LECTURE8:THEMOMENTMAPCONTENTS11.Propertiesofthemomentmap42.Existence and Uniqueness of the moment map73.Examples/Exercisesofmomentmaps94.Momentmapingaugetheory1.PROPERTIES OF THE MOMENT MAP Hamiltonian actions.Suppose a Lie group G acts smoothly on M.For simplicity we shall alwaysassume G is connected. Associated to each vector X e g = Lie(G) one has asmooth vector field on M defined bydXm(m) =exp(tX) -m.dt|t=0Now suppose M is symplectic with symplectic form w. The G-action is called.symplectic if each element g eG acts on M by symplectomorphismsEquivalently, for each X E g, Xm is symplectic, i.e. Lxmw = 0..weakly Hamiltonian if each Xm is Hamiltonian, with Hamiltonian functionsdepending nicely on X. More precisely, there exists functions μX on M,depending linearly on X e g, so thatdμX = txmw.There are two equivalent ways to group these functions μX's together:-Thecomomentmapisthelinearmapμ":g→C~(M),μ(X)=μX- The moment map is the smooth mapμ: M →g*,μ()= (μ(),X)Remark.One can regard the moment map μ as the dual map of the linearmap μ* restricted to M, where M is viewed as a subset of Co(M)* via(m,f) := f(m).1
LECTURE 8: THE MOMENT MAP Contents 1. Properties of the moment map 1 2. Existence and Uniqueness of the moment map 4 3. Examples/Exercises of moment maps 7 4. Moment map in gauge theory 9 1. Properties of the moment map ¶ Hamiltonian actions. Suppose a Lie group G acts smoothly on M. For simplicity we shall always assume G is connected. Associated to each vector X ∈ g = Lie(G) one has a smooth vector field on M defined by XM(m) = d dt t=0 exp(tX) · m. Now suppose M is symplectic with symplectic form ω. The G-action is called • symplectic if each element g ∈ G acts on M by symplectomorphisms. – Equivalently, for each X ∈ g, XM is symplectic, i.e. LXM ω = 0. • weakly Hamiltonian if each XM is Hamiltonian, with Hamiltonian functions depending nicely on X. More precisely, there exists functions µ X on M, depending linearly on X ∈ g, so that dµX = ιXM ω. There are two equivalent ways to group these functions µ X’s together: – The comoment map is the linear map µ ∗ : g → C ∞(M), µ∗ (X) = µ X. – The moment map is the smooth map µ : M → g ∗ , µX(·) = hµ(·), Xi. Remark. One can regard the moment map µ as the dual map of the linear map µ ∗ restricted to M, where M is viewed as a subset of C ∞(M) ∗ via hm, fi := f(m). 1
2LECTURE8:THEMOMENTMAP. Hamiltonian if it is weakly Hamiltonian, and the moment map is equivariantwith respect to the G-action on M and the coadjoint G-action on g*. Wecall (M,w,G,μ) a Hamiltonian G-space.