Hamiltonian flows, cotangent lifts, and momentum maps

Jordan Bell
April 3, 2014

1 Symplectic manifolds

Let (M,ω) and (N,η) be symplectic manifolds. A symplectomorphism F:MN is a diffeomorphism such that ω=F*η. Recall that for xM and v1,v2TxM,

(F*η)x(v1,v2)=ηF(x)((TxF)v1,(TxF)v2);

TxF:TxMTF(x)N. (A tangent vector at xM is pushed forward to a tangent vector at F(x)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

2 Symplectic gradient

If (M,ω) is a symplectic manifold and HC(M), using the nondegeneracy of the symplectic form ω one can prove that there is a unique vector field XHΓ(M) such that, for all xM,vTxM,

ωx(XH(x),v)=(dH)x(v).

This can also be written as

iXHω=dH,

where

(iXω)(Y)=(Xω)(Y)=ω(X,Y).

We call XH the symplectic gradient of H. If XΓ(M) and X=XH for some HC(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

XH=i=1nHpiqi-Hqipi.

We have, because dqiqj=δij, dpipj=δij, dqipj=0 and dpiqj=0, and because dqjdpj=-dpjdqj,

iXHω = i=1ndqidpij=1n(Hpjqj-Hqjpj)
= i=1ndqidpi(Hpiqi-Hqipi)
= i=1nHpidpi+Hqidqi
= dH.

3 Flows

Let M be a smooth manifold. Let D be an open subset of M×, and for each xM suppose that

Dx={t:(x,t)D}

is an open interval including 0. A flow on M is a smooth map ϕ:DM such that if xM then ϕ0(x)=x and such that if xM, sDx,tDϕs(x) and s+tDx, then

ϕt(ϕs(x))=ϕs+t(x).

For xM, define ϕx:DxM by ϕx(t)=ϕt(x). The infinitesimal generator of a flow ϕ is the vector field V on M defined for xM by

Vx=ddt|t=0ϕx(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 ×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,ω) be a symplectic manifold. We say that a vector field X on M is symplectic if

Xω=0,

where Xω is the Lie derivative of ω 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 ϕ:M× is the flow of X, then for all t, the map ϕt:MM is a symplectomorphism.

Let HC(M), and let ϕ be the flow of the vector field XH. If (x,s) is in the domain of the flow ϕ, we have

ddt|t=sH(ϕx(t)) = (dϕx(s)H)((ϕx)(s))
= (dϕx(s)H)(XH(ϕx(s)))
= ωϕx(s)(XH(ϕx(s)),XH(ϕx(s)))
= 0.

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

ddt(Hϕt) = ddt(ϕt*H)
= ϕt*(XHH)
= ϕt*((iXHω)(XH))
= ϕt*(ω(XH,XH))
= ϕt*(0)
= 0.

It is a fact that if HdR1(M)={0} (i.e. if α is a 1-form on M and dα=0 then there is some fC(M) such that α=df) then every symplectic vector field on M is Hamiltonian. In particular, if M is simply connected then HdR1(M)={0}, and hence if M is simply connected then every symplectic vector field on M is Hamiltonian.

5 Poisson bracket

For f,gC(M), we define {f,g}C(M) for xM by

{f,g}(x)=ωx(Xf(x),Xg(x)).

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

{f,g}=ω(Xf,Xg).

We have

{f,g}=Xfg=(df)Xg.

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 Xg.

We have

{f,g} = ω(Xf,Xg)
= i=1n(dqidpi)j=1n(fpjqj-fqjpj)k=1n(gpkqk-gqkpk)
= i=1n(fpidpi+fqidqi)k=1n(gpkqk-gqkpk)
= i=1n-fpigqi+fqigpi.

If xM and vTxM, then vf is the directional derivative in the direction v. If v=i=1naiqi+bipi and fC(M) then

vf=i=1naifqi+bifpi.

If X is a vector field on M then XfC(M), defined for xM by

(Xf)(x)=Xxf.

If τ is a covariant tensor field and X is a vector field, the Lie derivative of τ along the flow of X is defined as follows: if ϕ is the flow of X, then

(Xτ)(x)=ddt|t=0(ϕt*τ)(x),

and so if τ is a function fC(M), then

(Xf)(x)=ddt|t=0(ϕt*f)(x)=ddt|t=0f(ϕt(x))=Xxf=(Xf)(x).

Thus if X is a vector field and fC(M), then Xf=Xf.

For f,gC(M),

X{f,g}ω = d{f,g}
= d(Xgf)
= d(Xgf)
= Xg(df)
= Xg(Xfω)
= (XgXf)ω+XfXgω
= [Xg,Xf]ω+Xf0
= [Xg,Xf]ω
= -[Xf,Xg]ω.

