TITLE: Tautological 1-form on the cotangent bundle
QUESTION [28 upvotes]: 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.

REPLY [2 votes]: EASY METHOD:
Suppose that $(p,\omega_p)\in Q \equiv T^*(M)$, where $p \in M, \omega_p \in T_p^*(M)$, represents a point in the cotangent bundle $Q$ whose fibres are the cotangent linear spaces at each point in the base manifold. We have the canonical fibre bundle projection,
\begin{align*}
      \pi: \quad Q &\to M \\
                 (p,\omega_p) &\mapsto p \
\end{align*}
which induces the pullback of 1-forms,
\begin{align*}
      \pi^*: \quad \Omega (M) &\to \Omega (Q) \\
                 \omega &\mapsto \pi^*(\omega)\equiv \theta. \
\end{align*}
In local coordinates, if we write $ \omega = \sum_i \alpha_i \ \text{d}x^i $, then it is straightforward to compute the local representation of the pullback:
$$ \theta = \pi^*(\sum_i \alpha_i \ \text{d}x^i) = \sum_i (\alpha_i\circ\pi) \ \pi^*(\text{d}x^i)\ ``=" \sum_i \alpha_i \ \text{d}x^i$$
Note the quotation marks I used to emphasize that one should be careful about abuse of notation and use them smartly for convenience knowing that it should not be taken too literally.
ALTERNATIVE METHOD:
For a vector field  $X \in \mathfrak{X}(Q)$, one could locally decompose it into a component parallel to the fibre $T_p^*(M)$ and another parallel to the base $M$ for each $p \in M$ :
$$
X = \sum_j (X_{\parallel}^j\circ\pi)\frac{\partial}{\partial x^j} + \sum_i X_{\perp}^i \frac{\partial}{\partial \alpha^i}
$$ 
where we have again considered $ \omega = \sum_i \alpha_i \ \text{d}x^i $. By this decomposition, we mean that
$$ \pi_*(X)= \sum_j X_{\parallel}^j \ \frac{\partial}{\partial x^j}$$
And therefore, we have that
\begin{align*} 
\theta_{(p,\omega_p)}(X) = \pi_{(p,\omega_p)}^*(\omega_p)(X) &= \omega_p (\pi_{*_{(p,\omega_p)}}(X)) \\
&= \sum_i \alpha_i(p)\ \text{d}x^i (\sum_j X_{\parallel}^j(p) \ \frac{\partial}{\partial x^j}) \\
&= \sum_i (\alpha_i\circ\pi)(p,\omega_p)(X_{\parallel}^i\circ\pi)(p,\omega_p)\
\end{align*}
Thus: $$ \theta (X) = \sum_i (\alpha_i\circ\pi) (X_{\parallel}^i\circ\pi) = \sum_i (\alpha_i\circ\pi) \ \pi^*(\text{d}x^i)(X) $$
which is the same result that we convinced ourselves of before.