LECTURE 8: THE MOMENT MAP

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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 X M (m) = d dt 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, X M is symplectic, i.e. L XM ω = 0. weakly Hamiltonian if each X M 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 dµ X = ι XM 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, µ X ( ) = µ( ), X. 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 m, f := f(m). 1

2 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 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 µ X is an integral of motion of the Hamilton system (M, ω, f). Proof. Let Ξ f be the Hamiltonian vector field associated to f, then {f, µ X }(m) = ω(ξ f, X M )(m) = ι XM df(m) = L XM f(m) = d dt 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, µ Y }(m) = Y M (µ X )(m) = d dt µ(exp(ty ) m), X = d dt Ad exp(ty )µ(m), X = d dt µ(m), Ad exp( ty ) X = d dt µ(m), exp( tady )X = µ(m), [X, Y ] = µ [X,Y ] (m), where for the second equality we used the fact exp(ty ) m = exp(ty M )(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

3 LECTURE 8: THE MOMENT MAP 3 be written as a product of elements of the form exp(x). 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: As a result, to prove Let f : M 1 M 2 be a smooth map. For i = 1, 2 let Y i be a smooth vector field on M i, and let ρ i t be the flow of Y i. Lemma 1.3. If df(y 1 ) = Y 2 f, then f ρ 1 t = ρ 2 t f. Sketch of proof: Both sides define integral curves of the vector field Y 2 passing the point f(m) at t = 0. Let X g be the vector field on g generating the flow Ad exp(tx). Lemma 1.4. For any ξ g and any Y g, X g (ξ), Y = ξ, [X, Y ]. Proof. Differentiate both sides of the following formula at t = 0: Ad exp(tx)ξ, Y = ξ, Ad exp( tx) Y As a consequence, the theorem is proved if we can show dµ(x M ) = X g µ. To prove this we calculate for any Y g = (g ), dµ(x M (m)), Y = Y dµ(x M (m)) = d(y µ)(x M (m)) = X M (Y µ)(m) = X M ( 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. So It follows X M (µ Y ) = {µ Y, µ X } = µ [Y,X] = µ, [X, Y ] = X g (µ), Y. This is exactly what we wanted. Change of Lie groups. dµ(x M (m)), Y = X g (µ(m)), Y. 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 4 LECTURE 8: THE MOMENT MAP Proof. Let X k. Use the fact ϕ(exp(tx)) = exp(dϕ(tx)) we get It follows that for ν = (dϕ) T µ. X M (m) = d dt ϕ(exp(tx)) m = (dϕ(x)) M (m). d ν, X = d (dϕ) T µ, X = d µ, dϕ(x) = ι (dϕ(x))m ω = ι XM 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 = dι T 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 µ 1 and µ 2 are both moment maps for this action. What is the difference µ 1 µ 2? Instead of working on the moment maps µ 1 and µ 2, 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 X M. 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 c, X = c X. Note that in this case the two moment maps are related by µ 1 = µ 2 + 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, µ Y 1 } {µ X 2, µ Y 2 } = {µ X 2 + c X, µ Y 2 + c Y } {µ X 2, µ Y 2 } = 0.

5 It follows that the constant LECTURE 8: THE MOMENT MAP 5 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 gc = 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 = H 1 (g, R). As a consequence, Corollary 2.2. Let G be a compact Lie group with H 1 (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 H 1 (M, R) = 0, then any symplectic action on M is Hamiltonian. Proof. We have seen that under the condition H 1 (M) = 0, any symplectic vector field is a Hamiltonian vector field. We first choose a basis {X 1,, X d } of g. For each X i we can find a function µ X i on M with ι (Xi ) M ω = dµ X i. The functions µ X i are only unique up to constants, and we fix the constant by requiring µ X i ω n = 0. For any X g, one can write and we define M X = a i X i, µ X = a i µ X i. This defines a linear map µ : g C (M) with ι XM ω = dµ X, in other words, the G-action is a weak-hamiltonian action.

6 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, µ Y } = c X,Y on M. The function c X,Y is actually a constant since dc X,Y = dµ [X,Y ] d{µ X, µ Y } = ι [X,Y ]M ω ι [XM,Y M ]ω = 0. On the other hand, by definition µ [X,Y ] ω n = 0. By problem (6) in problem set 2, {µ X, µ Y }ω n = 0. It follows that c X,Y = 0. This completes the proof. M The first criteria is natural since we have seen that under the condition H 1 (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 H 1 (g, R) = 0 is equivalent to [g, g] = g. So any X M can be written as a summation of vector fields of the form [Y M, Z M ], 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 ι XM ω = dµ X 0. Again the function c X,Y = µ [X,Y ] 0 {µ X 0, µ Y 0 } is a constant function on M. (But we can t require M µx ω 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 H 2 (g, R) = 0, one can find an element b C 1 (g, R) = g so that In other words, Now we define c = db. c(x, Y ) = db(x, Y ) = b([x, Y ]). µ : g C (M), X µ X 0 + b(x).

7 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, µ Y 0 } = {µ (X), µ (Y )}. Finally This completes the proof. ι XM ω = dµ X 0 = dµ (X). According to the Whitehead lemma, H 1 (g, R) = H 2 (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 (t 1, t 2 ) = (t 1 + θ, t 2 ) violates both assumptions and is not Hamiltonian. Some linear examples. 3. Examples/Exercises of moment maps 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, Ω 0 ). The linear symplectic group Sp(2n) = {A GL(2n) A Ω 0 = Ω 0 } = {A J 0 = A T J 0 A}. acts on (R 2n, Ω 0 ) in the natural way. The Lie algebra of Sp(2n) is sp(2n) = {A gl(2n) A t J 0 = J 0 A}. For any x R 2n and any X sp(2n) gl(2n), X M (x) = d dt exp(tx)x = Xx One can check that the map µ : R 2n sp(2n) defined by is the moment map of this action. µ(x), X = 1 2 Ω 0(X(x), x) 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.

8 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. 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: As a consequence, we have and thus Observation 2: For any X g, π(g p) = g m = g π(p). dπ g p dg p = dg m dπ p dg p dπ g p = dπ p dg m. dgp 1 (X M (g p)) = (Ad g 1X) M (p). This follows from direct computation: (Ad g 1X) M (p) = d dt exp(tad g 1X) p = d dt g 1 exp(tx)g p = dgp 1 (X M (g p)) Theorem 3.1. The induced action γ is a Hamiltonian action with moment map µ : T M g given by µ(p), X = η, dπ p (X M ) Proof. Let α be the tautological 1-form on T M. Recall that for any p = (m, η) M, α p = (dπ p ) η. So (g α) p = (dg p ) α g p = (dg p ) (dπ g p ) (dg 1 ) mη = dπ pη = α p, i.e. α is invariant under the G-action. It follows that L XM α = 0 for all X g. So by Cartan s magic formula, dι XM α = ι XM ω.

9 LECTURE 8: THE MOMENT MAP 9 So the X component of the moment map is µ X = ι XM α. So µ(p), X = µ X (p) = ι XM α(p) = (dπ p)η, X M = η, (dπ) p X M. The G-equivariance follows µ(g p), X = (dg 1 ) nη, (dπ) g p X M = η, (dg 1 ) n (dπ g p )X M = η, dπ p dg 1 p (X M ) = η, dπ p (Ad g 1X) M (p) = µ(p), Ad g 1X = Ad gµ(p), X. student presentation 4. Moment map in gauge theory