-If G is abelian, then g*Rn,and the condition"μ is G-equivariant" isreduced to a simpler statement “μ is G-invariant" Noether principle.Recall that an integral of motion for a Hamiltonian system (M,w,f) is a s-mooth function that Poisson commutes with f.Now let (M,w,G,μ) be a weakly-Hamiltonian G-space and f E Co(M)G a G-invariant smooth function. The Noetherprinciple asserts that symmetries give rise to integral of motions.Theorem 1.1 (Noether). Suppose (M,w,G,μ) is a weakly-Hamiltonian G-space,and f ECoo(M)G a smooth G-invariant function.Then for any X Eg,the functionμ is an integral of motion of the Hamilton system (M,w, f).Proof.Let E, be theHamiltonian vector field associated to f,thend(f,μI(m) =w(三f,Xm)(m) =lxudf(m) = Cxuf(m) =f(exp(tX)·m)=0dt口because f is G-invariant.Moment map v.s. comoment map.A natural question is: How to describe Hamiltonian action via the comomentmap?Proposition 1.2. The moment map μ is G-equivariant if and only if the comomentmap μ*: (g,[,l)-→ (C(M), (,) is a Lie algebra homomorphism.Proof. First assume the G-action is Hamiltonian. Then for any X,Y E g,d(μX,μYI(m) = YM(μ)(m) =(μ(exp(tY) ·m),X)dtd(Adexp(ty)(m), X)dtd《μ(m), Adexp(-ty)X)dtd《μu(m),exp(-tadY)X)dt= 《μ(m),[X, Y]) = μ[x,Y(m),where for the second equality we used the fact exp(tY):m = exp(tYm)(m).Conversely, suppose μ* is a Lie algebra homomorphism. Since G is connectedand the exponential map exp is a local diffeomorphism, any element g of G can
2 LECTURE 8: THE MOMENT MAP • Hamiltonian if it is weakly Hamiltonian, and the moment map is equivariant with respect to the G-action on M and the coadjoint G-action on g ∗ . We call (M, ω, G, µ) a Hamiltonian G-space. – If G is abelian, then g ∗ ' R n , and the condition “µ is G-equivariant” is reduced to a simpler statement “µ is G-invariant”. ¶ Noether principle. Recall that an integral of motion for a Hamiltonian system (M, ω, f) is a smooth function that Poisson commutes with f. Now let (M, ω, G, µ) be a weaklyHamiltonian G-space and f ∈ C ∞(M) G a G-invariant smooth function. The Noether principle asserts that symmetries give rise to integral of motions. Theorem 1.1 (Noether). Suppose (M, ω, G, µ) is a weakly-Hamiltonian G-space, and f ∈ C ∞(M) G a smooth G-invariant function. Then for any X ∈ g, the function µ X is an integral of motion of the Hamilton system (M, ω, f). Proof. Let Ξf be the Hamiltonian vector field associated to f, then {f, µX}(m) = ω(Ξf , XM)(m) = ιXM df(m) = LXM f(m) = d dt t=0 f(exp(tX)·m) = 0 because f is G-invariant. ¶ Moment map v.s. comoment map. A natural question is: How to describe Hamiltonian action via the comoment map? Proposition 1.2. The moment map µ is G-equivariant if and only if the comoment map µ ∗ : (g, [·, ·]) → (C ∞(M), {·, ·}) is a Lie algebra homomorphism. Proof. First assume the G-action is Hamiltonian. Then for any X, Y ∈ g, {µ X, µY }(m) = YM(µ X)(m) = d dt t=0 hµ(exp(tY ) · m), Xi = d dt t=0 hAd∗ exp(tY )µ(m), Xi = d dt t=0 hµ(m), Adexp(−tY )Xi = d dt t=0 hµ(m), exp(−tadY )Xi = hµ(m), [X, Y ]i = µ [X,Y ] (m), where for the second equality we used the fact exp(tY ) · m = exp(tYM)(m). Conversely, suppose µ ∗ is a Lie algebra homomorphism. Since G is connected and the exponential map exp is a local diffeomorphism, any element g of G can
