# Get Exterior differential systems and Euler-Lagrange PDEs PDF By Bryant R., Griffiths P., Grossman D.

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Alternatively, one can simply pull everything up to a double cover of M on which I has a global generator, and little will be lost. 3. Step 1: Definition of the map η. The map in question is given by η(v) = v Π for v ∈ gΠ ⊂ V(M ). Note that locally v Π = (v θ)Ψ − θ ∧ (v Furthermore, the condition Lv Π = 0 gives 0=v so that v dΠ + d(v Ψ), so that v Π) = d(v Π lies in EΛ . Π), Π is closed, and gives a well-defined class η(v) ∈ H n(EΛ ). Step 2: η is injective. Write θ)Ψ − θ ∧ (v η(v) = (v Ψ). Suppose that this n-form is cohomologous to zero in H n(EΛ ); that is, (v θ)Ψ − θ ∧ (v Ψ) = = d(θ ∧ α + dθ ∧ β) −θ ∧ dα + dθ ∧ (α + dβ).

Note that the prescription is especially simple when v ∈ gΛ ⊆ g[Λ] , for then we can take γ = 0. Example. Let Ln+1 = {(t, y1 , . . , yn )} ∼ = Rn+1 be Minkowski space, and 2n+3 1 n+1 let M = J (L , R) be the standard contact manifold, with coordinates (t, yi , z, pa) (where 0 ≤ a ≤ n), θ = dz − p0 dt − pi dyi . For a Lagrangian, take Λ= 1 2 2 ||p|| + F (z) dt ∧ dy for some “potential” function F (z), where dy = dy 1 ∧ · · · ∧ dyn and ||p||2 = −p20 + p2i is the Lorentz-signature norm. The local symmetry group of this functional is generated by two subgroups, the translations in Ln+1 and the linear isometries SOo (1, n); as we shall see in Chapter 3, for certain F (z) the symmetry group of the associated Poincar´e-Cartan form is strictly larger.

This says that ϕ ∼ 0 in H n(I), and our proof is complete. It is important in practice to have a local formula for a representative in Ωn−1(M ), closed modulo EΛ , for the proper conservation law η(v). 13) and also, for a given v ∈ g[Λ] , Lv Λ ≡ dγ (mod I). 15) 20 ´ CHAPTER 1. LAGRANGIANS AND POINCARE-CARTAN FORMS is satisfactory. First, compute dϕ = ≡ ≡ ≡ (−Lv Λ + v dΛ) + d((v θ)β) + dγ v (Π + d(θ ∧ β)) + d((v θ)β) (mod I) η(v) + Lv (θ ∧ β) (mod I) η(v) (mod I). Now we have dϕ = η(v) + Ξ for some closed Ξ ∈ I n.