Let $N$ be a Riemannian manifold, and $\widetilde{N}\subset N$ a closed submanifold. If we look at the total spaces of the tangent bundles, we get that $T\widetilde{N}$ is a submanifold of $TN$.


If we now use the metric to identify $\widetilde{M}=T^\*\widetilde{N}$ with $T\widetilde{N}$, and $M=T^\*N$ with $TN$, this gives an embedding of $\widetilde{M}$ into $M$.


Both of $\widetilde{M}$ and $M$ are of course symplectic manifolds, with canonical symplectic forms. Is there a coordinate-free way of seeing that this is a symplectic embedding, i.e. that the canonical symplectic form on $M$ restricts to the canonical symplectic form on $\widetilde{M}$?



Cotangent bundles to Riemannian manifolds and submanifolds
May I please call the submanifold $M$ instead of $\tilde N$?


For a smooth manifold $N$, $T^\ast N$ does not just have a canonical symplectic form $\omega$. It also has a canonical $1$-form $\alpha$ such that $d\alpha=\omega$. For $w\in T^\ast N$, the element $\alpha\_w\in T^\ast\_w(T^\ast N)$ can be defined in a coordinate-free way by giving its value at $v\in T\_w(T^\ast N)$:
$$
\langle \alpha\_w,v\rangle=\langle w,\pi\_\ast v\rangle
$$
where $\pi\_\ast :T\_w(T^\ast N)\to T\_{\pi(w)} N$ is induced by the bundle projection $\pi:T^\ast N\to N$.


In the presence of a Riemannian structure on $N$ this becomes:


For $w\in T N$, the element $\alpha\_w\in T^\ast\_w(T N)$ is defined by giving its value at $v\in T\_w(T N)$:
$$
\langle \alpha\_w,v\rangle=w\cdot \pi\_\ast v
$$
Here "$\cdot$" means Riemannian inner product, and $\pi:T N\to N$ is the bundle projection.


If the $\alpha $ for $N$ restricts to the $\alpha$ for $M$ then also the $\omega$ restricts to the $\omega$. And surely this is the case; for $w\in TM$ and $v\in T\_w(TM)$ the inner product $w\cdot \pi\_\ast v$ is the same whether calculated in $M$ or in $N$.