Since the symplectic form ω is nondegenerate, if Xω=Yω then X=Y, so

X{f,g}=-[Xf,Xg].

It follows that C(M) is a Lie algebra using the Poisson bracket as the Lie bracket.

The set Γ(M) of vector fields on M are a Lie algebra using the vector field commutator [,]. 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 HdR1(M); this is why if HdR1(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 π:T*QQ, π(q,p)=q. For x=(q,p)T*Q, we have

dxπ:TxT*QTqQ.

Let

θx=(dxπ)*(p)=pdxπ:TxT*Q.

Thus θ:T*QT*T*Q. θ is called the tautological 1-form on T*Q.

If (Q1,,Qn) are coordinates on an open subset U of Q, Qi:U, then for each qU we have that dqQiTq*U=Tq*Q, 1in, are a basis for Tq*Q and Qi|q, 1in, are a basis for TqQ. For each pTq*Q,

p=i=1np(Qi|q)dqQi.

On T*U, define coordinates (q1,,qn,p1,,pn) by

qi(q,p)=Qi(q),

and

pi(q,p)=p(Qi|q).

On T*U we can write θ using these coordinates: for x=(q,p)T*Q,

θx=pdxπ=i=1npi(x)dxqi.

Thus, on T*U,

θ=i=1npidqi.

Let ω=-dθ. We have, on T*U,

ω = -di=1npidqi
= -i=1n(dpidqi+pid(dqi))
= -i=1ndpidqi
= i=1ndqidpi.

T*Q is a symplectic manifold with the symplectic form ω.

7 Cotangent lifts

Let Q be a smooth manifold and let F:QQ be a diffeomorphism. Define

F:T*QT*Q

for x=(q,p) by

F(q,p)=(F(q),(dF(q)(F-1))*(p)).

We call F:T*QT*Q the cotangent lift of F:QQ. It is a fact that it is a diffeomorphism. It is apparent that the following diagram commutes:

The pull-back of θ by F satisfies, for x=(q,p)T*Q and (ζ,η)=F(q,p)T*Q,

((F)*θ)x = (dxF)*(θF(x))
= (dxF)*((dF(x)π)*(η))
= (dx(πF))*(η)
= (dx(Fπ))*(η)
= (dxπ)*((dπ(x)F)*(η))
= (dxπ)*((dqF)*(η))
= (dxπ)*(p)
= θx.

Thus (F)*θ=θ, i.e. F pulls back θ to θ. The “naturality of the exterior derivative”44 4 For each k, Ωk is a contravariant functor, and if f:MN, then the functor Ωk sends f to f*:Ωk(N)Ωk(M). d is a natural transformation from the contravariant functor Ωk to the contravariant functor Ωk+1. is the statement that if G is a smooth map and η is a differential form then G*(dη)=d(G*η). Hence, with ω=dθ,

(F)*ω=(F)*(dθ)=d((F)*θ)=d(θ)=ω,

so F pulls back the symplectic form ω to itself. Thus F:T*MT*M is a symplectomorphism.

Let Diff(Q) be the set of diffeomorphisms QQ. Diff(Q) is a group. Let G be a group and let τ:GDiff(Q) be a homomorphism. Define τ:GDiff(T*Q) by (τ)g=(τg):T*QT*Q. τ:GDiff(T*Q) is a homomorphism, and for each gG, (τ)g:T*QT*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:MN then TF:TMTN satisfies, for XΓ(M) and fC(N),55 5 In words: TF pushes forward a vector field on M to a vector field on N.

((TF)X)(f)=X(fF),

i.e. for xM and vTxM,

((TxF)v)(f)=v(fF),

the directional derivative of fFC(M) in the direction of the tangent vector v.

Let G be a Lie group and for gG define Lg:GG by Lgh=gh. If X is a vector field on G, we say that X is left-invariant if

(ThLg)(Xh)=Xgh

for all g,hG. That is, X is left-invariant if

(TLg)(X)=X

for all gG.

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

(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:Lie(G)TeG by ϵ(X)=Xe, where eG is the identity element. It can be shown that this is a linear isomorphism. Hence, if vTeG then there is a unique left-invariant vector field X on G such that, for all gG,

Vg=(TeLg)(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×. For XLie(G), we call the unique integral curve of X that passes through e the one-parameter subgroup generated by X. Thus, for any vTeG there is a unique one-parameter subgroup γ:G such that

γ(0)=e,γ(0)=v.

We define exp:Lie(G)G by exp(X)=γ(1), where γ is the one-parameter subgroup generated by X. This is called the exponential map. Thus texp(tX) is the one-parameter subgroup generated by X.

