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Symplectic Vector Spaces, Lagrangian Subspaces, and Liouville's Theorem

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Symplectic Vector Spaces, Lagrangian Subspaces, and Liouville's Theorem SYMPLECTIC VECTOR SPACES, LAGRANGIAN SUBSPACES, AND LIOUVILLE’S THEOREM CONNER JAGER Celestial Mechanics Junior Seminar - Nicolas Templier Abstract. The purpose of this paper is to prove Liouville’s theorem on volume-preserving flows from a symplectic linear algebra point of view. Along the way, we will prove a series of properties about symplectic vector spaces that elucidate the usefulness of this approach with regard to the correspondence between the configuration space and the symplectic phase space. Though only briefly discussed, this paper motivates how symplectic manifolds provide the natural abstraction of the phase space of a dynamical system through the simplified linear example: symplectic vector spaces. 0. Motivation In the study of celestial mechanics, there arises a natural connection between symplectic manifolds and the phase space of a dynamical system. Intuitively, we may think of the two defining properties of symplectic manifolds as follows: (1) each point is associated with a vector in the tangent space of some embedded manifold and a covector in the cotangent space, and (2) it is equipped with a 2-form satisfying certain “natural” constraints. Indeed, in classical mechanics, we generate systems of objects under force that are naturally symplectic in nature: (1) each object is associated with a state consisting of a position in some embedded configuration space and momentum, and (2) we have a set of laws governing how the forces act upon the object (the Hamiltonian function). In fact, as pointed out in Cohn’s article “Why symplectic geometry is the natural setting for classical mechanics,” symplectic manifolds provide precisely this desired abstraction of phase spaces. A fundamental example of this intimate connection is the classic n-body problem, where we seek to determine the orbits of n objects in R3 by their mutual gravitational interactions. By identifying the initial phase space, that is, identifying a vector in the symplectic vector space R6n characterizing the initial position and momentum for each of the n bodies, and determining the energy (Hamiltonian) function, can construct a 6 dimensional phase flow that fully characterizes how the phase space (ie. position and momentum) evolves through time [C, A]. Understanding the properties of symplectic manifolds provides a clean way to analyze these dynamical systems. Though the manifold under which the system of study operates need not be Euclidean, as the n- body problem elucidates, very often the system involves the motion of particles through three space, and thus the manifold is Euclidean [C]. As such, this paper will limit its scope to symplectic vector spaces. This is a very strong simplifying assumption since, as we will later prove, every symplectic vector space is isomorphic to the R2n and therefore is characterized fully by its dimension. 1 2CONNERJAGER 1. Symplectic Vector Spaces It is, of course, important to first establish some basic definitions and principles concerning sym- plectic vector spaces. Definition 1.1. A symplectic vector space is a (finite-dimensional) vector space V equipped with a bilinear form ω (called the symplectic form of V ) that is antisymmetric: ω(u, v)= ω(v, u) • − non-degenerate: if u = 0, then there exists a vector v such that ω(u, v) = 0. • If we also assume the characteristic of the field of V is = 2, we can deduce that ω(v, v) = 0 for all vectors v V from the antisymmetry of ω. For the purpose of this paper, we will assume that the ∈ field of V is R. [HZ] Example 1.1. The canonical example of a symplectic vector space is R2n with the bilinear form ω (u, v)= Ju,v ,where , is the Euclidean inner product (ie. dot product) and J is the 2n 2n 0 · · × matrix: 0 I J = n I 0 − n We can easily verify that ω0 is antisymmetric and non-degenerate: Let u =(u ,...,u ),v =(v ,...,v ). • 1 2n 1 2n ω (u, v)=u v + + u v u v u v 0 n+1 1 ··· 2n n − 1 n+1 −···− n 2n ω (v, u)=v u + + v u v u v u = ω (u, v) 0 n+1 1 ··· 2n n − 1 n+1 −···− n 2n − 0 Let u =0 = u =(u ,...,u )whereu = 0 for some i.Thus,ω (u, e ) = u =0 = • ⇒ 1 2n i | 0 i | i ⇒ ω (u, e ) = 0 for the standard basis vector e . 