3LECTURE8:THEMOMENTMAPbewrittenas aproduct of elements of theform exp(X).As a result, to proveG-equivariance it is enough to proveμ(exp(tX) - m) = Adexp(tx)μ(m).We shall use the following two results from manifold theory:. Let f : Mi → M2 be a smooth map. For i = 1,2 let Y, be a smooth vectorfield on M, and let p be the flow of Y.Lemma 1.3. If df(Yi) =Y,o f, then f o pt = po fSketch of proof:Both sides define integral curves of the vectorfield Y2passing口the point f(m) at t = 0. Let Xg be the vector field on g* generating the flow Adexp(tx):Lemma 1.4.For anyEg*and anyY Eg,(Xg-(s), Y) = -(s,[X,Y]),Proof. Differentiate both sides of the following formula at t = O:(Adexp(tx)E,Y) = (E, Adexp(-tx)Y)口As a consequence, the theorem is proved if we can showdμ(XM) =Xg+oμ.To prove this we calculate for any Y E g = (g*)*,(dμ(XM(m)), Y) = Yodμ(XM(m)) = d(Yoμ)(XM(m)) = XM(Yoμ)(m) = XM((μ(m),Y)),where the second equality follows from the fact that Y is linear as a function on g*Now use the assumption that μ* is a Lie algebra homomorphism. SoXm(μ) = (μ,μ*) = μY,X) = -(μ,[X, Y)) = (Xg-(μ), Y),It follows(dμ(Xm(m)), Y) = (Xg*(μ(m), Y)口This is exactly what we wanted. Change of Lie groups.Proposition 1.5. Suppose (M,w,G,μ)is a Hamiltonian G-space. Let :K→Gbe a Lie group homomorphism. Then the induced K-action on M defined bykm :=p(k)·mis a Hamiltonian action with moment map v=(dp)Toμ
LECTURE 8: THE MOMENT MAP 3 be written as a product of elements of the form exp(X). As a result, to prove G-equivariance it is enough to prove µ(exp(tX) · m) = Ad∗ exp(tX)µ(m). We shall use the following two results from manifold theory: • Let f : M1 → M2 be a smooth map. For i = 1, 2 let Yi be a smooth vector field on Mi , and let ρ i t be the flow of Yi . Lemma 1.3. If df(Y1) = Y2 ◦ f, then f ◦ ρ 1 t = ρ 2 t ◦ f. Sketch of proof: Both sides define integral curves of the vector field Y2 passing the point f(m) at t = 0. • Let Xg∗ be the vector field on g ∗ generating the flow Ad∗ exp(tX) . Lemma 1.4. For any ξ ∈ g ∗ and any Y ∈ g, hXg ∗ (ξ), Y i = −hξ, [X, Y ]i. Proof. Differentiate both sides of the following formula at t = 0: hAd∗ exp(tX) ξ, Y i = hξ, Adexp(−tX)Y i As a consequence, the theorem is proved if we can show dµ(XM) = Xg ∗ ◦ µ. To prove this we calculate for any Y ∈ g = (g ∗ ) ∗ , hdµ(XM(m)), Y i = Y ◦dµ(XM(m)) = d(Y ◦µ)(XM(m)) = XM(Y ◦µ)(m) = XM(hµ(m), Y i), where the second equality follows from the fact that Y is linear as a function on g ∗ . Now use the assumption that µ ∗ is a Lie algebra homomorphism. So XM(µ Y ) = {µ Y , µX} = µ [Y,X] = −hµ, [X, Y ]i = hXg ∗ (µ), Y i. It follows hdµ(XM(m)), Y i = hXg ∗ (µ(m)), Y i. This is exactly what we wanted. ¶ Change of Lie groups. Proposition 1.5. Suppose (M, ω, G, µ) is a Hamiltonian G-space. Let ϕ : K → G be a Lie group homomorphism. Then the induced K-action on M defined by k · m := ϕ(k) · m is a Hamiltonian action with moment map ν = (dϕ) T ◦ µ.
