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I have a question about the action-angle theorem on p. 283 in Arnold's textbook on classical mechanics.(I added the link to this book in the last part of this question)

If you don't have the book or need information, then please leave a comment and I will try my best.

$\textbf{Theorem:}$ The theorem says that the transformation $(p,q) \mapsto (I,\phi)$ is symplectic, where $I$ are the action variables and $\phi$ the action angles.

He says that he will only sketch the proof which might be the source of confusion.

I will state the proof up to the point that causes the problems and explain what exactly causes the troubles.

$\textbf{Proof: }$ So first we consider the $1$-form $pdq$ on the manifold $M_f:=\{(p_1,..,p_n,q_1,..,q_n)=:(p,q) \in M; F_1(p,q)=f_1,...,F_n(p,q)=f_n\}$ where $F_1,..,F_n$ have linearly independent derivatives and $M$ is a symplectic manifold of dimension $2n$.

addendum: It can be shown that $\omega|_{M_f} = 0$ and he also assumed that $\frac{\partial I}{\partial f}|_{M_f}$ is invertible in a previous proof.

Therefore, $S(x)= \int_{x_0}^{x} pdq|_{M_f}$ is invariant under deformations of paths $(x_0 \rightarrow x)$ (by Stokes' theorem).

addendum: It can be shown that if $M_f$ is connected and compact it is diffeomorphic to a torus.

Still, $S$ is multiple-valued as when we integrate around one circle $\gamma_i$ of this torus, we get a period $\Delta_i (S)= \int_{\gamma_i} dS = 2 \pi I_i. $

Now he continues by saying: Let $x_0$ be a point on $M_f$, in a neighbourhood of which the $n$ variables $q$ are coordinates of $M_f$ such that the submanifold $M_f \subset \mathbb{R}^{2n}$ is given by $n$-equations of the form $p= p(I,q)$, $q(x_0)=q.$

In a simply connected neighborhood of the point $q_0$ a single-valued function is defined

$S(I,q) = \int_{q}^{q} p(I,q) dq.$

Finally, he remarks that: It is not difficult to verify that these formulas actually give a canonical transformation, not only in a neighborhood of the point under consideration but also "in the large" in a neighborhood of $M_f$.

$\textbf{Question:}$ Now my question is: Why is it possible to take $(I,q)$ as coordinates, i.e. what is the argument that explains why the coordinates $q$ can be taken as even global coordinates (as Arnold actually says that it is easy to conclude that this also holds true in the "large " around $M_f$) in a nbh of $M_f$?

EDIT: For those of you who don't have the book, you can download the pdf from this link and go to page 300 (according to the pdf).click me.

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  • $\begingroup$ @ChrisGerig Could you elaborate? Just to clarify this, the question is not why $M_f$ is a manifold (reg. value theorem), but why we can take $(I,q)$ as (apparently even global) coordinates in the nbh. of $M_f$? $\endgroup$ Commented Sep 12, 2015 at 20:08
  • $\begingroup$ The Inverse Function Theorem will give coordinates $(0,q)$ on $M_f\subset \mathbb{R}^{2n}$. We also know that $(I,\phi)$ are the coordinates for the neighborhood of $M_f$ (in $\mathbb{R}^{2n}$). That's what Arnold is using. $\endgroup$ Commented Sep 12, 2015 at 22:43
  • $\begingroup$ @ChrisGerig mhmm, I don't think I fully understand. Do you mind turning this into a complete answer and explain what you exactly use there?- I agree that we know that $(I,\phi)$ are coordinates, as for fixed $I$ we define $M_f$ and $M_f$ is nothing but a torus, which we can then parametrize via the angle variables. Thus, if we have another type of coordinates on $M_f$ (for instance your $q$ variables, then this would be okay, but I currently don't see what you mean). $\endgroup$ Commented Sep 12, 2015 at 23:02

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