Generating Functions

Method of Generating Functions

Let $X$ and $Y$ be two smooth manifolds, and let $M = T^\ast X, N = T^\ast Y$ with corresponding symplectic forms $\omega_M$ and $\omega_N$.

Question: How can we produce symplectomorphisms $\phi: M \to N$?

The most important construction from classical mechanics is the method of generating functions. I will outline this method, shameless stolen from Ana Cannas da Silva's lecture notes.

Suppose we have a smooth function $f \in C^\infty(X \times Y)$. Then its graph $\Gamma$ is a submanifold of $M \times N$: $\Gamma = \{ (x,y, df_{x,y}) \in M \times N \}$. Since $M \times N$ is a product, we have projections $\pi_M, \pi_N$, and this allows us to write the graph as
$$\Gamma = \{ (x, y, df_x, df_y) \}$$
Now there is a not-so-obvious trick: we consider the twisted graph $\Gamma^\sigma$ given by
$$\Gamma^\sigma =  \{(x,y, df_x, -df_y) \}$$
Note the minus sign.

Proposition If $\Gamma^\sigma$ is the graph of a diffeomorphism $\phi: M \to N$, then $\phi$ is a symplectomorphism.

Proof By construction, $\Gamma^\sigma$ is a Lagrangian submanifold of $M \times N$ with respect to the twisted symplectic form $\pi_M^\ast \omega_M - \pi_N^\ast \omega_N$. It is a standard fact that a diffeomorphism is a symplectomorphism iff its graph is Lagrangian with respect to the twisted symplectic form, so we're done.

Now we have:

Modified question: Given $f \in C^\infty(M \times N)$, when is its graph the graph of a diffeomorphism $\phi: M \to N$?

Pick coordinates $x$ on $X$ and $y$ on $Y$, with corresponding momenta $\xi$ and $\eta$. Then if $\phi(x,\xi) = (y,\eta)$, we obtain
$$\xi = d_x f, \ \eta = -d_y f$$
Note the simlarity to Hamilton's equations. By the implicit function theorem, we can construct a (local) diffeomorphism $\phi$ as long as $f$ is sufficiently non-degenerate.

Different Types of Generating Functions

We now concentrate on the special case of $M = T^\ast \mathbb{R} = \mathbb{R} \times \mathbb{R}^\ast$. Note that this is a cotangent bundle in two ways: $T^\ast \mathbb{R} \cong T^\ast \mathbb{R}^\ast$. Hence we can construct local diffeomorphisms $T^\ast \mathbb{R} \to T^\ast \mathbb{R}$ in four ways, by taking functions of the forms

$$f(x_1, x_2), \ f(x_1, p_2), \ f(p_1, x_2), \ f(p_1, p_2)$$

Origins from the Action Principle, and Hamilton-Jacobi

Suppose that we have two actions

$$S_1 = \int p_1 \dot{q}_1 - H_1 dt, \ S_2 = \int p_2 \dot{q}_2 - H_2 dt$$

which give rise to the same dynamics. Then the Lagrangians must differ by a total derivative, i.e.

$$p_1 \dot{q}_1 - H_1 = p_2 \dot{q}_2 - H_2  + \frac{d f}{dt}$$

Suppose that $f = -q_2 p_2 + g(q_1, p_2, t)$. Then we have

$$p_1 \dot{q}_1 - H_1 = -q_2 \dot{p}_2 - H_2 + \frac{\partial g}{\partial t} + \frac{\partial g}{\partial q_1}\dot{q}_1 + \frac{\partial g}{\partial p_2} \dot{p_2}$$

Comparing coefficients, we find
$$p_1 = \frac{\partial g}{\partial q_1}, \ q_2 = \frac{\partial g}{\partial p_2}, \ H_2 = H_1 + \frac{\partial g}{\partial t}$$

