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3 (Hamiltonian vector field). Suppose (M, ω) is a symplectic manifold, and suppose H be a function H : M → R. Then the Hamiltonian vector field induced by H is the unique (see next paragraph) vector field XH ∈ X (M ) determined by the condition dH = ι XH ω. In the above, ι is the contraction mapping ι X : Ωr M → Ωr−1 M defined by (ιX ω)(·) = ω(X, ·). To see that XH is well defined, let us consider the mapping X → ω(X, ·). By non-degeneracy, it is injective, and by the rank-nullity theorem, it is surjective, so the Hamiltonian vector field X H is uniquely determined.

Then for all x ∈ U , t ∈ I, we have (Φ∗t ω)x = ωx , where Φt = Φ(t, ·). Proof. As Φ0 = idM , we know that the relation holds, when t = 0. Therefore, let us fix x ∈ U , a, b ∈ Tx M , and consider the function r(t) = (Φ∗t ω)x (a, b) with t ∈ I. Then r (t) = = = d r(s + t) ds s=0 d ∗ (Φs ω)y (a , b ) ds LXH ω y (a , b ), s=0 where y = Φt (x), a = (DΦt )(a), b = (DΦt )(b), and the last line is the definition of the Lie derivative. Using Cartan’s formula, L X = ιX ◦ d + d ◦ ιX , we have LXH ω = ιXH dω + dιXH ω = ιXH 0 + ddH = 0, so r (t) = 0, and r(t) = r(0) = ωx (a, b).

The cotangent bundle T ∗ M \ {0} of manifold M is a symplectic manifold with a symplectic form ω given by ω = dθ = dxi ∧ dξi , where θ is the Poincar´e 1-form θ ∈ Ω1 T ∗ M \ {0} . ∂ i ∂ i ∂ Proof. It is clear that ω is closed. If X = a i ∂x i + b ∂ξ , and Y = v ∂xi + i wi ∂ξ∂ i , then ω(X, Y ) = a · w − b · v. By setting v = w we obtain a = b, and by setting w = 0 we obtain a = b = 0. The next example shows that we can always formulate Hamilton’s equations on the cotangent bundle. This motivates the name for X H .