mma
May19-11, 02:34 AM
The canonical symplectic form on T^*M is the exterior derivative of the tautological 1-form:
\omega=d\alpha
where \alpha_p(X):=p(d\pi(X)) is the tautological 1-form.
Let Y \in T_pT^*M a vertical vector, that is d\pi(Y)=0.
It's trivial to prove using canonical coordinates that for all X \in T_pT^*M
\omega(X,Y) = y(d\pi(X))
where y \in T_{\pi(p)}^*M such that for any differentiable function f: T^*M \to \mathbb R Y(f)=\left. \frac{df(p+ty)}{dt}\right|_{t=0}.
But how can it be proved in a coordinate-free manner?
\omega=d\alpha
where \alpha_p(X):=p(d\pi(X)) is the tautological 1-form.
Let Y \in T_pT^*M a vertical vector, that is d\pi(Y)=0.
It's trivial to prove using canonical coordinates that for all X \in T_pT^*M
\omega(X,Y) = y(d\pi(X))
where y \in T_{\pi(p)}^*M such that for any differentiable function f: T^*M \to \mathbb R Y(f)=\left. \frac{df(p+ty)}{dt}\right|_{t=0}.
But how can it be proved in a coordinate-free manner?