Now suppose that the coordinates $(q_2, p_2)$ are chosen so that Hamilton's equations become
$$\dot{q_2} = 0, \ \dot{p}_2 = 0$$
Then we must have $H_2 = 0$, i.e.
$$H_1 + \frac{\partial g}{\partial t} = 0$$
Now we also have $\partial H_2 / \partial p_2 = 0$, so this tells us that $g$ is independent of $p_2$, i.e. $g = g(q_1, t)$. Since $p_1 = \partial g / \partial q_1$, we obtain
$$\frac{\partial g}{\partial t} + H_1(q_1, \frac{\partial g}{\partial q_1}) = 0$$
This is the Hamilton-Jacobi equation, usually written as
$$\frac{\partial S}{\partial t} + H(x, \frac{\partial S}{\partial x}) = 0$$
Note the similarity to the Schrodinger equation! In fact, one can derive the Hamilton-Jacobi equation from the Schrodinger equation by taking a wavefunction of the form
$$\psi(x,t) = A(x,t) \exp({\frac{i}{\hbar} S(x,t)})$$
and expanding in powers of $\hbar$. This also helps to motivate the path integral formulation of quantum theory.

KAM I

In this post I want to sketch the idea of KAM, following these lecture notes.

Integrable Systems

I don't want to worry too much about details, so for now we'll define an integrable system to be a Hamiltonian system $(M, \omega, H)$ for which we can choose local Darboux coordinates $(I, \phi)$ with $I \in \mathbb{R}^N$ and $\phi \in T^N$, such that the Hamiltonian is a function of $I$ only. Defining $\omega_j := \partial H / \partial I_j$, Hamilton's equations then read
$$\begin{align} \dot{I}_j &= 0, \\\ \dot{\phi}_j &= \omega_j(I). \end{align}$$
Hence we obtain linear motion on the torus as our dynamics. Note in particular that the sets $\{I = \mathrm{const}\}$ are tori, and that the dynamics are constrained to these tori. We call these tori "invariant".


Now suppose that our Hamiltonian $H$ is of the form
$$H(I, \phi) = h(I) + f(I, \phi)$$
with $f$ "small". What can be said of the dynamics? Specifically, do there exist invariant tori? KAM theory lets us formulate this question in a precise way, and gives an explicit quantitative answer (as long as $f$ is nice enough, and small enough).

I want to sketch the idea of the KAM theorem, completely ignoring analytical details.

Constructing the Symplectomorphism

Suppose we could find a symplectomorphism $\Phi$: (I, \phi) \mapsto (\tilde{I}, \tilde{\phi})\) such that $H(I, \phi) = H(\tilde{I}$. Then our system would still be integrable (just in new action-angle coordinates), and we'd be done. There are two relatively easy ways of constructing symplectomorphisms: integrating symplectic vector fields, and generating functions. In the lecture notes, generating functions are used, so let's take a minute to discuss them.

Proposition Let $\Sigma(\tilde{I}, \phi)$ be a smooth function and suppose that the transformation
$$I = \frac{\partial \Sigma}{\partial \phi}, \ \tilde{\phi} = \frac{\partial \Sigma}{\partial \tilde{I}}$$
can be inverted to produce a diffeomorphism $\Phi: (I, \phi) \mapsto (\tilde{I}, \tilde{\phi})$. Then $\Phi$ is a symplectomorphism.

Proof
$$dI = \frac{\partial^2 \Sigma}{\partial \phi \partial \tilde{I}} d \tilde{I}$$
$$d\tilde{\phi} = \frac{\partial^2 \Sigma}{\partial \phi \partial \tilde{I}} d\phi$$
Hence
$$dI \wedge d\phi = \frac{\partial^2 \Sigma}{\partial \phi \partial \tilde{I}} d \tilde{I} \wedge d\phi = d\tilde{I} \wedge d\tilde{\phi}.$$

We want a symplectomorphism $\Phi$ such that
$$H \circ \Phi(\tilde{I}, \tilde{\phi}) = \tilde{h}(\tilde{I}$$
If $\Phi$ came from a generating function $\Sigma$, then we have
$$H(\frac{\partial \Sigma}{\partial \phi}, \phi) = \tilde{h}(\tilde{I})$$
Expanding things, we have
$$h(\frac{\partial \Sigma}{\partial \phi}) + f(\frac{\partial \Sigma}{\partial \phi}, \phi) = \tilde{h}(\tilde{I}).$$

