Tautological 1-form on the cotangent bundle

I'm trying to understand a little bit about symplectic geometry, in particular the tautological 1-form on the cotangent bundle. I'm following Ana Canas Da Silva's notes.

On page 10 she describes the coordinate free definitions and gives an exercise to find the expression in the local coordinates $\sum_{i=1}^n \xi_i dx_i$. I've tried to do this exercise but can't seem to be able to do it which is really annoying since everywhere I look it is said to be trivial and as a consequence never formally proved.

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The link in this post is broken; the document in question is presumably currently at math.ist.utl.pt/~acannas/Books/lsg.pdf. – E.P. Dec 4 '13 at 3:15

1 Answer

Let $(x^i)$ be local coordinates on our base manifold $M$ and let $(x^i, \xi_j)$ be the induced coordinates on the cotangent bundle $T^* M$. Let $\pi : T^*M \to M$ be the projection $(x^i, \xi_j) \mapsto (x^i)$. It induces a $C^\infty (M)$-linear map on $1$-forms, which I will write as $\pi^* : \Omega^1 (M) \to \Omega^1 (T^* M)$. In coordinates, this sends a $1$-form $\phi = \phi_i \, \mathrm{d} x^i$ (summation convention) to $(\phi_i \circ \pi) \, \mathrm{d} x^i$. As usual this induces a $\mathbb{R}$-linear map on the fibres, namely $\pi^*_{(x, \xi)} : T^*_x M \to T^*_{(x, \xi)} (T^* M)$, sending the covector $p$ to the covector $(p, 0)$. (We must be careful and distinguish between covectors and $1$-forms here, to avoid confusion.)

The tautological $1$-form on $T^* M$ is defined to be $\pi^*_{(x, \xi)} \xi$ at each point $(x, \xi)$ in $T^* M$. Why does this formula even make sense? Well, $\xi$ by definition is an element of $T_x^* M$, so it typechecks. Thus the point $(x, \xi)$ is mapped to the covector $(\xi, 0)$ in $T^*_{(x, \xi)} (T^* M)$, and so the tautological $1$-form in coordinates is given by $$\xi_i \, \mathrm{d} x^i$$ as claimed. (The coefficient of $\mathrm{d} \xi_j$ is $0$, of course.)

(Perhaps the reason no-one likes writing this out in full is because the tautological nature of the construction makes it quite confusing, unless one keeps track of the types of all the expressions involved.)

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Thanks a lot !! – Zorba le Grec Dec 6 '12 at 9:29