TITLE: Nash isometric embedding theorem with keeping the symplectic structures of our ambient spaces
QUESTION [5 upvotes]: I apologize in advance if this question has an obvious answer.
Let $(M,g)$ be a Riemannian manifold.
Then the tangent bundle $TM$ carries a natural symplectic structure $\omega_g$. In fact  $\omega_g$ is  the  pull  back of the  canonical  symplectic  structure of the  cotangent bundle  via  the  obvious  diffeomorphism  between $TM$  and  $T^* M$ which is  defined by the inner product $g$.
The standard structure of $T\mathbb{R}^n=\mathbb{R}^n \times \mathbb{R}^n$ is denoted by $\omega$.

For every Riemannian manifold $(M,g)$, is there an isometric embedding $j$ of $M$ in some $\mathbb{R}^n$ such that $j^*(\omega)=\omega_g?$

REPLY [8 votes]: This is always the case (using "naturality" as Paul Bryan suggested in the comments).
Let $f: M \to N$ be a smooth map between Riemannian manifolds $(M,g)$ and $(N, h)$. Let $g^\flat: TM \to T^*M$ denote the musical isomorphism induced by $g$ (and similarly $h^\flat$).
Now, $f$ is an isometry if and only if $h_{f(m)} (T_m f (X), T_m f (\cdot)) = g_m (X, \cdot)$ holds for all $X \in T_m M$. In other words, $g^\flat (X) = f^* (h^\flat (T_m f(X)))$ or shorter $$g^\flat = T^* f \circ h^\flat \circ Tf,$$ where $T^*f: T^* N \to T^* M$ is the cotangent lift of $f$ (often called a point transformation).
Using the fact that cotangent lifts are always symplectic maps, we have
$$\omega_g = (g^\flat)^* \omega_M = (Tf)^* (h^\flat)^* \omega_N = (Tf)^* \omega_h,$$
where $\omega_M$ and $\omega_N$ denote the canonical symplectic forms on $T^* M$ and $T^* N$, respectively.