4LECTURE8:THEMOMENTMAPProof. Let X E t. Use the fact (exp(tX) = exp(dp(tX)) we getdXm(m) =l=og(exp(tX) m = (d(X)s(m).It follows thatforv= (dp)Toμ.d(v, X) = d((dp)T oμ, X) = d(μ, dp(X)) = t(dp(x)xw= 1xmw.To prove v is equivariant,one only need to prove v* is a Lie algebra homomor-phism. But by definition,v*=μ"odp,so μ* is a Lie algebra homomorphism since both dp and μ are Lie algebra homo口morphisms.As a consequence we seeCorollary 1.6. If (M,w,G,μ) is a Hamiltonian G-space and t : H G a Liesubgroup. Then the induced H-action on M is Hamiltonian with moment mapV=diToμ.2.EXISTENCE AND UNIQUENESS OF THE MOMENT MAPI Uniqueness.Suppose Lie group G acts in a Hamiltonian way on (M,w), we would like toknow how unique is the moment map. In other words, suppose μi and μ2 are bothmoment maps for this action. What is the difference μi-μ2?Instead of working on the moment maps μi and μ2, we works on the comomentmaps μ and μ. By definition for each X g, μ and μ are both Hamiltonianfunctions for the same vector field Xm.It follows that the difference(X)-(X)=-=cXis locally constant, and thus a constant on M (we will always assume that M isconnected). Obviously cX depends linearly in X. So we get an element c E g* with(c, X) = cX.Note that in this case the two moment maps are related byμ= μ2 + c,in other words, they differed by a constant in g*.Since μi and μ are both Lie algebra homomorphisms, for any X, Y e g,ex= x=(,-(,)=+c,+c-,)= 0
4 LECTURE 8: THE MOMENT MAP Proof. Let X ∈ k. Use the fact ϕ(exp(tX)) = exp(dϕ(tX)) we get XM(m) = d dt|t=0ϕ(exp(tX)) · m = (dϕ(X))M(m). It follows that for ν = (dϕ) T ◦ µ. dhν, Xi = dh(dϕ) T ◦ µ, Xi = dhµ, dϕ(X)i = ι(dϕ(X))M ω = ιXM ω. To prove ν is equivariant, one only need to prove ν ∗ is a Lie algebra homomorphism. But by definition, ν ∗ = µ ∗ ◦ dϕ, so µ ∗ is a Lie algebra homomorphism since both dϕ and µ are Lie algebra homomorphisms. As a consequence we see Corollary 1.6. If (M, ω, G, µ) is a Hamiltonian G-space and ι : H ,→ G a Lie subgroup. Then the induced H-action on M is Hamiltonian with moment map ν = dιT ◦ µ. 2. Existence and Uniqueness of the moment map ¶ Uniqueness. Suppose Lie group G acts in a Hamiltonian way on (M, ω), we would like to know how unique is the moment map. In other words, suppose µ1 and µ2 are both moment maps for this action. What is the difference µ1 − µ2? Instead of working on the moment maps µ1 and µ2, we works on the comoment maps µ ∗ 1 and µ ∗ 2 . By definition for each X ∈ g, µ X 1 and µ X 2 are both Hamiltonian functions for the same vector field XM. It follows that the difference µ ∗ 1 (X) − µ ∗ 2 (X) = µ X 1 − µ X 2 = c X is locally constant, and thus a constant on M (we will always assume that M is connected). Obviously c X depends linearly in X. So we get an element c ∈ g ∗ with hc, Xi = c X. Note that in this case the two moment maps are related by µ1 = µ2 + c, in other words, they differed by a constant in g ∗ . Since µ ∗ 1 and µ ∗ 2 are both Lie algebra homomorphisms, for any X, Y ∈ g, c [X,Y ] = µ [X,Y ] 1 − µ [X,Y ] 2 = {µ X 1 , µY 1 } − {µ X 2 , µY 2 } = {µ X 2 + c X, µY 2 + c Y } − {µ X 2 , µY 2 } = 0.
