Identity for projection and inclusion of the canonical one-form on the cotangent bundle

Question

I'm currently reading through Introduction to Mechanics and Symmetry by Marsden and Ratiu, specifically the section on Cotangent bundles. I'm trying to do the following exercise:

Let $N$ be a submanifold of $M$ and denote by $\Theta_N$ and $\Theta_M$ the canonical one-forms on the cotangent bundles $\pi_N: T^*N\to N$ and $\pi_M: T^*M\to M$, respectively. Let $\pi: (T^*M)|N\to T^*N$ be the projection defined by $\pi(\alpha_n)=\alpha_n|T_nN$, where $n\in N$ and $\alpha\in T^*_nM$. Show that $\pi^*\Theta_N=i^*\Theta_M$ where $i:(T^*M)|N\to T^*M$ is the inclusion.

I believe we have $\pi^*\Theta_N(v)=\Theta_N(\pi_* v)$. I believe that $\pi_* v$ just turns $v\in T_\alpha(T^*M)$ into an element of $T_\alpha(T^*N)$ by removing whatever part of it was in $T_nM$. Is it true that $i_* v$ does the same thing, i.e. it takes a vector in $T_\alpha(T^*M)$ and turns it into a vector in $T_\alpha(T^*N)$? If not, how would I go about showing the identity in question?

Answer 1

Firstly, to answer your questions:

I believe that $\pi_*v$ just turns $v\in T_\alpha(T^*M)$ into an element of $T_\alpha(T^*N)$ by removing whatever part of it was in $T_nM$.

I don't really know what this means - there's no `part' of $\pi_*v$ in $T_nM$.

Is it true that $i_*v$ does the same thing, i.e. it takes a vector in $T_\alpha(T^*M)$ and turns it into a vector in $T_\alpha(T^*N)$?

Not quite. In Marsden and Ratiu's notation, the vector would be $T_\alpha i(v)$ ($i_*$ is reserved for vector fields). Then we have $$ i:(T^*M)\vert N\to T^*M \implies T_\alpha i :T_\alpha ((T^*M)\vert N)\to T_{i(\alpha)}(T^*M). $$

The way to prove this result is to use the definitions of the canonical one-forms, along with the identity $$ \pi_N\circ \pi = \pi_M\circ i $$ (both sides map $\alpha_n\in (T^*M)\vert N$ to $n\in N$). Then for any $\alpha \in (T^*M)\vert N$ and $v_\alpha \in T_\alpha((T^*M)\vert N)$ \begin{align*} (\pi^*\Theta_N)_\alpha(v_\alpha) &= (\Theta_N)_{\pi(\alpha)}(T_\alpha\pi(v_\alpha)) \\ &= \pi(\alpha)(T_{\pi(\alpha)}\pi_N(T_\alpha\pi(v_\alpha))) \qquad\text{by the definition of $\Theta_N$}\\ &= (\alpha\vert TN)(T_\alpha(\pi_N\circ \pi)(v_\alpha)) \qquad\text{by the chain rule}\\ &= i(\alpha)(T_\alpha(\pi_M\circ i)(v_\alpha)) \qquad\text{by the above identity}\\ &= i(\alpha)(T_{i(\alpha)}\pi_M(T_\alpha i(v_\alpha))) \qquad \text{by the chain rule}\\ &= (\Theta_M)_{i(\alpha)}(T_\alpha i(v_\alpha)) \qquad\text{by the definition of $\Theta_M$}\\ &= (i^*\Theta_M)_\alpha(v_\alpha). \end{align*} Hence $\pi^*\Theta_N = i^*\Theta_M$.