Symplectic structure and isomorphisms

Question

In his book Mathematical Methods of Classical Mechanics, V.I. Arnold writes

To each vector $\xi$, tangent to a symplectic manifold $(M^{2n},\omega^2)$ at the point $\mathbf{x}$, we associate a 1-form $\omega^1_\xi$ on $TM_\mathbf{x}$ by the formula $$\omega^1_\xi(\boldsymbol{\eta})=\omega^2(\boldsymbol{\eta},\xi)\quad\forall\boldsymbol{\eta}\in TM_\mathbf{x}$$

I see how $\omega^2$ furnishes an isomorphism $\xi\rightarrow \omega^1_\xi$. But then Arnold has the example

In $\mathbb{R}^{2n}=\{(\mathbf{p},\mathbf{q})\}$ we will identify vectors and 1-forms using the Euclidean structure $(\mathbf{x},\mathbf{x})=\mathbf{p}^2+\mathbf{q}^2$. Then the correspondence $\xi\rightarrow\omega^1_\xi$ determines a transformation $\mathbb{R}^{2n}\rightarrow \mathbb{R}^{2n}$.

By "Euclidean structure" I presume he is talking about the Euclidean metric. But I don't see how this isomorphism induces the transformation $\mathbb{R}^{2n}\rightarrow \mathbb{R}^{2n}$ or furthermore how to determine the matrix of this transformation.

And help would be greatly appreciated.

Answer 1

Same as with the symplectic form: $\omega(v) = (u_\omega,v)$ defines the isomorphism between 1-forms and vector fields. When the metric is Euclidean the dual basis to an orthonormal basis corresponds to the basis itself.

Answer 2

The linear transformation is the following composition of linear maps:

• Go from $R^n$ to $T_m M$ using the natural identification
• Go from $T_m M$ to $T^*_m M$ with the symplectic form
• Go from $T^*_m M$ to $T_m M$ using the inverse isomorphism given by the metric
• Go again back to $R^n$

(here $M = R^n$ and $m = (q,p)$ By the way, this transformation of course just coincides with the usual symplectic matrix if the symplectic form is the standard one $\omega = d q^i \wedge d p_i$.