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May 12, 2015

Action-Angle Variables

This semester, I taught the Graduate Mechanics course. As is often the case, teaching a subject leads you to rethink that you thought you understood, sometimes with surprising results.

The subject for today’s homily is Action-Angle variables.

Let (,ω)(\mathcal{M},\omega) be a 2n2n-dimensional symplectic manifold. Let us posit that \mathcal{M} had a foliation by nn-dimensional Lagrangian tori (a torus, TMT\subset M, is Lagrangian if ω| T=0\omega|_T =0). Removing a subset, SS\subset \mathcal{M}, of codimension codim(S)2codim(S)\geq 2, where the leaves are singular, we can assume that all of the leaves on =\S\mathcal{M}'=\mathcal{M}\backslash S are smooth tori of dimension nn.

The objective is to construct coordinates φ i,K i\varphi^i, K_i with the following properties.

  1. The φ i\varphi^i restrict to angular coordinates on the tori. In particular φ i\varphi^i shifts by 2π2\pi when you go around the corresponding cycle on TT.
  2. The K iK_i are globally-defined functions on \mathcal{M} which are constant on each torus.
  3. The symplectic form ω=dK idφ i\omega= d K_i\wedge d \varphi^i.

From 1, it’s clear that it’s more convenient to work with the 1-forms dφ id\varphi^i, which are single-valued (and closed, but not necessarily exact), rather than with the φ i\varphi^i themselves. In 2, it’s rather important that the K iK_i are really globally-defined. In particular, an integrable Hamiltonian is a function H(K)H(K). The K iK_i are the nn conserved quantities which make the Hamiltonian integrable.

Obviously, a given foliation is compatible with infinitely many “integrable Hamiltonians,” so the existence of a foliation is the more fundamental concept.

All of this is totally standard.

What never really occurred to me is that the standard construction of action-angle variables turns out to be very closely wedded to the particular case of a cotangent bundle, =T *M\mathcal{M}=T^*M.

As far as I can tell, action-angle variables don’t even exist for foliations of more general symplectic manifolds, \mathcal{M}.

Any contagent bundle, T *MT^*M, has a canonical 1-form, θ\theta, on it. The standard symplectic structure is ω=dθ\omega = d\theta. The construction of the action-variables requires that we choose a homology basis, γ i\gamma_i, for each torus, in a fashion that is locally-constant1, as we move between tori of the foliation. The K iK_i are then defined as
(1)K i=12π γ iθK_i = \frac{1}{2\pi}\int_{\gamma_i} \theta
Note that, because the torus is Lagrangian, the values of K iK_i are independent of the particular choices of path chosen to represent γ i\gamma_i. Having constructed the K iK_i, the closed 1-forms
(2)dφ i=i /K iωd\varphi^i = i_{\partial/\partial K_i}\omega
Great! Except that, for a general symplectic manifold, there’s no analogue of θ\theta. In particular, it’s trivial to construct examples of symplectic manifolds, foliated by Lagrangian tori, for which no choice of action variables, K iK_i, exist. As a simple example, take (,ω)=(T 4,dθ 1dθ 3+dθ 2dθ 4) (\mathcal{M},\omega)= (T^4, d\theta_1\wedge d\theta_3 +d\theta_2\wedge d\theta_4) Obviously, we can foliate this by Lagrangian tori (taking TT to be the subsets {θ 1,θ 2=const}\{\theta_1, \theta_2=\text{const}\}). But the corresponding action variables don’t exist. We’d happily choose K i=θ iK_i=\theta_i, for i=1,2i=1,2, but those aren’t single-valued functions on \mathcal{M}. You could try to use functions that are actually single-valued (e.g., K i=sin(θ i)K_i=\sin(\theta_i)), but then the corresponding 1-forms, η i\eta^i, in ω=dK iη i\omega = dK_i\wedge\eta^i, don’t have 2π×2\pi\timesintegral periods (heck, they’re not even closed!).

Surely, there’s some sort of cohomological characterization of when Action-Angle variables exist. The situation feels a lot like the characterization of when symplectomorphisms (vector fields that preserve the symplectic form) are actually Hamiltonian vector fields2.

And, even when the obstruction vanishes, how do we generalize the construction (1), (2) to more general symplectic manifolds?