Fact: If (TF)X=Y and X has flow ϕ and Y has flow η, then

ηtF=Fϕt

for all t in the domain of ϕ. Hence

Lgϕt=ϕtLg.

Hence the flow ϕ of a left-invariant vector field X satisfies

gexp(tX) = Lgexp(tX)
= Lg(ϕte)
= ϕt(Lge)
= ϕt(g).

9 Coadjoint action

First we’ll define the adjoint action of G on 𝔤=TidGG. For gG, define Ψg:GG by Ψg(h)=ghg-1; Ψg is an automorphism of Lie groups. Define

Adg:𝔤𝔤

by

Adg=TidGΨg;

since Ψg is an automorphism of Lie groups, it follows that Adg is an automorphism of Lie algebras. We can also write Adg as

Adg(ξ)=ddt|t=0(gexp(tξ)g-1).

The adjoint action of G on 𝔤 is

gξ=Adg(ξ).

For each gG, one proves that there is a unique map Adg*:𝔤*𝔤* such that for all l𝔤*,ξ𝔤,

(Adg*l)(ξ)=l(Adg(ξ)).

The coadjoint action of G on 𝔤* is

gl=Adg-1*(l).

10 Momentum map

Let (M,ω) be a symplectic manifold, let G be a Lie group, and let σ:GDiff(M) be a homomorphism such that for each g in G, σg is a symplectomorphism.

Let 𝔤=TidGG, and define ρ:𝔤Γ(M) by

ρ(ξ)(x)=ddt|t=0σexp(tξ)(x)TxM,ξ𝔤,xM;

tσexp(tξ)(x) is M and at t=0 the curve passes through x, so indeed ρ(ξ)(x)Tx(M). ρ 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,ω) is a map μ:M𝔤* such that, for xM, vTxM and ξ𝔤,

((Txμ)v)ξ=ωx(ρ(ξ)(x),v), (1)

where

Txμ:TxMTμ(x)𝔤*=𝔤*,

and such that if gG and xM then

μ(σg(x))=gμ(x), (2)

where gμ(x) is the coadjoint action of G on 𝔤*, defined in section §9; we say that μ is equivariant with respect to the coadjoint action of G on 𝔤*.

11 Angular momentum

Let G=SO(3)={A3×3:ATA=I,det(A)=1}. The Lie algebra of SO(3) is

𝔤=𝔰𝔬(3)={a3×3:a+aT=0}.

Let Q=3, and define τ:GDiff(Q) by τg(q)=gq.

Let θ be the tautological 1-form on T*Q and let ω=-dθ. (T*Q,ω) is a symplectic manifold and τ:GDiff(T*Q) is a homomorphism such that for each gG, (τ)g is a symplectomorphism. For gG, (q,p)T*Q,

(τ)g(q,p) = (τg)(q,p)
= (τgq,(dτgq(τg-1))*p)
= (τgq,(dτgq(τg-1))*p)
= (τgq,p(dτgqτg-1))
= (τgq,pτg-1)
= (gq,pg-1)
= (gq,pgT).

Hence for ξ𝔤 and (q,p)T*Q,

ρ(ξ)(q,p) = ddt|t=0(τ)exp(tξ)(q,p)
= ddt|t=0(exp(tξ)q,pexp(tξT))
= (ξq,pξT)
= (ξq,-pξ).

Define V:𝔤3 by

V(0-ξ3ξ2ξ30-ξ1-ξ2ξ10)=(ξ1ξ2ξ3).

One checks that ξq=V(ξ)×q and pξ=pT×V(ξ).

For (q,p)T*Q, (v,w)T(p,q)T*Q, and ξ𝔤, we have

ω(q,p)(ρ(ξ)(q,p),(v,w)) = ω(q,p)((ξq,-pξ),(v,w))
= j=13dqjdpj((ξq,-pξ),(v,w))
= j=13((ξq)jdpj+(pξ)jdqj)(v,w)
= j=13wj(ξq)j+vj(pξ)j
= w(V(ξ)×q)+v(pT×V(ξ)).

Define μ:T*Q𝔤* by μ(q,p)(ξ)=(q×pT)V(ξ). I claim that μ 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 gG and ξ𝔤, Adgξ=gξg-1, and hence, for (q,p)T*Q,

(gμ(q,p))ξ = (Adg-1*μ(q,p))ξ
= μ(q,p)(Adg-1ξ)
= μ(q,p)(g-1ξg)
= (q×pT)V(g-1ξg).

On the other hand,

μ((τ)g(q,p))ξ = μ(gq,pgT)ξ
= ((gq)×(pgT)T)V(ξ)
= ((gq)×(gpT))V(ξ)
= (g(q×pT))V(ξ)
= (q×pT)(gTV(ξ))
= (q×pT)(g-1V(ξ)).