3
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ This tells me what I already know. I don't understand why $\mathbb{R}^{2n}\rightarrow\mathbb{R}^{2n}$ happens and how to determine the matrix of this transformation. $\endgroup$ – Ryan Unger Feb 18 '15 at 13:57
  • $\begingroup$ the matrix is basically written in words in my answer: it is just the identity matrix, of you are implying the matrix of the symplectic form, this is just the standard symplectic form on $\mathbb R^2$ tensor with the $n\times n$ identity matrix. $\endgroup$ – Phoenix87 Feb 18 '15 at 14:05
  • $\begingroup$ He gives the answer in the $\mathbf{p},\mathbf{q}$ basis: \begin{pmatrix} 0 & \mathbb{1} \\ -\mathbb{1} & 0 \end{pmatrix} $\endgroup$ – Ryan Unger Feb 18 '15 at 14:11
  • $\begingroup$ I see how this is symplectic, but what exactly does the transformation mean? $\endgroup$ – Ryan Unger Feb 18 '15 at 14:13
  • 1
    $\begingroup$ it's just an isomorphism between 1-forms and vector fields given by a non-degenerate skew-symmetric form. Recall that there is no natural isomorphism between a vector space and its dual, so one usually needs extra structure to define one in a "canonical way". On a symplectic manifold the canonical way is to use the symplectic form (if the manifold happens to be Riemannian as well, or just Riemannian, another possible "canonical" choice is the musical isomorphism). $\endgroup$ – Phoenix87 Feb 18 '15 at 14:16
0
$\begingroup$

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 $\omega = d q^i \wedge d p_i$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.