Update:

Just to be clear, there are plenty of examples where you can construct action-angle variables for foliations of symplectic manifolds which are not cotangent bundles. An easy example is (,ω)=(S 2,rdrdθ(1+r 2) 2) (\mathcal{M},\omega) = \left(S^2, \frac{r dr\wedge d\theta}{{(1+r^2)}^2}\right) where (K,φ)=(12(1+r 2),θ)(K,\varphi)= \left(-\frac{1}{2(1+r^2)},\theta\right) are action-angle variables for the obvious foliation by circles. This example “works” because once you remove the singular leaves (at r=0,r=0,\infty), ω\omega becomes cohomologically trivial on \mathcal{M}' and we can then use the standard construction. [ω]=0H 2(\S,)[\omega]=0\in H^2(\mathcal{M}\backslash S,\mathbb{R}) sounds like a sufficient condition for constructing action-angle variables. But is it necessary?

1I’m pretty sure we need them to be globally-constant over \mathcal{M}'. I’ll assume there’s no obstruction to doing that.

2If you’re not familiar with that story, note that Xω=0 \mathcal{L}_X \omega = 0 is tantamount to the condition that i Xωi_X\omega is a closed 1-form. If it happens that it is an exact 1-form, i Xω=df i_X\omega = d f then X={f,}X = \{f,\cdot\} is a Hamiltonian vector field. The obstruction to writing XX as a Hamiltonian vector field is, thus, the de Rham cohomology class, [i Xω]H 1(,)[i_X\omega]\in H^1(\mathcal{M},\mathbb{R}).

In the example at hand, that’s exactly what is going on. Any single-valued function, H(θ 1,θ 2)H(\theta_1,\theta_2), is an “integrable” Hamiltonian for the above foliation. But the symmetries, X 1=θ 3X_1=\frac{\partial}{\partial\theta_3} and X 2=θ 4X_2=\frac{\partial}{\partial\theta_4} are not Hamiltonian vector fields. Hence, there are no corresponding conserved action variables.

Posted by distler at May 12, 2015 11:49 AM

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Re: Action-Angle Variables

Which textbook did you use?

Posted by: Rick on May 12, 2015 4:53 PM | Permalink | Reply to this

Textbook

José and Saletan. Not without its issues, but not bad …

Posted by: Jacques Distler on May 12, 2015 5:41 PM | Permalink | PGP Sig | Reply to this

Re: Textbook

I thought the gold standard graduate mechanics textbook was Goldstein. Is there another book you would prefer to use, either as the primary textbook or a companion textbook? Sorry for being off topic.

Posted by: Rick on May 12, 2015 6:09 PM | Permalink | Reply to this

Re: Textbook

Some nonlinear combination of José-Saletan and Arnold would be the ideal text for a graduate-level course.

Posted by: Jacques Distler on May 12, 2015 7:18 PM | Permalink | PGP Sig | Reply to this

Re: Action-Angle Variables

Maybe I am missing your point here - see the penultimate sentence.


I wouldn’t say this is a very standard definition of action-angle variables. Or at least, with that definition you won’t necessarily have them.

If you just have a symplectic manifold, all you have is Darboux’s theorem. I.e., locally near each point the manifold looks like a cotangent bundle, with the standard symplectic form.


If you have an integrable system - i.e., a map h: M -> B from M to some “base” space B, with Lagrangian fibres, then by Liouville-Arnold the fibres, if compact, are tori. Otherwise they are products of a torus and a vector space (assuming the fibres are connected.) Now,
you have AA coordinates in a neighborhood of each (smooth) fibre.
I.e., for each fibre, there is a ball downstairs, so that over its preimage you can do what you describe (i.e., locally there is a section). If the base is contractible, the K_i’s are globally defined, else they are defined on a neighborhood of a fibre (that neighborhood looks like preimage under h of a contractible open set downstairs).

Of course, even if you start with some base space like B= R^n, once you remove the singular fibres, you may get a non-trivial topology there.
So there can be monodromy issues, when comparing the angle coordinates over adjacent patches and going around a singular fibre.

On the other hand, the integrable system morphism (“hamiltonians”), defining the map h, say
h= (h_1, … h_n): M -> B= R^n is globally defined. The K_i’s are functions of the h_i’s.

There are global issues - e.g., there is an obstruction to the existence of section. You can have a look at Duistermaat, “Global action-angle coordinates” (1980), but see also Arnold’s book.

