# Canonical transformations on cotangent bundle

We got as part of homework for Mechanics this exercise. As the course was a little chaotic I barely got grip of some notions and I feel a bit lost so any solution would be welcomed.

Let V be a finite dimensional manifold with cotangent bundle T*(V): By α denote the canonical 1-form on T*(V), while ω := dα. Given a diffeomorphism Φ : T*(V) -> T*(V); show that the following two statements are equivalent:

1. Φ is canonical, i.e.Φ*ω = ω
2. Locally there exists a function S : T*(V) ->\mathbb{R}; such that dS = Φ*α - α

$$\Phi^*\omega=\omega$$ is equivalent to $$\Phi^*d\alpha=d\Phi^*\alpha=d\alpha$$, this is equivalent to $$d(\Phi^*\alpha-\alpha)=0$$. Let $$U$$ be a contractible open neighboorhood diffeomorphic to a ball, the Poincare lemma implies that the fact that the restriction of $$\Phi^*\alpha-\alpha$$ on a contractible open subset $$U$$ is closed is equivalent to the existence of a function $$S$$ defined on $$U$$ such that $$dS=\Phi^*\alpha-\alpha$$.
Let $$U$$ be an open subset diffeomorphic to an open ball, suppose that there exists a function $$S$$ defined on $$U$$ such that $$dS=\Phi^*\alpha-\alpha$$, $$0=d(dS)=d(\Phi^*\alpha-\alpha)$$.
This implies that $$d(\Phi^*\alpha-\alpha)=0$$ and $$d(\Phi^*\alpha)=\Phi^*d\alpha=d\alpha$$, since $$\omega=d\alpha$$, we deduce that $$\Phi^*\omega=\omega$$.