Notes on symplectic geometry and group actions

Notes on symplectic geometry and group actions Peter Hochs November 5, 2013 Contents 1 Example of Hamiltonian mechanics 1 2 Proper actions and Hausdorff quotients 4 3 N particles in R 3 7 4 Commuting actions 9 1 Example of Hamiltonian mechanics Consider a point particle of mass m moving in 3-dimensional Euclidean space R 3. Let q = (q 1, q 2, q 3 ) be the position coordinates of the particle. Suppose the particle is acted upon by an external force field F : R 3 R 3 that is determined by a potential function V C (R 3 ), by ( V F = grad V = q 1, V q 2, V ) q 3. (1) Then the motion of the particle, as a function of time t, is given by a curve γ in R 3, determined by the differential equation F (γ(t)) = mγ (t), (2) which is Newton s second law F = ma. Let δ(t) := mγ (t) be the momentum of the particle at time t as it moves along the curve γ. Then (1) and (2) may be rewritten as γ (t) = 1 m δ(t); δ (t) = grad V (γ(t)). (3) 1

Given this system of equations, the particle s trajectory is determined uniquely if both its position q := γ(t 0 ) and momentum p := δ(t 0 ) at a time t 0 are given. This motivates the definition of the phase space of the particle as R 6 = R 3 R 3, consisting of all possible positions q = (q 1, q 2, q 3 ) and momenta p = (p 1, p 2, p 3 ) the particle can have. A point in phase space, called a state, determines the motion of the particle, through Newton s law (3). To rewrite (3) in a way that will clarify the link between classical and quantum mechanics, consider the Hamiltonian function H C (R 6 ), given by the total energy of the particle: H(q, p) := 1 2m 3 ( p j ) 2 + V (q). (4) Furthermore, for two functions f, g C (R 6 ), we define the Poisson bracket {f, g} := 3 f g p j q j f g q j p j C (R 6 ). (5) One can check that the Poisson bracket is a Lie bracket on C (R 6 ), and that it has the derivation property that for all f, g, h C (R 6 ), {f, gh} = g{f, h} + {f, g}h. (6) The reason why we consider this bracket is that it allows us to restate (3) as follows. Write γ(t) = ( γ 1 (t), γ 2 (t), γ 3 (t) ) ; δ(t) = ( δ 1 (t), δ 2 (t), δ 3 (t) ). Then (3) is equivalent to the system of equations ( γ j ) (t) = {H, q j }(γ(t), δ(t)); ( δ j ) (t) = {H, p j }(γ(t), δ(t)), (7) for j = 1, 2, 3, where q j, p j C (R 6 ) denote the coordinate functions. Renaming the curves q(t) := γ(t) and p(t) := δ(t), we obtain the more familiar form ( q j ) = {H, q j }; ( p j ) = {H, p j (8) }. 2

Here q j and p j denote both the components of the curves q and p and the coordinate functions on R 6, making (8) shorter and more suggestive, but mathematically less clear than (7). To describe the curves γ and δ in a different way, we note that the linear map f {H, f}, from C (R 6 ) to itself, is a derivation by (6). Hence it defines a vector field ξ H on R 6, called the Hamiltonian vector field of H. Let e tξ H : R 6 R 6 be the flow of this vector field over time t. That is, d dt f ( e tξ H (q, p) ) = ξ H (f)(q, p) = {H, f}(q, p) t=0 for all f C (R 6 ) and (q, p) R 6. Then, if γ(0) = q and δ(0) = p, conditions (7) simply mean that (γ(t), δ(t)) = e tξ H (q, p). (9) An observable in this setting is by definition a smooth function of the position and the momentum of the particle, i.e. a function f C (R 6 ). The Hamiltonian function and the Poisson bracket allow us to write the time evolution equation of any observable f as the following generalisation of (7): d ( ) f(γ(t), δ(t)) = {H, f}(γ(t), δ(t)). (10) dt Here γ and δ are curves in R 3 satisfying (7). This time evolution equation for f can be deduced from the special case (7) using the chain rule. In (10), the state (γ, δ) of the system changes in time, whereas the observable f is constant. To obtain a time evolution equation that resembles the quantum mechanical version more closely, we define the time-dependent version f C (R R 6 ) of f, by f(t, q, p) := f(e tξ H (q, p)) =: f t (q, p). Then by (9), equation (10) becomes f = {H, f t }. (11) t t Motivated by this example of one particle in R 3 moving in a conservative force field, we define a classical mechanical system to be a triple (M, {, }, H), where M is a smooth manifold called the phase space (replacing R 6 in the preceding example), {, } is a Lie bracket on C (M) satisfying (6) for all f, g, h C (M), and H is a smooth function on M, called 3