LECTURE8:THEMOMENTMAP5It follows that the constantc e [g, g)° = H'(g, R),Conversely, for any c e [g, g]° and any moment map μ, it is easy to see that the mapμ+c is a moment map for the same action, where the equivariance follows from thefact that Ad,c = 0 for any c e [g, g]°. (Check the last statement).Inconclusion,wegetTheorem 2.1.Any two moment maps of the same Hamiltonian action differ by aconstant in [g,glo =H'(g,R).As a consequence,Corollary 2.2. Let G be a compact Lie group with H'(g, R) = 0, then the momentmaps for any Hamiltonian G-action is unique.On the other extreme, since [g, g]° = g* if G is an abelian Lie group, we getCorollary 2.3.If (M,w,Tn,μ)is aHamiltonian Tn-system, then for any c E g,μ+c is a moment map for the Tn-action."Existence".In this subsection we dress at the following question: Under what condition wecan assert that any symplectic action is a Hamiltonian action? We will give twoindependent criteria, one on the manifold M and one on the Lie group G.Theorem 2.4. Suppose (M,w) is a connected compact symplectic manifold withH'(M,R)-o, then any symplectic action on M is Hamiltonian.Proof. We have seen that under the condition H'(M)= O, any symplectic vectorfield is a Hamiltonian vector field.We first choose a basis [Xi,..., Xa} of g. For each X, we can find a functionμX on M with t(x)mw = dμx. The functions μX are only unique up to constants,and wefix the constant by requiringuXwn=0.For any X Eg, one can writeX=aiXiand wedefineux-EaiuxiThis defines a linear map μ* : g -→ C(M) withixmw= dμX,in other words, the G-action is a weak-Hamiltonian action
LECTURE 8: THE MOMENT MAP 5 It follows that the constant c ∈ [g, g] 0 = H 1 (g, R). Conversely, for any c ∈ [g, g] 0 and any moment map µ, it is easy to see that the map µ + c is a moment map for the same action, where the equivariance follows from the fact that Ad∗ g c = 0 for any c ∈ [g, g] 0 . (Check the last statement). In conclusion, we get Theorem 2.1. Any two moment maps of the same Hamiltonian action differ by a constant in [g, g] 0 = H1 (g, R). As a consequence, Corollary 2.2. Let G be a compact Lie group with H1 (g, R) = 0, then the moment maps for any Hamiltonian G-action is unique. On the other extreme, since [g, g] 0 = g ∗ if G is an abelian Lie group, we get Corollary 2.3. If (M, ω, T n , µ) is a Hamiltonian T n -system, then for any c ∈ g ∗ , µ + c is a moment map for the T n -action. ¶ “Existence”. In this subsection we dress at the following question: Under what condition we can assert that any symplectic action is a Hamiltonian action? We will give two independent criteria, one on the manifold M and one on the Lie group G. Theorem 2.4. Suppose (M, ω) is a connected compact symplectic manifold with H1 (M, R) = 0, then any symplectic action on M is Hamiltonian. Proof. We have seen that under the condition H1 (M) = 0, any symplectic vector field is a Hamiltonian vector field. We first choose a basis {X1, · · · , Xd} of g. For each Xi we can find a function µ Xi on M with ι(Xi)M ω = dµXi . The functions µ Xi are only unique up to constants, and we fix the constant by requiring Z M µ Xiω n = 0. For any X ∈ g, one can write X = XaiXi , and we define µ X = Xaiµ Xi . This defines a linear map µ ∗ : g → C ∞(M) with ιXM ω = dµX, in other words, the G-action is a weak-Hamiltonian action