(I thought this is kind of standard in physics, e.g. Gaoitto-Moore-Neitzke mention these issues in various specific cases.)

Or maybe your question is about foliations which don’t have the structure of an integrable system? BTW, if you have a lagrangian foliation, there’s at least a connection along it.

Posted by: dalpezzo on May 13, 2015 11:38 AM | Permalink | Reply to this

Re: Action-Angle Variables

Maybe I am missing your point here

I think you may be.

I wouldn’t say this is a very standard definition of action-angle variables. Or at least, with that definition you won’t necessarily have them.

I learned this definition from Arnold’s book. So if it’s non-standard, blame him.

If you have an integrable system - i.e., a map h: M -> B from M to some “base” space B, with Lagrangian fibres, then by Liouville-Arnold the fibres, if compact, are tori.

That’s just assuming the answer I wish to derive. Perhaps I should have made some obvious remarks:

  • Given a foliation of \mathcal{M}, as above, if I can find action variables, K iK_i, then it is obvious that they endow \mathcal{M} with the structure of a fibration (K 1,,K n):B n(K_1,\dots,K_n)\colon \mathcal{M}\to B\subset\mathbb{R}^n where the fibers are Lagrangian (and, by my assumptions, compact though not necessarily connected).
  • My K iK_i are not functions of your h ih_i; they are your h ih_i. By construction, they generate translations along the fibers.

The question I was asking was:

  • Given a foliation, when does it give rise to such a fibration?

Clearly, there are examples (I gave one above) where you have a foliation, but no fibration. But it’s also clear that there are examples, which are not cotangent bundles, where you do get a fibration.

There are global issues - e.g., there is an obstruction to the existence of section. You can have a look at Duistermaat, “Global action-angle coordinates” (1980), but see also Arnold’s book.

Duistermaat’s paper looks very interesting. I’ll have to read it carefully, but I think he starts with the assumption that a fibration exists, and then examines global issues to do with the existence of sections.

(I thought this is kind of standard in physics, e.g. Gaoitto-Moore-Neitzke mention these issues in various specific cases.)

Hence (in part) my interest.

Or maybe your question is about foliations which don’t have the structure of an integrable system?

Exactly.

BTW, if you have a lagrangian foliation, there’s at least a connection along it.

Indeed.

Posted by: Jacques Distler on May 13, 2015 2:50 PM | Permalink | PGP Sig | Reply to this

Re: Action-Angle Variables

I don’t mean to be impolite, but Arnold’s conventions disagree with what you claim they are:
have a look at section Chapter 10, section 50 A (p.279) “Description of the action-angle coordinates”. The first sentence after the first equation: “We will now look at a neighborhood of the manifold M_f (a fibre!)”, tells you that your coordinates are “global” only after you restrict the base sufficiently. Next, in the last paragraph on that page he underlines that the K_i’s need not be the h_i’s, but just functions of them (the may happen to coincide, but they don’t have to). And these dependences may happen to be non-trivial and locally constant on the base.

All I am saying is that while the fibration is “globally defined”, the AA coordinates are defined over small patches in the base (and this is the case with Hitchin system, of course).

Some of the global issues - when you already have a fibration - are also discussed in Ch. IV, sect.44 of Guillemin-Sternberg (“Symplectic techniques in physics”).

The general case, when you only have a foliation, is non-trivial… I believe (in the smooth category) Ping Xu has some work on this, and also on reduction, etc. using symplectic groupoids (some of it with Weinstein).
There is a short talk on the issues with leaf spaces of a foliation, why groupoids enter, and what are the alternatives - by Moerdijk, “Models for the leaf spaces of a foliation”.

Posted by: dalpezzo on May 14, 2015 3:44 AM | Permalink | Reply to this

Re: Action-Angle Variables

The first sentence after the first equation: “We will now look at a neighborhood of the manifold Mf (a fibre!)”, tells you that your coordinates are “global” only after you restrict the base sufficiently. Next, in the last paragraph on that page he underlines that the Ki’s need not be the h_i’s, but just functions of them (the may happen to coincide, but they don’t have to).

My K iK_is are Arnold’s I iI_is, which he defines locally, in Chapter 10, section 50C (p.282). As he remarks later in the chapter, the construction of the action variables works more generally on cotangent bundles.