the Hamiltonian function. The bracket {, } is called a Poisson bracket, and the pair (M, {, }) is a Poisson manifold. In this course, we will consider symplectic manifolds, a special kind of Poisson manifolds. Given a classical mechanical system, the dynamics of any observable f C (M) is determined by the classical time evolution equation (11). For more extensive treatments of the Hamiltonian formulation of classical mechanics, see [1, 2]. 2 Proper actions and Hausdorff quotients In this section, we prove the following fact. Theorem 2.1. Let X be a locally compact Hausdorff topological space, and let G be a topological group acting continuously and properly on X. Then the orbit space X/G is Hausdorff. The proof of this theorem is based on the following characterisation of the Hausdorff property. Lemma 2.2. A topological space X is Hausdorff if and only if the diagonal is closed in X X. (X) := {(x, x); x X} X X Proof. Suppose X is Hausdorff. We will show that the complement (X X)\ (X) is open. Indeed, let (x, y) (X X)\ (X). Then x y, and by the Hausdorff property, there are open neighbourhoods U and V of x and y, repsectively, which are disjoint. That means that the open neighbourhood U V of (x, y) in X X does not intersect the diagonal, i.e. U V (X X) \ (X). We conclude that (X X) \ (X) is open. Next, suppose (X) is closed in X X. Then (X X) \ (X) is open. Let x y X be given. Since (x, y) (X X) \ (X), there is an open neighbourhood W of (x, y) in (X X) \ (X). By definition of the product topology, there are open neighbourhoods U and V of x and y, respectively, such that (x, y) U V W (X X) \ (X). Since U V (X X)\ (X), U and V are disjoint. Hence X is Hausdorff. 4

The key property of locally compact Hausdorff spaces that we will use in the proof of Theorem 2.1, is the following criterion for closedness in such spaces. Proposition 2.3. Let X be a localy compact Hausdorff space, and A X a subset. Then A is closed if and only if for all compact subsets C X, the intersection A C is compact. Proof. One implication is easy to prove: if A is closed, then for all compact C X, the intersection A C is a closed subset of the compact set C, and hence compact. So conversely, suppose that A C is compact for all compact C X. We will show that X \ A is open. Let x X \ A. By local compactness of X, there is an open neighbourhood U of X whose closure U is compact. By assumption, the intersection A U is compact. Since X is Hausdorff and x A, we can choose open neighbourhoods V a of all points a A and corresponding neighbourhoods W a of X, such that for all a A, V a W a =. Then {V a } a A U is an open cover of the compact set A U, and hence has a finite subcover {V a1,..., V an }. Write W := n W aj U. Then W is an open neighbourhood of x. We will show that W A is empty, which completes the proof. Note that W A = n W aj U A n n W aj W aj (U A) n V aj Since W aj V aj = for all j, the latter set is empty.. 5

We now turn to a proof of Theorem 2.1. By Lemma 2.2, it is enough to show that the diagonal (X/G) X/G X/G is closed. By definition of the topology on X/G X/G = (X X)/(G G), this exactly means that q 1 ( (X/G)) X X is closed. Here q : X X (X X)/(G G) is the quotient map q(x, y) = (Gx, Gy), for x, y X. An important step in the proof of Theorem 2.1 is the following relation between q 1 ( (X/G)) and the map given by ψ(g, x) = (x, gx). ψ : G X X X Lemma 2.4. The set q 1 ( (X/G)) is the image im(ψ) of ψ. Proof. Note that q 1 ( (X/G)) = {(x, y) X X; q(x, y) (X/G)} = {(x, y) X X; Gx = Gy}. Now Gx = Gy exactly means that there is a g G such that y = gx. Hence which is the image of ψ. q 1 ( (X/G)) = {(x, gx); x X, g G}, Proof of Theorem 2.1. It remains to show that q 1 ( (X/G)) = im(ψ) is closed. By Proposition 2.3, this is equivalent to the property that im(ψ) C is compact for all compact subsets C X X. Now im(ψ) C = ψ(ψ 1 (C)). By properness of the action, ψ 1 (C) is compact. Therefore, by continuity of the action, ψ(ψ 1 (C)) is compact. 6

3 N particles in R 3 Consider a classical mechanical system of N particles moving in R 3. We will now describe the symplectic reduction of the phase space M = ( T R 3) N of this system by the action of the Euclidean motion group G = R 3 O(3). First, consider the action on M of the translation subgroup R 3 of G. The total linear momentum of the system defines a momentum map for this action. The reduced phase space for this restricted action is M 0 = ( T R 3N) 0 = T (R 3N /R 3 ). Let V be the (3N 3)-dimensional vector space R 3N /R 3. As coordinates on V, one can take the difference vectors q i q j, i < j. These coordinates satisfy the relations (q i q j ) + (q j q k ) = q i q k, i < j < k. Other possible coordinates are q i := q i N c j q j, i = 1,... N, for any set of coefficients {c j } with sum 1. The coordinates then satisfy the single relation N c i q i = 0. i=1 A physically natural choice for the c j is c j := m j N k=1 m, k where m j is the mass of particle j. The coordinates q i are then related by n m i q i = 0. i=1 Thus, the reduced phase space may be interpreted as the space of states of the N particle system in which the centre of mass is at rest in the origin. 7

