Problem: Let $M$ be a smooth manifold. Let $\omega$ be the canonical symplectic form on $T^{*} M$. Prove that there exists a unique $\alpha \in \Omega^{1} (T^{*} M)$ with the following properties:
(i) $d \alpha = \omega$
(ii) $(\iota_{T_x^{*} M})^{*} \alpha = 0$ for all $x \in M$, where $\iota_{T_x^{*} M}: T_x^{*} M \rightarrow T^{*} M$ is the inclusion
(iii) $\iota^{*} \alpha = 0$, where $\iota : M \rightarrow T^{*} M$ is the inclusion of $M$ as the zero section
Attempt: We will follow the notation of Cannas da Silva. We let $(x_1, \ldots, x_n)$ be coordinates on $U \subset M$. The associated cotangent coordinates on $T^{*} U$ are then $x_1, \ldots, x_n, \xi_1, \ldots, \xi_n$. The canonical symplectic form on $T^{*} M$ is then $\omega = \sum_i dx_i \wedge d \xi_i$.
I think the only option is to define $\alpha := - \sum_i \xi_i dx_i$. Then clearly we have $d \alpha = \omega$.
Now I had trouble to prove (ii) and (iii) and uniqueness.
For (iii), If we write $$ \iota (M) := M_0 = \left\{ (x, \xi) \in T^{*} M \mid \xi_x = 0 \ \text{in} \ T_x^{*} M \right\} $$ then it is clear that $\alpha = - \sum_i \xi_i dx_i$ vanishes on $M_0 \cap T^{*} U$. So that means that $\iota^{*} (\alpha) = 0$. Indeed, let $x \in M$ and let $v_x \in T_x M$. Then $$ (\iota^{*} \alpha)_x (v_x) = \alpha_{\iota(x)} ( (\iota_{*})_x v_x) = 0 $$ because $\iota(x) = (x,0)$, i.e. $\iota : M \rightarrow T^{*} (M)$ is the inclusion of $M$ as the zero section. Is this reasoning correct?
For (ii), I have no idea. Also, I did not know how to show uniqueness.
Help is appreciated.
There is a more general result that will also answer the uniqueness question. Following Lee, p. 570, we take smooth local coordinates $(x^i)$ for $M$. Now note that a point $(q,\varphi)\in T^*M$ can be expressed in (local) coordinates on the cotangent bundle as $(x^1,\cdots, x^n,\xi_1,\cdots , \xi_n)$ where $\varphi=\sum \xi_idx^i.$
Now, suppose $\sigma:M\to T^*M:\ (x^i)\mapsto (x^i,\sigma_i(x))$ in these coordinates. This means in particular, that $x^i\circ \sigma(x)=x^i$ and $\xi_i\circ \sigma(x)=\sigma_i(x)$
Putting all this together, we get
$\sigma(x)^*(\alpha)=\sigma(x)^*\left(\sum \xi_i dx^i\right)=\sum (\xi_i\circ \sigma(x)) d(x^i\circ \sigma(x) )=\sum \sigma_i(x) dx^i=\sigma(x)$
This is true for all $q\in M$ so we conclude that $\sigma^*(\alpha)=\sigma.$ And since this is true for all $\sigma,\ \alpha $ is unique.