If $f$ is small, then we might expect $\Phi$ to be close to the identity, and hence $\Sigma$ ought to be close to the generating function for the identity (which is $\langle I, \phi \rangle$). So we take
$$\Sigma(\tilde{I}, \phi) = \langle \tilde{I}, \phi \rangle + S(\tilde{I}, \phi)$$
where $S$ should be "small". So we linearize the equation in $S$:
$$\langle \omega(\tilde{I}), \frac{\partial S}{\partial \phi} \rangle + f(\tilde{I}, \phi) = \tilde{h}(\tilde{I}) - h(\tilde{I})$$

Now we can expand $S$ and $f$ in Fourier series and solve coefficient-wise. This gives a formal solution $S(\tilde{I}, \phi)$ of the equation
$$\langle \omega, \frac{\partial S}{\partial \phi} \rangle + f(\tilde{I}, \phi) = 0.$$

Getting it to Work

Unfortunately, the Fourier series for $S$ has no chance to converge, so instead we take a finite truncation. If we assume $f$ is analytic, its Fourier coefficients decay exponentially fast, so this provides a very good approximate solution to the linearized equation (and we can give an explicit bound in terms of a certain norm of $f$). Call this function $S_1$. We then use $S_1$ to construct a symplectomorphism $\Phi_1$.

Now we take
$$H_1(I, \phi) = H \circ \Phi_1(I, \phi) = h_1(I) + f_1(I, \phi).$$
Some hard analysis then shows that $h - h_1$ is small, and $f_1$ is much smaller than f.

The Induction Step

The above arguments sketch a method to put the system "closer" to an integrable form. By carefully controlling $\epsilon$'s and $\delta$'s, one then shows that iterated sequence $\Phi_1, \Phi_2 \circ \Phi_1, \ldots$ converges to some limiting symplectomorphism $\Phi_\infty$.

Circle Diffeomorphisms I

This is the first of a series of posts based on these lecture notes on KAM theory. For now I just want to outline section 2, which is a toy model of KAM thoery.

Circle Diffeomorphisms

We consider a map $\phi: \mathbb{R} \to \mathbb{R}$ defined by
$$\phi(x) = x + \rho + \eta(x)$$
where $\rho$ is its rotation number and $\eta(x)$ is "small".

Define $S_\sigma$ to be the strip $\{ |\mathrm{Im} z|<\sigma\} \subset \mathbb{C}$ and let $B_\sigma$ be the space of holomorphic functions bounded on $S_\sigma$ with sup norm $\|\cdot\|_\sigma$.

Goal: Show that if $\|\eta\|_\sigma$ is sufficiently small, then there exists some diffeomorphism $H(x)$ such that
$$H^{-1} \circ \phi \circ H (x) = x + \rho$$
i.e. that $\phi$ is conjugate to a pure rotation.

Linearization

The idea is that if $\eta$ is small, then $H$ should be close to the identity, so we suppose that
$$H(x) = x + h(x)$$
where $h(x)$ is small. Plugging this into the equation above and discarding higher order terms yields
$$h(x+\rho) - h(x) = \eta(x)$$
Since $\eta$ is periodic, we Fourier transform both sides to obtain an explicit formula for the Fourier coefficients of $h(x)$. We have to show several things:

1. The Fourier series defining $h(x)$ converges in some appropriate sense.

2. The function $H(x) = x + h(x)$ is a diffeomorphism.

3. The composition $\tilde{\phi} = H^{-1} \circ \phi \circ H$ is closer to a pure rotation than $\phi$, in the sense that
$$\tilde{\phi}(x) = x + \rho + \tilde{\eta}(x)$$
where $\|\tilde{\eta}\| \ll \|\eta\|$.

Newton's Method

Carrying out the analysis, one finds that for appropriate epsilons and deltas, if $\eta \in B_\sigma$ then $H \in B_{\sigma - \delta}$ and that $\|\tilde{\eta}\|_{\sigma-\delta} \leq C \|\eta\|_\sigma^2$. By carefully choosing the deltas, we can iterate this procedure (composing the $H$'s) to obtain a well-defined limit $H_\infty \in B_{\sigma/2}$ such that
$$H_\infty^{-1} \circ \phi \circ H_\infty (x) = x + \rho,$$
as desired.

So in fact the idea of the proof is extremely simple, and all of the hard work is in proving some estimates.