Next, we consider the action on M of the connected component G 0 = R 3 SO(3) of G. The action of G 0 on M is not free, and the momentum map µ has singular points. Indeed, a point m = (q, p) in M is singular if and only if its stabiliser group G 0 m has dimension at least 1. Note that G 0 m is the group of elements (v, A) R 3 SO(3) such that Aq + v = q (12) Ap = p. (13) The group of elements (v, A) for which condition (12) holds is at least onedimensional if and only if the vectors q 1,..., q N R 3 are collinear, i.e. they lie on a single line l in R 3. This line does not have to pass through the origin, and some points q i may coincide. Given such a q, the group of A SO(3) for which (13) holds is at least one-dimensional if and only if all vectors p i lie on the line through the origin, parallel to l. In other words, the set M sing of singular points in M is the set of (q, p) R 3N R 3N, for which there exist vectors b, v R 3, and sets of coefficients {α i } and {β i }, such that q i = b + α i v (14) p i = β i v. (15) The set of singular values of µ in g is the image µ(m sing ). By (14) and (15), this set equals {( ( N µ(m sing ) = β i) ( N v, β i) } b v ); β i R, b, v R 3 i=1 i=1 = {(P, L) R 3 R 3 ; P L}. In particular, (0, 0) is a singular value. We now turn to the action of the whole group G. A point m = (q, p) has nontrivial stabiliser in G if and only if the q i are coplanar (i.e. they lie in a plane W in R 3 ) and the p i lie in the plane W 0 through the origin, parallel to W. Indeed, if R O(3) is reflection in W 0, then (q1 Rq 1, R) G m. Note that q 1 may be replaced by any point in W, and that R SO(3). 8

We define M M to be the open dense submanifold of points with trivial stabiliser in G. That is, M is the set of points (q, p) M such that there is no plane W 0 in R 3 containing the origin, and no vector b R 3, such that for all i, q i W := W 0 + b p i W 0. The set of points q in R 3N on which G acts with trivial stabiliser is the set R 3N = {q R 3N ; There is no plane in R 3 containing all q i.} Now ( ) T R3N 0 = T ( R3N ) /G ( ) = T Ṽ /O(3). Here Ṽ := R 3N /R 3. The coordinates q i q j that we used on V reduce to the coordinates q i q j 2, i < j. (16) on Ṽ /O(3). There are ( ) N 2 of these coordinates, and the dimension of Ṽ /O(3) is 3N 6. So if N 5 (so that ( ) N 2 > 3N 6), then there are relations between the coordinates (16). In any case, it is clear that a function (observable) on the reduced phase space T R3N) corresponds to ( 0 a function on T R3N that only depends on the relative distances between the particles. 4 Commuting actions To study actions on P(C n ) induced by unitary representations in C n+1, we consider the following situation. Let G be a Lie group, and let H, H < G be closed subgroups. Suppose G acts properly on a symplectic manifold (M, ω), and assume that the action is Hamiltonian, with momentum map µ : M g. Suppose that the restricted actions of H and H on M commute. By Exercise 11.2 in the exercise set, the action of H on M is Hamiltonian, with momentum map µ : M µ g ĩ h, 9

where ĩ : h g is the inclusion map. Consider the symplectic reduction M ξ of the action of H on M, at the regular value ξ h of µ. Because the actions of H and H on M commute, the action of H on M induces an action of H on the symplectic reduction M ξ. Proposition 4.1 (Commuting actions). The action of H on M ξ is Hamiltonian, with momentum map µ : M ξ h induced by the composition µ 1 (ξ) ι M µ g i h, where ι : µ 1 (ξ) M and i : h g are the inclusion maps. Proof. We may assume that H = G, for otherwise we can apply Exercise 11.2 mentioned above. Our claim is that for all X g, d(µ X) = X M ξ ω. Here the symplectic form ω is determined by ι ω = π ω. The maps ι and π are the inclusion and quotient maps in Because the linear map µ 1 (ξ) ι M π µ 1 (ξ)/ H ξ π : Ω( M ξ ) Ω( µ 1 (ξ)) is injective, it is enough to prove that Note that π ( d(µ X) ) = π ( X M ξ ω ). (17) π ( d(µ X) ) = d ( π µ ) X = d (µ X ι) which proves (17). = ι dµ X = ι ( X M ω) = X µ 1 (ξ) ι ω = X µ 1 (ξ) π ω = π ( X M ξ ω ), 10

References [1] R. Abraham and J. E. Marsden. Foundations of mechanics. Addison Wesley, Redwood City, second edition, 1985. [2] V. I. Arnold. Mathematical methods of classical mechanics. Springer, New York, 1978. 11