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LECTURE 8: the MOMENT MAP Contents 1. Properties of The 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 2 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 !. The G-action is called • symplectic if each element g 2 G acts on M by symplectomorphisms. { Equivalently, for each X 2 g, XM is symplectic, i.e. LXM ! = 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 2 g, so that X dµ = ιXM !: There are two equivalent ways to group these functions µX 's together: { The comoment map is the linear map µ∗ : g ! C1(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 C1(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; !; 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; !; f) is a s- mooth function that Poisson commutes with f. Now let (M; !; G; µ) be a weakly- Hamiltonian G-space and f 2 C1(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; !; G; µ) is a weakly-Hamiltonian G-space, and f 2 C1(M)G a smooth G-invariant function. Then for any X 2 g, the function µX is an integral of motion of the Hamilton system (M; !; f). Proof. Let Ξf be the Hamiltonian vector field associated to f, then X d ff; µ g(m) = !(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; [·; ·]) ! (C1(M); {·; ·}) is a Lie algebra homomorphism. Proof. First assume the G-action is Hamiltonian. Then for any X; Y 2 g, X Y X d fµ ; µ g(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 2 g∗ and any Y 2 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 2 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 ) = fµ ; µ g = 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; !; G; µ) is a Hamiltonian G-space. Let ' : K ! G be a Lie group homomorphism. Then the induced K-action on M defined by k · m := '(k) · m is a Hamiltonian action with moment map ν = (d')T ◦ µ. 4 LECTURE 8: THE MOMENT MAP Proof. Let X 2 k. Use the fact '(exp(tX)) = exp(d'(tX)) we get d X (m) = j '(exp(tX)) · m = (d'(X)) (m): M dt t=0 M It follows that for ν = (d')T ◦ µ. T dhν; Xi = dh(d') ◦ µ, Xi = dhµ, d'(X)i = ι(d'(X))M ! = ιXM !: To prove ν is equivariant, one only need to prove ν∗ is a Lie algebra homomor- phism. But by definition, ν∗ = µ∗ ◦ d'; so µ∗ is a Lie algebra homomorphism since both d' and µ are Lie algebra homo- morphisms. As a consequence we see Corollary 1.6. If (M; !; 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; !), 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 2 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 2 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 2 g, [X;Y ] [X;Y ] [X;Y ] c = µ1 − µ2 X Y X Y = fµ1 ; µ1 g − fµ2 ; µ2 g X X Y Y X Y = fµ2 + c ; µ2 + c g − fµ2 ; µ2 g = 0: LECTURE 8: THE MOMENT MAP 5 It follows that the constant 0 1 c 2 [g; g] = H (g; R): Conversely, for any c 2 [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 2 [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; !; Tn; µ) is a Hamiltonian Tn-system, then for any c 2 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; !) 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 fX1; ··· ;Xdg of g. For each Xi we can find a function Xi Xi Xi µ on M with ι(Xi)M ! = dµ . The functions µ are only unique up to constants, and we fix the constant by requiring Z µXi !n = 0: M For any X 2 g, one can write X X = aiXi; and we define X X Xi µ = aiµ : This defines a linear map µ∗ : g ! C1(M) with X ιXM ! = dµ ; in other words, the G-action is a weak-Hamiltonian action.
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