm{♯}}:T^{*}Q\to T^{*}Q$. ${\tau }^{\mathrm{♯}}:G\to \text{Diff}(T^{*}Q)$ is a homomorphism, and for each $g\in G$, ${({\tau }^{\mathrm{♯}})}_g:T^{*}Q\to T^{*}Q$ 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 $X\in {\mathrm{\Gamma }}^{\mathrm{\infty }}(M)$ and $f\in C^{\mathrm{\infty }}(N)$,

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

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 $f\circ F\in C^{\mathrm{\infty }}(M)$ 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

$$(T{L}_g)(X)=X$$

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 $ϵ:\text{Lie}(G)\to T_{e}G$ by $ϵ(X)=X_e$, 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 $G\times ℝ$. For $X\in \text{Lie}(G)$, 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 $\gamma :ℝ\to G$ such that

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

We define $\exp :\text{Lie}(G)\to G$ 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 $\varphi$ and $Y$ has flow $\eta$, then

$${\eta }_t\circ F=F\circ {\varphi }_t$$

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

$$L_g\circ {\varphi }_t={\varphi }_t\circ L_g.$$

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

$g\exp (tX)$ $=$ $L_g\exp (tX)$

$=$ $L_g({\varphi }_{t}e)$

$=$ ${\varphi }_t(L_{g}e)$

$=$ ${\varphi }_t(g).$

9 Coadjoint action

First we’ll define the adjoint action of $G$ on $𝔤=T_{{\text{id}}_G}G$. For $g\in G$, define ${\mathrm{\Psi }}_g:G\to G$ by ${\mathrm{\Psi }}_g(h)=gh{g}^{-1}$; ${\mathrm{\Psi }}_g$ is an automorphism of Lie groups. Define

$${\text{Ad}}_g:𝔤\to 𝔤$$

by

$${\text{Ad}}_g=T_{{\text{id}}_G}{\mathrm{\Psi }}_g;$$

since ${\mathrm{\Psi }}_g$ is an automorphism of Lie groups, it follows that ${\text{Ad}}_g$ is an automorphism of Lie algebras. We can also write ${\text{Ad}}_g$ as

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

The adjoint action of $G$ on $𝔤$ is

$$g\cdot \xi ={\text{Ad}}_g(\xi ).$$

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

$$({\text{Ad}}_g^{*}l)(\xi )=l({\text{Ad}}_g(\xi )).$$

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

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

10 Momentum map

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

Let $𝔤=T_{{\text{id}}_G}G$, and define $\rho :𝔤\to {\mathrm{\Gamma }}^{\mathrm{\infty }}(M)$ by

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

$t\mapsto {\sigma }_{\exp (t\xi )}(x)$ is $ℝ\to M$ 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 $𝔤$ on $M$. Each element of $G$ acts on $M$ as a symplectomorphism, each element of $𝔤$ acts on $M$ as a vector field.

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

$$((T_x\mu )v)\xi ={\omega }_x(\rho (\xi )(x),v),$$ (1)

where

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

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

11 Angular momentum

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

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

Let $Q={ℝ}^3$, and define $\tau :G\to \text{Diff}(Q)$ 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 ${\tau }^{\mathrm{♯}}:G\to \text{Diff}(T^{*}Q)$ is a homomorphism such that for each $g\in G$, ${({\tau }^{\mathrm{♯}})}_g$ is a symplectomorphism. For $g\in G$, $(q,p)\in T^{*}Q$,

${({\tau }^{\mathrm{♯}})}_g(q,p)$ $=$ ${({\tau }_g)}^{\mathrm{♯}}(q,p)$

$=$ $({\tau }_{g}q,{(d_{{\tau }_{g}q}({\tau }_g^{-1}))}^{*}p)$

$=$ $({\tau }_{g}q,{(d_{{\tau }_{g}q}({\tau }_{g^{-1}}))}^{*}p)$

$=$ $({\tau }_{g}q,p\circ (d_{{\tau }_{g}q}{\tau }_{g^{-1}}))$

$=$ $({\tau }_{g}q,p\circ {\tau }_{g^{-1}})$

$=$ $(gq,p{g}^{-1})$

$=$ $(gq,p{g}^T).$

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

$\rho (\xi )(q,p)$ $=$ ${{\displaystyle \frac{d}{dt}}|}_{t=0}{({\tau }^{\mathrm{♯}})}_{\exp (t\xi )}(q,p)$

$=$ ${{\displaystyle \frac{d}{dt}}|}_{t=0}(\exp (t\xi )q,p\exp (t{\xi }^T))$

$=$ $(\xi q,p{\xi }^T)$

$=$ $(\xi q,-p\xi ).$

Define $V:𝔤\to {ℝ}^3$ by

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 $\xi \in 𝔤$, we have

${\omega }_{(q,p)}(\rho (\xi )(q,p),(v,w))$ $=$ ${\omega }_{(q,p)}((\xi q,-p\xi ),(v,w))$

$=$ ${\displaystyle \sum _{j=1}^3}d{q}_j\wedge d{p}_j((\xi q,-p\xi ),(v,w))$

$=$ ${\displaystyle \sum _{j=1}^3}\left({(\xi q)}_{j}d{p}_j+{(p\xi )}_{j}d{q}_j\right)(v,w)$

$=$ ${\displaystyle \sum _{j=1}^3}{w}_j{(\xi q)}_j+v_j{(p\xi )}_j$

$=$ $w\cdot (V(\xi )\times q)+v\cdot (p^T\times V(\xi )).$

Define $\mu :T^{*}Q\to {𝔤}^{*}$ 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 $\xi \in 𝔤$, ${\text{Ad}}_g\xi =g\xi g^{-1}$, and hence, for $(q,p)\in T^{*}Q$,

$(g\cdot \mu (q,p))\xi$ $=$ $\left({\text{Ad}}_{g^{-1}}^{*}\mu (q,p)\right)\xi$

$=$ $\mu (q,p)({\text{Ad}}_{g^{-1}}\xi )$

$=$ $\mu (q,p)(g^{-1}\xi g)$

$=$ $(q\times p^T)\cdot V(g^{-1}\xi g).$

On the other hand,

$\mu ({({\tau }^{\mathrm{♯}})}_g(q,p))\xi$ $=$ $\mu (gq,p{g}^T)\xi$

$=$ $((gq)\times {(p{g}^T)}^T)\cdot V(\xi )$

$=$ $((gq)\times (g{p}^T))\cdot V(\xi )$

$=$ $(g(q\times p^T))\cdot V(\xi )$

$=$ $(q\times p^T)\cdot (g^{T}V(\xi ))$

$=$ $(q\times p^T)\cdot (g^{-1}V(\xi )).$