0 i i Definition 1.2. The symplectic orthogonal complement of some subspace E V is defined as ⊆ E⊥ := v V : ω(u, v)=0, u E { ∈ ∀ ∈ } It is critical to note that in general, E and E⊥ need not be complements and E⊥ need not itself be a symplectic vector space. As a counterexample, consider when E = span(v) for some nonzero vector v V . For any x, y E,x = c v, y = c v and thus ∈ ∈ 1 2 ω(x, y)=c c ω(v, v)=0 1 2 · (In this case, E E and ω is degenerate on E.) ⊆ ⊥ From these observations, we will define special classes of subspaces of a symplectic vector space. [L] SYMPLECTIC VECTOR SPACES, LAGRANGIAN SUBSPACES, AND LIOUVILLE’S THEOREM 3 Definition 1.3. Let V be a symplectic vector space. A subspace E V is called ⊆ isotropic if E E • ⊆ ⊥ Lagrangian if E = E • ⊥ symplectic if ω =0 • |E We will now prove the following theorem that allows gives us an isomorphism between any sym- 2n plectic vector space and the canonical symplectic space (R ,ω0). We will make use of the following propositions. The proofs of these propositions are straightforward, and they are left to the reader as an exercise (see [L]). Proposition 1.1. dim V =dimE +dimE⊥ Proposition 1.2. (E⊥)⊥ = E Proposition 1.3. Let E V . The following conditions are equivalent: ⊆ (1) E is symplectic (2) E⊥ is symplectic (3) E E = 0 ∩ ⊥ { } (4) V = E E ⊕ ⊥ Theorem 1.1. Given a symplectic vector space (V,ω), there exists a basis e1,...,en,f1,...,fn of V such that ω(ei,ej)=0 ω(fi,fj)=0 ω(ei,fj)=δij Proof. Choose any vector e V . By the non-degeneracy of ω, there exists some vector f V such ∈ ∈ that ω(e, f )=c = 0. Using the linearity of ω in the second component, we can normalize f to f = f /c such that ω(e, f) = 1. Suppose e and f were linearly dependent. We could then write f = λe,butthenω(e, f)=λω(e, e) = 0, a contradiction. Thus, e and f span a subspace E V ⊆ of dimension 2. By our construction, E is symplectic: every non-zero vector v E has a non-zero ∈ component of either e or f,soω applied to v and f or e, respectively, is non-zero and thus ω is |E non-degenerate. If dim V = 2, we are done; if not, we use Proposition 1.3 and induction to generate the desired basis on E .SinceV = E E and ω(v, w) = 0 for v E,w E , this completes ⊥ ⊕ ⊥ ∈ ∈ ⊥ the proof. We leave it to the reader to show that the desired conditions on the basis hold, and that the matrix representation of ω with respect to this basis is J. One important consequence of Theorem 1.1 is that the symplectic vector spaces have even dimension. [HZ] 4CONNERJAGER 2. Lagrangian Subspaces The goal of this section is to highlight the importance of Lagrangian subspaces. As we will see, we can think of the configuration space as an embedded subspace of the symplectic phase space. This embedded space is Lagrangian. The following theorem, which shows how we can generate a symplectic vector space from a given vector space, proves this point. Theorem 2.1. For any vector space V with dual V , the vector space V V is symplectic, with ∗ ⊕ ∗ symplectic form ω((v ,f ), (v ,f )) = f (v ) f (v ) 1 1 2 2 2 1 − 1 2 Proof. It suffices to show that ω is non-degenerate and skew-symmetric since the bilinearity is inherited from the linearity of f ,f . Choose any non-zero (v, f) V V .Ifv = 0, then there 1 2 ∈ ⊕ ∗ must be some linear functional g V such that g(v) = 0. The vector (0,g) V V satisfies the ∈ ∗ ∈ ⊕ ∗ non-degeneracy condition: ω((v, f), (g, 0)) = g(f) f(0) = g(f) =0 − If v = 0, then f =0 = f(w) = 0 for some w V . The vector (w, 0) V V satisfies the ⇒ ∈ ∈ ⊕ ∗ non-degeneracy condition: ω((v, f), (w, 0)) = 0 f(w)= f(w) =0 − − Finally, we note that ω((v ,f ), (v ,f )) = f (v ) f (v )= ω((v ,f ), (v ,f )) 2 2 1 1 1 2 − 2 1 − 1 1 2 2 Corollary 2.1. E = V 0 and E =0 V are Lagrangian subspaces of the symplectic vector ⊕ n∗ n ⊕ ∗ space V V = E E . ⊕ ∗ ⊕ Proof. ( ) Let ω be the symplectic form on V V as defined above. Choose any v V . ⊆ ⊕ ∗ ∈ ω((v, 0), (v, 0)) = 0(v) 0(v)=0 = (v, 0) E⊥ − ⇒ ∈ ( ) Let (w, f) E . For any v V , ⊇ ∈ ⊥ ∈ ω((w, f), (v, 0)) = 0(w) f(v)=0 = f(v)=0 v V = f =0 = (w, f) E − ⇒ ∀ ∈ ⇒ ⇒ ∈ Thus, E = E⊥. A similar argument shows E is also Lagrangian. Corollary 2.1 tells us that not only can we generate a symplectic vector space from a generic vector space, but the generating vector space will be embedded in the symplectic space as a Lagrangian subspace. [L] We will now prove the converse. Theorem 2.2. Any symplectic vector space (V,ω) contains a Lagrangian subspace E V such ⊆ that V = E E .
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