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

d XM (m) = exp(tX) · m. dt t=0 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 µX on M, depending linearly on X ∈ g, so that X dµ = ιXM ω. There are two equivalent ways to group these functions µX ’s together: – The comoment map is the linear map µ∗ : g → C∞(M), µ∗(X) = µX . – 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 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∗ ' Rn, 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 s- mooth function that Poisson commutes with f. Now let (M, ω, G, µ) be a weakly- Hamiltonian 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 µX is an integral of motion of the Hamilton system (M, ω, f).

Proof. Let Ξf be the Hamiltonian vector field associated to f, then

X d {f, µ }(m) = ω(Ξf ,XM )(m) = ιXM df(m) = LXM f(m) = f(exp(tX)·m) = 0 dt t=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,

X Y X d {µ , µ }(m) = YM (µ )(m) = hµ(exp(tY ) · m),Xi dt t=0

d ∗ = hAdexp(tY )µ(m),Xi dt t=0

d = hµ(m), Adexp(−tY )Xi dt t=0

d = hµ(m), exp(−tadY )Xi dt t=0 = 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 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) = Adexp(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 i field on Mi, and let ρt be the flow of Yi. 1 2 Lemma 1.3. If df(Y1) = Y2 ◦ f, then f ◦ ρt = ρt ◦ 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 Adexp(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: ∗ hAdexp(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 Y Y X [Y,X] XM (µ ) = {µ , µ } = µ = −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. 4 LECTURE 8: THE MOMENT MAP

Proof. Let X ∈ k. Use the fact ϕ(exp(tX)) = exp(dϕ(tX)) we get d X (m) = | ϕ(exp(tX)) · m = (dϕ(X)) (m). M dt t=0 M It follows that for ν = (dϕ)T ◦ µ. T dhν, Xi = dh(dϕ) ◦ 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 homomor- phism. But by definition, ν∗ = µ∗ ◦ dϕ, so µ∗ is a Lie algebra homomorphism since both dϕ and µ are Lie algebra homo- morphisms.  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 ∗ ∗ X X maps µ1 and µ2. By definition for each X ∈ g, µ1 and µ2 are both Hamiltonian functions for the same vector field XM . It follows that the difference ∗ ∗ X X X µ1(X) − µ2(X) = µ1 − µ2 = c is locally constant, and thus a constant on M (we will always assume that M is connected). Obviously cX depends linearly in X. So we get an element c ∈ g∗ with hc, Xi = cX . 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 µ1 and µ2 are both Lie algebra homomorphisms, for any X,Y ∈ g, [X,Y ] [X,Y ] [X,Y ] c = µ1 − µ2 X Y X Y = {µ1 , µ1 } − {µ2 , µ2 } X X Y Y X Y = {µ2 + c , µ2 + c } − {µ2 , µ2 } = 0. LECTURE 8: THE MOMENT MAP 5

It follows that the constant 0 1 c ∈ [g, g] = H (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 ∗ 0 fact that Adgc = 0 for any c ∈ [g, g] . (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, Tn, µ) is a Hamiltonian Tn-system, then for any c ∈ g∗, µ + c is a moment map for the Tn-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 Xi Xi Xi µ on M with ι(Xi)M ω = dµ . The functions µ are only unique up to constants, and we fix the constant by requiring Z µXi ωn = 0. M For any X ∈ g, one can write X X = aiXi, and we define X X Xi µ = aiµ . This defines a linear map µ∗ : g → C∞(M) with X ιXM ω = dµ , in other words, the G-action is a weak-Hamiltonian action. 6 LECTURE 8: THE MOMENT MAP

It remains to prove that µ∗ is a Lie algebra homomorphism. We consider the function µ[X,Y ] − {µX , µY } = cX,Y on M. The function cX,Y is actually a constant since X,Y [X,Y ] X Y dc = dµ − d{µ , µ } = ι[X,Y ]M ω − ι−[XM ,YM ]ω = 0. On the other hand, by definition R µ[X,Y ]ωn = 0. By problem (6) in problem set 2, Z {µX , µY }ωn = 0. M X,Y It follows that c = 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 1 2 H (g, R) = H (g, R) = 0, then every symplectic G-action is Hamiltonian.