The angle variables, φ i\varphi^i are only defined locally over BB. For a general symplectic manifold, there is a non-vanishing Chern class (as defined by Duistermatt). But, for \mathcal{M} a cotangent bundle (or, more generally, whenever ω\omega is exact), the Chern class vanishes and the only non-triviality is the possible monodromies among the φ i\varphi^i.

In general, there is monodromy for the angle-variables, φ i\varphi^i. But there is no monodromy in the definition of the K iK_is.

The spherical pendulum is Duistermatt’s prime example (in the paper you cited) of the monodromy phenomenon. There we can be really explicit, so let’s see what happens, shall we?

The Hamiltonian is H=12[p ϕ 21u 2+(1u 2)p u 2]+au H =\tfrac{1}{2}\left[\frac{p_\phi^2}{1-u^2}+(1-u^2)p_u^2\right]+a u The K iK_i are as follows. K 1=p ϕ K_1 = p_\phi For K 2K_2, there are two cases

(1)K 2=2aπ 1 min(H/a,1)H/au1u 2du,forp ϕ=0K_2=\frac{\sqrt{2a}}{\pi}\int_{-1}^{\min(H/a,1)} \sqrt{\frac{H/a-u}{1-u^2}}du,\qquad\text{for}\, p_\phi=0

and

(2)K 2=2aπ u min u max(uu min)(u maxu)(u 0u)du1u 2,for|p ϕ|>0K_2 = \frac{\sqrt{2a}}{\pi}\int_{u_{\text{min}}}^{u_{\text{max}}} \sqrt{(u-u_{\text{min}})(u_{\text{max}}-u)(u_0-u)} \frac{du}{1-u^2},\qquad\text{for}\, |p_\phi|\gt 0

where 1<u minu max<1,u 0>1 -1\lt u_{\text{min}}\leq u_{\text{max}}\lt 1,\quad u_0\gt 1 are the roots of the cubic 2(Hau)(1u 2)p ϕ 2=0 2(H-a u)(1-u^2) -p_\phi^2 =0

It remains to be checked that nothing wonky happens to (2) in the limit p ϕ0p_\phi\to 0. In that limit, u min1,u maxmin(H/a,1),u 0max(H/a,1) u_{\text{min}}\to -1,\quad u_{\text{max}}\to \min(H/a,1),\quad u_0\to \max(H/a,1) so (2) goes smoothly over to (1).

Again, the angle-variables undergo monodromy (specifically, the angle variable φ 2\varphi^2 shifts by a multiple of φ 1=ϕ\varphi^1=\phi. But the action variables are globally defined (for =T *M\mathcal{M}=T^*M).

Posted by: Jacques Distler on May 15, 2015 11:51 AM | Permalink | PGP Sig | Reply to this

Re: Action-Angle Variables

Sorry for being slow with replying - I’ve been catching up on things, but hopefully will be able to write more later today. The point is that there are two obstructions to the existence of global action-angle coordinates. One - the (Duistermaat-)Chern class - measures whether you can choose a global section of (M,ω)B(M,\omega)\to B or not. The second obstruction is the monodromy. I was mostly concerned with the first one, while you are talking about the monodromy. In the spherical pendulum example the first obstruction vanishes, but the second does not.

Posted by: dalpezzo on May 19, 2015 12:08 PM | Permalink | Reply to this

Re: Action-Angle Variables

Sorry for being slow with replying - I’ve been catching up on things, but hopefully will be able to write more later today. The point is that there are two obstructions to the existence of global action-angle coordinates. One - the (Duistermaat-)Chern class - measures whether you can choose a global section of (M,ω)B(M,\omega)\to B or not. The second obstruction is the monodromy. I was mostly concerned with the first one …

The Duistermaat Chern class vanishes for ω\omega exact (in particular, for \mathcal{M} a cotangent bundle, with its standard symplectic structure).

In the spherical pendulum example the first obstruction vanishes, but the second does not.

As I just said, the vanishing of the Chern class is much more general than this example.

The monodromy is interesting. But I think it’s important that there’s no “monodromy” of the action-variables.

Posted by: Jacques Distler on May 19, 2015 3:05 PM | Permalink | PGP Sig | Reply to this

Re: Action-Angle Variables

Interesting

http://muse.jhu.edu/journals/posc/summary/v023/23.2.ritson.html

Posted by: Rick on May 13, 2015 10:04 PM | Permalink | Reply to this

People are still publishing crap like that?