6LECTURE8:THEMOMENTMAPIt remains to prove that μ* is a Lie algebra homomorphism. We consider thefunctionμx, - (ux,μ] = exyon M. The function cX,Y is actually a constant sincedex,Y = dμ(x, - d(μx,μ) = [x,xw - L-[xm,Ym) = 0.On the other hand, by definition J μx,Ylwn = 0. By problem (6) in problem set 2,(μ,μ"w" = 0.口It follows that cx,Y =0.This completes theproof.The first criteria is natural since we have seen that under the condition H'(M) =O, any symplectic vector field is Hamiltonian. Our second criteria is on G and is notat all obvious at the first glance:Theorem 2.5.Let G be a connected Lie group withH'(g, R) = H2(g, R) = 0,then every symplectic G-action is Hamiltonian.Proof. First observe that H'(g, R) = 0 is equivalent to [g, g] = g. So any XM can bewritten as a summation of vector fields of the form [YM, ZM], which is Hamiltoniansince the Lie bracket of any two symplectic vector fields is Hamiltonian.Now repeat the proof of the proceeding theorem, we can find smooth functionsμd, depending linearly in X, so that txmw = dμ. Again the functionx =x- (,)is a constant function on M. (But we can't require JM μwn = O because M couldbe noncompact. And unlike the previous theorem, this μ's do not glue to a momentmap in general.)Obviously c(X,Y) := cx,Y is bi-linear and anti-symmetric, and thus defines anelement c e C2(g, R). Moreover, according to the Jacobi identities for [-, ] and forf,] we getdc(X,Y, Z) = -c([X,Y], Z) + c([X, Z], Y) - c([Y, Z], X) = 0In other words, c is an element in z?(g,R). Since H?(g, R) = 0, one can find anelement b E Ci(g, R) = g* so thatc = db.In other words,c(X, Y) = db(X,Y) = -b([X,Y)Now we defineμ* : g →C(M), X →略 +b(X)
6 LECTURE 8: THE MOMENT MAP It remains to prove that µ ∗ is a Lie algebra homomorphism. We consider the function µ [X,Y ] − {µ X, µY } = c X,Y on M. The function c X,Y is actually a constant since dcX,Y = dµ[X,Y ] − d{µ X, µY } = ι[X,Y ]M ω − ι−[XM,YM]ω = 0. On the other hand, by definition R µ [X,Y ]ω n = 0. By problem (6) in problem set 2, Z M {µ X, µY }ω n = 0. It follows that c X,Y = 0. This completes the proof. The first criteria is natural since we have seen that under the condition H1 (M) = 0, any symplectic vector field is Hamiltonian. Our second criteria is on G and is not at all obvious at the first glance: Theorem 2.5. Let G be a connected Lie group with H 1 (g, R) = H 2 (g, R) = 0, then every symplectic G-action is Hamiltonian. Proof. First observe that H1 (g, R) = 0 is equivalent to [g, g] = g. So any XM can be written as a summation of vector fields of the form [YM, ZM], which is Hamiltonian since the Lie bracket of any two symplectic vector fields is Hamiltonian. Now repeat the proof of the proceeding theorem, we can find smooth functions µ X 0 , depending linearly in X, so that ιXM ω = dµX 0 . Again the function c X,Y = µ [X,Y ] 0 − {µ X 0 , µY 0 } is a constant function on M. (But we can’t require R M µ Xω n = 0 because M could be noncompact. And unlike the previous theorem, this µ X 0 ’s do not glue to a moment map in general.) Obviously c(X, Y ) := c X,Y is bi-linear and anti-symmetric, and thus defines an element c ∈ C 2 (g, R). Moreover, according to the Jacobi identities for [·, ·] and for {·, ·} we get dc(X, Y, Z) = −c([X, Y ], Z) + c([X, Z], Y ) − c([Y, Z], X) = 0. In other words, c is an element in Z 2 (g, R). Since H2 (g, R) = 0, one can find an element b ∈ C 1 (g, R) = g ∗ so that c = db. In other words, c(X, Y ) = db(X, Y ) = −b([X, Y ]). Now we define µ ∗ : g → C ∞(M), X 7→ µ X 0 + b(X).
