Contraction of the canonical symplectic form by vertical vectors

[mma]

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?

 

[mma]

Erratum.

The canonical symplectic form on T^*M is the exterior derivative of the tautological 1-form:

\omega=d\alpha​

should be

The canonical symplectic form on T^*M is the negative of the exterior derivative of the tautological 1-form:

\omega=-d\alpha​

Any idea?

 

[quasar987]

Your question is giving me a headache, but have you looked in the book by Anna Canna Da Silva (first or second chapter I think)? There, she proves many properties of the tautological 1-form in coordinate free form.

 

[mma]

Your question is giving me a headache, but have you looked in the book by Anna Canna Da Silva (first or second chapter I think)? There, she proves many properties of the tautological 1-form in coordinate free form.

Yes, I know (and like) da Silva's book, but I didn't find this in it. I give a better chance to the book of Libermann and Marle. Thanks anyway.

 

[quasar987]

I didn't know about this book until now. Thanks.