There is apparently a naturally defined one-form on the cotangent bundle of a smooth manifold $M$:
We have the cotangent bundle $\pi:T^*M \rightarrow M$; taking its derivative gives $d \pi:TT^*M \rightarrow TM$.
Then taking its pullback, we have: $d \pi^*:T^*M \rightarrow T^* T^* M$.
I fail to see how this is a one form on the cotangent space; which ought to be of the form '$\alpha:T(T^*M) \rightarrow R$'.
Where am I going wrong?
edit
After Martinis answer, I see that I missed the obvious; that is the above is just:
$d \pi^*:(T^*M) \rightarrow T^* (T^* M)$.
Which is just a section of the cotangent bundle on $T^*M$, and therefore a 1-form.
Let $p \in T^*M$ and $x = \pi(p)$. Then $d\pi^*_x \colon T^*_xM \to T^*_pT^*M$. Now let $\alpha_p = d\pi^*_xp$, $\alpha$ is a map $M \to T^*(T^*M)$, that is a one-form on $T^*M$.