Homework 3 Problem 3

#### Problem

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

(1)
\begin{align} \Theta (\beta) \cdot v = \langle \beta, (T \pi_Q) v \rangle \end{align}

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)
\begin{align} \alpha: Q \rightarrow T^*Q : q \mapsto (q,p(q)) \end{align}

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

(3)
\begin{align} \pi_Q: T^*Q \rightarrow Q : (q,p) \mapsto q \end{align}

The following theorem will be useful

##### Claim

For all $v \in T_qQ$

(4)
\begin{align} T_{\alpha(q)} \pi_Q \cdot T_q \alpha \cdot v = v \end{align}
##### 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

(5)
\begin{align} \left. \frac{d}{dt}\right|_{t=0} \alpha(c(t)) = (v,\dot{p}(q)}) \end{align}

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

(6)
\begin{align} T_{\alpha(q)} \pi_Q \cdot T_q \alpha \cdot v = T_{\alpha(q)} \pi_Q \cdot (v,\dot{p}(q)}) = v \end{align}

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)
\begin{align} \langle (\alpha^*\Theta)(q), v \rangle = \langle \Theta(\alpha(q)) , T_q \alpha \cdot v \rangle = \langle \alpha(q) , T_{\alpha(q)} \pi_Q \cdot T_q \alpha \cdot v \rangle = \langle \alpha(q) , v \rangle \end{align}

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

page revision: 1, last edited: 16 Apr 2007 14:39