Interesting.

No, not interesting … stupid.

Please take anything else you’d like to post, along these lines, over to Peter Woit’s blog, where it will be properly appreciated.

(If you post it over here, I will delete it.)

Posted by: Jacques Distler on May 13, 2015 11:00 PM | Permalink | PGP Sig | Reply to this

Re: Action-Angle Variables

What do you think of this paper?
http://arxiv.org/abs/1506.05277

Posted by: Felise on June 18, 2015 12:59 AM | Permalink | Reply to this

Re: Action-Angle Variables

I’m having confused about something. In equation (2), you define the 1-form dφ i=i /K iωd\varphi^i = i_{\partial/\partial K_i}\omega. But K iK_i is just a real-valued function, so how are you interpreting /K i\partial/\partial K_i as a vector field?

Posted by: Matthew Kvalheim on April 13, 2018 8:51 PM | Permalink | Reply to this

Re: Action-Angle Variables

Let me make sure I understand what your question is.

Everywhere that it’s defined, K i\frac{\partial}{\partial K_i} is obviously a vector field. Recall that a vector field, vv, is functional v:C (M)C (M)v\colon C^\infty(M)\to C^\infty(M) which is linear v(f+αg)=v(f)+αv(g),α v(f+\alpha g) =v(f) +\alpha v(g),\quad\forall \alpha\in \mathbb{R} and obeys a Leibnitz rule v(fg)=v(f)g+fv(g) v(f g)=v(f)g+f v(g)

Of course, you’re right: K i\frac{\partial}{\partial K_i} isn’t generally defined everywhere. The archetypical example is M= 2M=\mathbb{R}^2 (with its standard symplectic structure) and K=r 2K=r^2. The vector field v=Kv=\frac{\partial}{\partial K} is not defined at the origin and (hence) neither is the angular coordinate ϕ\phi.

This is not a problem. (Hint: I have just described for you the 1D harmonic oscillator – a Hamiltonian system with which you are doubtless familiar.)

Posted by: Jacques Distler on April 13, 2018 11:18 PM | Permalink | PGP Sig | Reply to this

Re: Action-Angle Variables

Thank you very much for taking the time to respond to me.

Indeed, I am familiar with vector fields as you describe. But I think you might be missing the point that I’m trying to make. Let me try to explain myself.

Given coordinates (x 1,,x n)(x^1,\ldots,x^n) on an nn-manifold, of course x 1\frac{\partial}{\partial x^1} defines a local tangent vector field. It acts on \mathbb{R}-valued functions ff by writing ff in terms of the coordinates (x 1,,x n)(x^1,\ldots,x^n) and then differentiating ff with respect to x 1x^1.

But now let (x 1,y 2,,y n)(x^1,y^2,\ldots, y^n) be new coordinates, where x 1x^1 is the same as the above but the y iy^i are different functions. Then x 1\frac{\partial}{\partial x^1} generally defines a local vector field which is different than the vector field in the above paragraph, despite being represented by the same symbol. Hence the vector field x 1\frac{\partial}{\partial x^1} implicitly depends on the set of functions used to complete x 1x^1 to a coordinate system. (Arnold makes this same point in footnote 81 on p. 258 of his classical mechanics book, 2nd ed.)

As a concrete example, in 2\mathbb{R}^2 in a neighborhood of any point not on either standard coordinate axis, we can use polar coordinates (r,θ)(r,\theta) but we can also use mixed coordinates (r,x)(r,x) where xx is the standard Cartesian coordinate. But r\frac{\partial}{\partial r} will be a different derivation depending on which coordinate system is used.

In light of the above, what I’m wondering is: don’t you need to complete the functions (K 1,,K n)(K^1,\ldots, K^n) to a coordinate system (K 1,,K n,z 1,,z n)(K^1,\ldots,K^n, z^1,\ldots,z^n) in order to interpret K i\frac{\partial}{\partial K^i} as a tangent vector? Or perhaps it is possible to do this in a different way?

Thanks again for your help.

Posted by: Matthew Kvalheim on April 13, 2018 11:57 PM | Permalink | Reply to this

Re: Action-Angle Variables

In light of the above, what I’m wondering is: don’t you need to complete the functions (K 1,,K n)(K^1,\ldots, K^n) to a coordinate system (K 1,,K n,z 1,,z n)(K^1,\ldots,K^n, z^1,\ldots,z^n) in order to interpret K i\frac{\partial}{\partial K^i} as a tangent vector?

