completely integrable system

Informal overview

Integrable systems have applications in mechanics. Say, for a physical system, you are given a symplectic manifold (that usually means some type of phase space) and a Hamiltonian function $H$.

The goal is usually to understand the “trajectories” of the system, meaning the integral curves of the Hamiltonian vector field (lets call it $X_H$), even better is to understand ALL the integral curves and get a picture of the flow ${\Phi }^H$ of $X_H$. Unfortunately, this is usually very hard.

In a special case, if enough conserved quantities can be identified, the flow (as a bunch of differential equations) becomes easier to solve (AKA integrate). Note that the Hamiltonian function itself is a conserved quantity by design. $H$ is constant on integral curves of $X_H$. So, if we can find $n-1$ other functions that are also constant along integral curves of $X_H$, that are “different” enough functions (here’s where the functional independence condition in the second part of the definition comes in), then we can find a chart that preserves the symplectic form that is MUCH simpler to solve. This comes with fancy tools like generating functions and action-angle coordinates. See @abraham2008 chapter 5.2 for more about the physics.

Definition

On symplectic manifold

Let $(M,\omega )$ be a symplectic manifold of dimension $2n$. A completely integrable system is a smooth map

$F=(f_1,\dots ,f_n):M⟶{\mathbb{R}}^n$

where each $f_i:M\to \mathbb{R}$ pairwise Poisson commute, i.e.

$\{f_i,f_j\}=01\le i,j\le n$

and $F:M\to {\mathbb{R}}^n$ is a smooth submersion almost everywhere. (In other words, the set of regular values is an open, dense subset.)

On Poisson manifold

The definition for a Poisson manifold $(M,\sigma )$ is almost the same, but instead we need $\frac{1}{2}(\dim M+\text{corank }\sigma )$ Poisson commuting, functionally independent functions.

As a maximal Poisson commuting subalgebra

In both the symplectic and Poisson commuting case, we want to define a maximal Poisson-commuting subalgebra of $C^{\infty }(M)$.

Symplectic case

Given a symplectic manifold $(M,\omega )$ with $\dim M=2n$, there can be at most $n$ Poisson commuting functions that are functionally independent.

Proof

Suppose there are $m$ smooth functions $f_1,\dots ,f_m:M⟶\mathbb{R}$ such that $\{f_i,f_j\}=0$.

If we pick any point $p\in M$, we can look at the vector subspace $W\subset T_{p}M$ defined as

$W=\text{span}\{(X_{f_1}{)}_p,\dots ,(X_{f_m}{)}_p\}$

Where $X_{f_{}i}$ denotes the Hamiltonian vector field for the function $f_i$.

Looking at the symplectic form on this subspace, ${\omega }_p((X_{f_i}{)}_p,(X_{f_j}{)}_p)=\{f_i,f_j\}(p)=0.$

Thus, by linearity of the symplectic form, ${\omega }_p{|}_W$ vanishes, so $W$ is an isotropic subspace.

This means that $\dim W\le n$.

Next, assume that the functions $\{f_1,\dots ,f_m\}$ are independent. Thus, $d{F}_p:M⟶{\mathbb{R}}^m$ is surjective (almost everywhere).

Note that for the standard basis $\{e_j\}$ of ${\mathbb{R}}^m$, $F=f_1\cdot e_1+\cdots +f_m\cdot e_m.$

Using linearity of the differential, we have

$$\begin{array}{rl}d{F}_p& =d(f_1\cdot e_1+\cdots +f_m\cdot e_m)\\ & =d(e_1\circ f_1{)}_p+\cdots +d(e_m\circ f_m{)}_p\\ & =d{e}_1\circ (d{f}_1{)}_p+\cdots +d{e}_m\circ (d{f}_m{)}_p\\ & =(d{f}_1{)}_p\cdot e_1+\cdots +(d{f}_m{)}_p\cdot e_m\end{array}$$

Written somewhat more clearly, $d{F}_p=\left[\begin{array}{c}(d{f}_1{)}_p\\ ⋮\\ (d{f}_m{)}_p\end{array}\right]:T_{p}M\to {\mathbb{R}}^m$ So this can only be surjective if $\{(d{f}_i{)}_p\}$ is a linearly independent set of $T_p^{\ast }M$.

Since $\omega$ is non-degenerate, we can look at the isomorphism

$$\begin{array}{rl}{\omega }^{\vee }:T_{p}M& ⟶T_p^{\ast }M\\ v& ⟼\omega (v,-)\end{array}$$

Note that since $X_{f_i}$ are Hamiltonian vector fields, ${\omega }^{\vee }(X_{f_i}{)}_p=\omega (X_{f_i},-)=({\iota }_{X_{f_i}}\omega {)}_p=(d{f}_i{)}_p$

So thus, the subspace $\text{span}\{(d{f}_1{)}_p,\dots ,(d{f}_m{)}_p\}\cong W$, and there can be at most $n$ independent functions.

Poisson manifold case

Given a Poisson manifold $(M,\sigma )$, there can be at most $\frac{1}{2}(\dim M+\text{corank }\sigma )$ Poisson commuting functions that are functionally independent.

Proof

By the Weinstein splitting theorem, we can pick small enough open sets centered at each point such that

$\chi (U)=V\times W\subset {\mathbb{R}}^{2s}\times {\mathbb{R}}^k$

where $V$ has the symplectic form induced from $\sigma$.

Since $V$ is an open set from a symplectic manifold there can only be $s$ independent functions that Poisson commute on $V$. $W$ is not symplectic, so we can look for $k$ independent functions that Poisson commute.

Note that we can always include independent Casimir functions, since they Poisson commute with every function, so they will automatically commute with the other $s$ functions already found. There can only be $k$ other independent functions since $\dim T_{p}W=k$.

Thus, there can be $\frac{1}{2}\dim V+\dim W=\frac{1}{2}(\dim M+\text{corank }\sigma )$ functions (for the largest dimension symplectic leaf).

Cool things

• There is a relationship between certain Hamiltonian torus actions and completely integrable systems shown here.