Weakly Hamiltonian Group Actions-Classical and Relativistic Mechanics-Lecture Handout, Exercises of Classical and Relativistic Mechanics

Download Weakly Hamiltonian Group Actions-Classical and Relativistic Mechanics-Lecture Handout and more Exercises Classical and Relativistic Mechanics in PDF only on Docsity! This is legitimate if G is a group of matrices.  So γ[r, s] = 0. But {γ(r), γ(s)} 6= 0, since γ(r) = p1 (momentum generates translations), and γ(s) = mq1 (mass time position generates boosts), so {γ(r), γ(s)} = n∑ i=1 ∂ ∂pi p1 ∂ ∂qi mq1 − ∂ ∂qi p1 ∂ ∂pi mq1 = m Here m ∈ C∞(X) is really the constant function equal to m at all points of X . But what vector field on X does this observable generate? What is vm = {m, ·}? It is zero! It generates this flow: φ:R×X → X (t, x) 7→ x All constant functions, or indeed, all locally constant functions f on any Poisson manifold give vf = 0. picture of phase space with two connected components The problem is that this diagram: g γ ""E EE EE EE EE α // Vect(X) C∞(X) β 99ssssssssss β is not 1-1; it sends all locally constant functions to 0, so we can have βγ[x, y] = [βγ(x), βγ(y)] even though γ[x, y] 6= [γ(x), γ(y)]. This also means that different choices of γ can make this diagram commute. Definition 2 If G is a Lie group acting on a Poisson manifold X: A:G×X → X we say A is weakly Hamiltonian if there exists a linear map γ: g→ C∞(X) such that α = βγ with γ not necessarily a Lie algebra homomorphism. 2 docsity.com