#### Problem

Let $Q$ be a finite-dimensional manifold. Show that the canonical one form $\Theta$ on $T^*Q$, defined such that

(1)for $\beta \in T^*Q$ and $v \in T_\beta T^*Q$, satisfies $\alpha^* \Theta = \alpha$ for any one form $\alpha$ on $T^* Q$.

#### Solution

We will think of $\alpha$ as a map from $Q$ to $T^*Q$. In coordinates

(2)$\pi_Q$ is the canonical projection from $T^*Q$ to $Q$. In coordinates, it looks like

(3)The following theorem will be useful

##### Claim

For all $v \in T_qQ$

(4)##### Proof

I will use the above coordinate expessions. Consider a curve $c$ on $Q$ such that $c(0) = q$ and $c'(0)=v$. Because $\alpha(c(t)) = (c(t), p(c(t))$,

in coordinates

Evaluating $T_{\alpha(q)} \pi_Q \cdot T_q \alpha \cdot v$ in coordinates

(6)We can use the above claim to prove the final result. Consider pairing the 1-form $\alpha^*\Theta(q)$ with an arbitray vector in $T_qQ$

(7)Because we made no assumptions about $q$ or $v$, we can say $\alpha^*\Theta = \alpha$