Indeed, you do.

But the symplectic form (more or less) tells you how to do that. Namely, given (K 1,,K n)(K_1,\ldots, K_n), we need to find (ϕ 1,,ϕ n)(\phi^1,\ldots,\phi^n) such that ω= idK idϕ i\omega = \sum_i d K_i\wedge d\phi^i. There’s still an ambiguity of the form ϕ iϕ i+f(K i)\phi^i\to \phi^i +f(K_i), which can’t be fixed by this procedure. But that’s to be expected …

Posted by: Jacques Distler on April 14, 2018 12:50 AM | Permalink | PGP Sig | Reply to this

Re: Action-Angle Variables

I agree that the action-angle variables we ultimately seek are not uniquely defined by the 3 properties mentioned in the post, and that in particular the angle variables have the ambiguity you just pointed out.

But in your specific construction, given that equation (2) defines the “coordinates” φ i\varphi^i (or rather the related closed 1-forms dφ id\varphi^i), it seems to me like we can’t use coordinates (K 1,,K n,φ 1,,φ n)(K^1,\ldots,K^n,\varphi^1,\ldots,\varphi^n) to give meaning to K i\frac{\partial}{\partial K^i} without circular reasoning.

Am I missing something?

Posted by: Matthew Kvalheim on April 14, 2018 1:15 AM | Permalink | Reply to this

Re: Action-Angle Variables

I’m not sure what you mean.

ω\omega is a fixed, known 2-form. Given ω\omega and the K iK_i, this defines a set of 1-forms dϕ id\phi^i (up to the aforementioned ambiguity). The vector field K i\frac{\partial}{\partial K_i} is then well-defined: i /K idK j=δ j i,i /K idϕ j=0 i_{\partial/\partial K_i} d K_j =\delta^{i}_{j},\quad i_{\partial/\partial K_i} d\phi^j = 0

I think what you are objecting to is the consistency condition: the partial derivative with respect K iK_i is defined by holding the other K jK_j and all of the ϕ i\phi^i fixed.

Or maybe I’m just not understanding…

Posted by: Jacques Distler on April 14, 2018 2:08 AM | Permalink | PGP Sig | Reply to this

Re: Action-Angle Variables

      ω\omega is a fixed, known 2-form.

I completely agree.

      Given ω\omega and the K iK_i, this defines a set of 1-forms dφ id\varphi_i (up to the aforementioned ambiguity).

I think this sentence is precisely where my confusion lies. How does this define the 1-forms dφ id\varphi_i? If you mean that it defines them via equation (2), equation (2) says that dφ id\varphi_i is the 2-form ω\omega contracted with K i\frac{\partial}{\partial K_i}. But since forms can only be contracted with vectors, then for this to make any sense, we must be able to interpret K i\frac{\partial}{\partial K_i} as a vector. But I do not see how we can do this at this point.

      The vector field K i\frac{\partial}{\partial K_i} is then well-defined:

(The following text is redundant in light of my last paragraph; I’m including it for extra clarity.) This sentence of yours seems to assert that the well-definedness of K i\frac{\partial}{\partial K_i} follows from the set of 1-forms dφ id\varphi_i being well-defined (or at least defined up to some ambiguity). But as I express in the previous paragraph, it seems like K i\frac{\partial}{\partial K_i} needs to be well-defined before the 1-forms dφ id\varphi_i are defined. Hence it seems like there is a chicken-and-egg problem here.

Does that explain myself any better? It’s possible I’m misunderstanding something basic…

Posted by: Matthew Kvalheim on April 14, 2018 3:28 AM | Permalink | Reply to this

Re: Action-Angle Variables

It seems to me that your complaint, here, is purely a local one, and it is the thing that is answered by the Carathéodory-Jacobi-Lie Theorem.

Posted by: Jacques Distler on April 14, 2018 12:12 PM | Permalink | PGP Sig | Reply to this

Re: Action-Angle Variables

Interesting. I didn’t know that result. Thanks for the reference.

Using that result, given (K 1,,K n)(K_1,\ldots,K_n), we can locally complete these to symplectic coordinates (K 1,,K n,φ 1,,φ n)(K_1,\ldots,K_n,\varphi^1,\ldots,\varphi^n). But then there is the question: why do the local 1-forms dφ 1d\varphi^1 patch together to yield well-defined global 1-forms?

