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.