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 $\omega_g$.

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 $j^*(\omega)=\omega_g?$

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 $j^*(\omega)=\omega_g?$

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$\omega_g$.

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 $j^*(\omega)=\omega_g)$

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  $\omega_g$.

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 $j^*(\omega)=\omega_g?$