LECTURE8:THEMOMENTMAP7Then the map μ* is linear, and is a Lie algebra homomorphism since([X, Y) =x + b([X,Y)) = - cXY = (, = ((X),μ(Y),Finallytxmw = dμx = dμ*(X).口This completes the proofAccording to theWhitehead lemma, H'(g,R)=H?(g,R)=0 for semi-simpleLie groups. It followsCorollary 2.6.If G is semi-simple, then any symplectic G-action is Hamiltonian.Remark. The example “si acts on T? via . (ti,t2) = (ti + ,t2)" violates bothassumptions and is not Hamiltonian.3. EXAMPLES/EXERCISES OF MOMENT MAPS Some linear examples.Erample.The s'action on S? byrotations described above is a Hamiltonian actionwith moment map μ(, z) = z.Erample (Linearaction).Consider (R2n, 2o). The linear symplectic groupSp(2n)=(AEGL(2n)|A*20=20} =(A/Jo=ATJoA)acts on (R2n, 2o) in the natural way. The Lie algebra of Sp(2n) isSp(2n) =[A E g(2n) [At Jo =-JoA)For any r E R2n and any X E sp(2n) C gl(2n),dXM(r) =exp(tX)r=XadttOne can check that the map μ: R2n→ sp(2n)* defined by(μ(c),X) = 20(X(c), a)is the moment map of this action.Erample. Let G= Rn acts on R2n by translationsr.(r,y)=(r+r,y)This is a Hamiltonian action with moment map μ(r,y)=y
LECTURE 8: THE MOMENT MAP 7 Then the map µ ∗ is linear, and is a Lie algebra homomorphism since µ˜([X, Y ]) = µ [X,Y ] 0 + b([X, Y ]) = µ [X,Y ] 0 − c X,Y = {µ X 0 , µY 0 } = {µ ∗ (X), µ∗ (Y )}. Finally ιXM ω = dµX 0 = dµ∗ (X). This completes the proof. According to the Whitehead lemma, H1 (g, R) = H2 (g, R) = 0 for semi-simple Lie groups. It follows Corollary 2.6. If G is semi-simple, then any symplectic G-action is Hamiltonian. Remark. The example “S 1 acts on T 2 via θ · (t1, t2) = (t1 + θ, t2)” violates both assumptions and is not Hamiltonian. 3. Examples/Exercises of moment maps ¶ Some linear examples. Example. The S 1 action on S 2 by rotations described above is a Hamiltonian action with moment map µ(θ, z) = z. Example (Linear action). Consider (R 2n , Ω0). The linear symplectic group Sp(2n) = {A ∈ GL(2n) | A ∗Ω0 = Ω0} = {A | J0 = A T J0A}. acts on (R 2n , Ω0) in the natural way. The Lie algebra of Sp(2n) is sp(2n) = {A ∈ gl(2n) | A t J0 = −J0A}. For any x ∈ R 2n and any X ∈ sp(2n) ⊂ gl(2n), XM(x) = d dt t=0 exp(tX)x = Xx One can check that the map µ : R 2n → sp(2n) ∗ defined by hµ(x), Xi = 1 2 Ω0(X(x), x) is the moment map of this action. Example. Let G = R n acts on R 2n by translations r · (x, y) = (x + r, y). This is a Hamiltonian action with moment map µ(x, y) = y.