1 Proof. First observe that H (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 X X µ0 , depending linearly in X, so that ιXM ω = dµ0 . Again the function X,Y [X,Y ] X Y c = µ0 − {µ0 , µ0 } R X n is a constant function on M. (But we can’t require M µ ω = 0 because M could X be noncompact. And unlike the previous theorem, this µ0 ’s do not glue to a moment map in general.) Obviously c(X,Y ) := cX,Y is bi-linear and anti-symmetric, and thus defines an element c ∈ C2(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 Z2(g, R). Since H2(g, R) = 0, one can find an element b ∈ C1(g, R) = g∗ so that c = db. In other words, c(X,Y ) = db(X,Y ) = −b([X,Y ]). Now we define ∗ ∞ X µ : g → C (M),X 7→ µ0 + b(X). LECTURE 8: THE MOMENT MAP 7

Then the map µ∗ is linear, and is a Lie algebra homomorphism since

[X,Y ] [X,Y ] X,Y X Y ∗ ∗ µ˜([X,Y ]) = µ0 + b([X,Y ]) = µ0 − c = {µ0 , µ0 } = {µ (X), µ (Y )}. Finally X ∗ ιXM ω = dµ0 = 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.

1 2 Remark. The example “S acts on T 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 S1 action on S2 by rotations described above is a Hamiltonian action with moment map µ(θ, z) = z. Example (Linear action). 2n Consider (R , Ω0). The linear symplectic group ∗ T Sp(2n) = {A ∈ GL(2n) | A Ω0 = Ω0} = {A | J0 = A J0A}. 2n acts on (R , Ω0) in the natural way. The Lie algebra of Sp(2n) is t sp(2n) = {A ∈ gl(2n) | A J0 = −J0A}. For any x ∈ R2n and any X ∈ sp(2n) ⊂ gl(2n),

d XM (x) = exp(tX)x = Xx dt t=0 One can check that the map µ : R2n → sp(2n)∗ defined by 1 hµ(x),Xi = Ω (X(x), x) 2 0 is the moment map of this action.

Example. Let G = Rn acts on R2n by translations r · (x, y) = (x + r, y). This is a Hamiltonian action with moment map µ(x, y) = y. 8 LECTURE 8: THE MOMENT MAP

Example. One can identify U(n) as a Lie subgroup of Sp(2n) via X −Y  Z = X + iY . YX It follows that the moment map of the canonical U(n) action on Cn = R2n is i ∗ Hamiltonian. Check: the moment map is µ(z) = 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 −1 ∗ g · (m, η) = (g · m, (dg )mη), ∗ where η ∈ TmM. 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 ∗ ∗ ∗ ∗ dgp ◦ dπg·p = dπp ◦ dgm. Observation 2: For any X ∈ g, −1 dgp (XM (g · p)) = (Adg−1 X)M (p). This follows from direct computation: d (Ad −1 X) (p) = | exp(tAd −1 X) · p g M dt t=0 g d = | g−1 exp(tX)g · p dt t=0 −1 = dgp (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 ∗ ∗ ∗ ∗ −1 ∗ ∗ (g α)p = (dgp) αg·p = (dgp) (dπg·p) (dg )mη = dπpη = 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. LECTURE 8: THE MOMENT MAP 9

X So the X component of the moment map is µ = ιXM α. So X ∗ hµ(p),Xi = µ (p) = ιXM α(p) = h(dπp)η, XM i = hη, (dπ)pXM i. The G-equivariance follows −1 ∗ hµ(g · p),Xi = h(dg )nη, (dπ)g·pXM i −1 = hη, (dg )n(dπg·p)XM i −1 = hη, dπp ◦ dgp (XM )i

= hη, dπp ◦ (Adg−1 X)M (p)i

= hµ(p), Adg−1 Xi ∗ = hAdgµ(p),Xi. 

4. Moment map in gauge theory student presentation