I think I finally found an answer in pp. 3-7 of Duistermaat’s paper (mentioned in another comment above). Duistermaat proves that the integral curves of each X K iX_{K_i} are periodic (I wish I understood this part more intuitively), where X K iX_{K_i} the Hamiltonian vector field of K iK_i. Hence the integral curves of these vector fields define circle-valued coordinates on each torus. But notice now that φ i=X K i\frac{\partial}{\partial \varphi^i} = X_{K_i}, where φ i\varphi^i is a local coordinate as described in the above paragraph. Hence the local coordinates φ i\varphi^i of the last paragraph are none other than the restrictions of the well-defined global circle-valued coordinates determined by the integral curves of the X K iX_{K_i}, so these local coordinates indeed patch together to yield well-defined circle-valued coordinates on each torus.

Posted by: Matthew Kvalheim on April 14, 2018 6:57 PM | Permalink | Reply to this

Re: Action-Angle Variables

But then there is the question: why do the local 1-forms dφ id\varphi^i patch together to yield well-defined global 1-forms?

They don’t. That’s the monodromy phenomenon discussed above, where the spherical pendulum gives a nice example.

Hence the local coordinates φ i\varphi^i of the last paragraph are none other than the restrictions of the well-defined global circle-valued coordinates determined by the integral curves of the X K iX_{K_i}, so these local coordinates indeed patch together to yield well-defined circle-valued coordinates on each torus.

They do, as you say, yield well-defined circle-valued coordinates on each torus, just not globally over all of \mathcal{M}.

Posted by: Jacques Distler on April 14, 2018 8:06 PM | Permalink | PGP Sig | Reply to this

Re: Action-Angle Variables

Sorry — when I said “global”, what I really meant was globally on defined on a torus or on a small neighborhood of a torus, not globally defined on all of phase space. That’s the part I’m trying to understand.

Posted by: Matthew Kvalheim on April 14, 2018 8:23 PM | Permalink | Reply to this

Re: Action-Angle Variables

I know this is extremely late, but I just stumbled on this post recently.

In the setting where your symplectic 2-form ω\omega is not exact, you can still perform action integrals, except they are integrals of ω\omega over 2-dimensional areas instead of 1-dimensional boundaries. That is: your homology basis γ i\gamma_i are the boundaries of disks, γ i=D i\gamma_i = \partial D_i. Using Stokes’ theorem, rewrite your action integral as K i=12π D iω, K_i = \frac{1}{2\pi} \int_{D_i} \omega \,, and since dω=0d\omega=0, this is insensitive to choice of disk D iD_i, so long as its boundary γ i\gamma_i is in the same homology class on that same Lagrangian submanifold.

There is an ambiguity in how to “cap” the loops, when they do not bound a disk. I think we are supposed to be talking about “relative homology,” where our action integrals are defined relative to some choice of “capping”. For example, consider S 2S^2 as a symplectic manifold, foliated by circle. We can choose to “cap” each circle by either the “Northern” disk or “Southern” disk, or instead, relative to some other (fixed) circle, e.g. the equator. The one-form dKdK agrees for every one of these definitions, even though KK is shifted by different constants.

Posted by: Leo Stein on May 5, 2022 10:29 AM | Permalink | Reply to this

non-exact

That is: your homology basis γ i\gamma_i are the boundaries of disks, γ i=D i\gamma_i = \partial D_i.

That sounds like an oxymoron. Homology cycles are precisely the ones which cannot be written as the boundary of something.

Think how this would work in the example I gave: (,ω)=(T 4,dθ 1dθ 3+dθ 2dθ 4)(\mathcal{M},\omega)=(T^4,d\theta_1\wedge d\theta_3+d\theta_2\wedge d\theta_4). Our Lagrangian tori are T 2T^2s (parametrized by θ 3,θ 4\theta_3,\theta_4, for fixed values of θ 1,θ 2\theta_1,\theta_2). The γ i\gamma_i are the one-cycles dual to dθ 3d\theta_3 and dθ 4d\theta_4. They are not boundaries of disks.

Your would-be example, of S 2S^2 foliated by circles, is precisely the one discussed in the Update above.

Posted by: Jacques Distler on May 5, 2022 11:24 AM | Permalink | PGP Sig | Reply to this

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