8LECTURE8:THEMOMENTMAPErample.One can identifyU(n)as aLie subgroup of Sp(2n)viaX-YZ-X+iwYXIt follows that the moment map of the canonical U(n) action on Cn = R2n isHamiltonian.Check:the momentmap is μ(z)=zz*From smooth action to Hamiltonian action.Erample (Liftingof smooth actionto cotangent bundle)LetT :GDiff(M)bea smoothaction ofa compactLiegroupGonasmoothmanifold M. The action induces a natural action : G →Diff(T*M) of G on T*Mbyg (m, n) = (g m, (dg-1)mn),where n E T*M. We shall denote p = (m, n) for simplicity.Observation 1: The projection map π : T*M → M is G-equivariant:元(g·p)=g·m=g ·元(p)As a consequence, we havedigpo dgp= dgm o diand thusdgp o drgp = dr, o dgmObservation 2: For any X E g,dg,'(Xm(g -p)) = (Adg-1X)m(p).Thisfollows fromdirectcomputation:d(Adg-1X)m(p) ==l/=-0 exp(tAdg-X) -pdl=0g- exp(tX)g - p= dgp'(Xm(g · p)Theorem3.1.The induced actionis aHamiltonian action withmoment mapμ: T*M -→ g* given by(μ(p), X) = (n, dp(XM)Proof.Letα be thetautological 1-form on T*M.Recall that for anyp= (m,)E M,ap=(dp)*n. So(g*α)p= (dgp)*ag-p=(dgp)*(dgp)*(dg-1)mn=dπ,n=αpi.e. Q is invariant under the G-action. It follows that CxaQ = O for all X E g. Soby Cartan's magic formula,dixQ= iXmw
8 LECTURE 8: THE MOMENT MAP Example. One can identify U(n) as a Lie subgroup of Sp(2n) via Z = X + iY X −Y Y X . It follows that the moment map of the canonical U(n) action on C n = R 2n is Hamiltonian. Check: the moment map is µ(z) = i 2 zz∗ . ¶ From smooth action to Hamiltonian action. Example (Lifting of smooth action to cotangent bundle). Let τ : G → Diff(M) be a smooth action of a compact Lie group G on a smooth manifold M. The action induces a natural action γ : G → Diff(T ∗M) of G on T ∗M by g · (m, η) = (g · m,(dg−1 ) ∗ mη), where η ∈ T ∗ mM. We shall denote p = (m, η) for simplicity. Observation 1: The projection map π : T ∗M → M is G-equivariant: π(g · p) = g · m = g · π(p). As a consequence, we have dπg·p ◦ dgp = dgm ◦ dπp and thus dg∗ p ◦ dπ∗ g·p = dπ∗ p ◦ dg∗ m. Observation 2: For any X ∈ g, dg−1 p (XM(g · p)) = (Adg−1X)M(p). This follows from direct computation: (Adg−1X)M(p) = d dt|t=0 exp(tAdg−1X) · p = d dt|t=0g −1 exp(tX)g · p = dg−1 p (XM(g · p)) Theorem 3.1. The induced action γ is a Hamiltonian action with moment map µ : T ∗M → g ∗ given by hµ(p), Xi = hη, dπp(XM)i Proof. Let α be the tautological 1-form on T ∗M. Recall that for any p = (m, η) ∈ M, αp = (dπp) ∗ η. So (g ∗α)p = (dgp) ∗αg·p = (dgp) ∗ (dπg·p) ∗ (dg−1 ) ∗ mη = dπ∗ p η = αp, i.e. α is invariant under the G-action. It follows that LXM α = 0 for all X ∈ g. So by Cartan’s magic formula, dιXM α = ιXM ω
9LECTURE 8:THEMOMENTMAPSo the X component of the moment map is μ= txmQ. So<μ(p), X) =μ*(p) = (xmQ(p) = ((d)n, Xm) = (n, (d元)pXm),The G-equivariance follows《μ(g p),X) = ((dg-1)mn,(d元)gpXm)=(n,(dg-1)n(dπgp)Xm)= (n,dpo dgp'(Xm))= (n,dp o (Adg-1X)M(p))= (μ(p), Adg-1X)= (Adgμ(p), X).口4.MOMENTMAPINGAUGETHEORYstudent presentation
LECTURE 8: THE MOMENT MAP 9 So the X component of the moment map is µ X = ιXM α. So hµ(p), Xi = µ X(p) = ιXM α(p) = h(dπ∗ p )η, XMi = hη,(dπ)pXMi. The G-equivariance follows hµ(g · p), Xi = h(dg−1 ) ∗ n η,(dπ)g·pXMi = hη,(dg−1 )n(dπg·p)XMi = hη, dπp ◦ dg−1 p (XM)i = hη, dπp ◦ (Adg−1X)M(p)i = hµ(p), Adg−1Xi = hAd∗ gµ(p), Xi. 4. Moment map in gauge theory student presentation