Test independence and Poisson commutativity

Are three conserved spectral quantities enough to establish integrability of three Toda particles? You will check their pairwise Poisson brackets and the rank of their differentials on the canonical six-dimensional phase space. The calculation establishes three commuting integrals that are independent on an open dense set, and shows why one vanishing Jacobian minor is inconclusive.

Required background. Compute $P$, $H$, and $J$ as in Extract spectral invariants, and differentiate functions of several variables. The entry repair reviews the bracket and the rank test.

Helpful background. The bracket bridge separates conservation from mutual involution and tests canonical coordinate changes. The Library derivation gives the argument for general finite $N$; this lesson makes the three-particle checks explicit. The Liouville–Arnold reference explains why a torus conclusion requires additional hypotheses.

Three integrals on six-dimensional phase space

Work with the dimensionless open Toda chain on $\mathbb R^6$, with canonical coordinates $(q_1,q_2,q_3,p_1,p_2,p_3)$. Put

$$u=e^{q_1-q_2}\gt0,\qquad v=e^{q_2-q_3}\gt0.$$

The three functions found from the Lax matrix are

$$\begin{aligned} P&=p_1+p_2+p_3,\\ H&=\frac12(p_1^2+p_2^2+p_3^2)+u+v,\\ J&=\frac13(p_1^3+p_2^3+p_3^3) +u(p_1+p_2)+v(p_2+p_3). \end{aligned}$$

The phase space has three canonical coordinate–momentum pairs. The Liouville test therefore requires three independent commuting integrals, including the Hamiltonian. It does not require six. Keep the following questions separate:

Question	Test in this Hamiltonian system
Is $F$ conserved?	$\{F,H\}=0$ for a function with no explicit time dependence
Are $F$ and $G$ in involution?	$\{F,G\}=0$
Are $P,H,J$ independent at a point?	The differentials $dP,dH,dJ$ have rank $3$ there

Conservation under one chosen flow does not usually give all pairwise brackets. Nor does having different names or different numerical values show independence. For example, $H$ and $H^2$ commute and are conserved, but $d(H^2)=2H\,dH$.

All derivatives below are taken on the full canonical phase space. The variables $(a_1,a_2,p_1,p_2,p_3)$ omit the common position coordinate and do not constitute three canonical pairs.

Entry check and repair

For one canonical pair, compute $\{qp,p^2\}$ with the convention $\{q,p\}=1$. Separately, if a $3\times6$ Jacobian has one $3\times3$ minor with nonzero determinant, what is its rank? What can you conclude if just one of its minors is zero?

Repair. The bracket is $\{f,g\}=f_qg_p-f_pg_q$. Therefore $\{qp,p^2\}=p(2p)-q(0)=2p^2$. For several canonical pairs, add one such term per pair:

$$\{f,g\}=\nabla_q f\mathbin{\cdot}\nabla_p g -\nabla_p f\mathbin{\cdot}\nabla_q g.$$

The Jacobian of $(P,H,J)$ has one row per function and one column per phase-space coordinate. A nonzero $3\times3$ minor proves rank $3$, the largest possible row rank. A single zero minor proves only that those particular three columns do not suffice. Rank below $3$ requires every $3\times3$ minor to vanish.

Momentum brackets from translation invariance

Because $P$ depends only on momenta and $\partial P/\partial p_i=1$,

$$\{P,F\}=-\sum_{i=1}^{3}\frac{\partial F}{\partial q_i}.$$

For the Hamiltonian,

$$\nabla_qH=(u,v-u,-v),$$

whose components sum to zero. Hence $\{P,H\}=0$.

The same cancellation works for $J$ because its coordinate dependence is entirely through $u$ and $v$. More explicitly,

$$\begin{aligned} \partial_{q_1}J&=u(p_1+p_2),\\ \partial_{q_2}J&=-u(p_1+p_2)+v(p_2+p_3),\\ \partial_{q_3}J&=-v(p_2+p_3). \end{aligned}$$

Their sum vanishes, giving $\{P,J\}=0$. This is the differential form of invariance under $q_i\mapsto q_i+c$. The remaining bracket $\{H,J\}$ is the guided calculation below. Its cancellation uses the mixed bond terms in $J$.

A worked independence test

Hold $q$ fixed and form the minor using the three momentum columns:

$$D=\det\frac{\partial(P,H,J)}{\partial(p_1,p_2,p_3)} =\det\begin{pmatrix} 1&1&1\\ p_1&p_2&p_3\\ p_1^2+u&p_2^2+u+v&p_3^2+v \end{pmatrix}.$$

At $q=(0,0,0)$ and $p=(0,1,3)$, it becomes

$$D=\det\begin{pmatrix}1&1&1\\0&1&3\\1&3&10\end{pmatrix} =1-(-3)-1=3.$$

This exact nonzero value proves independence at that point. Continuity proves independence in some neighborhood, but a single example by itself does not prove independence on a dense set. For that stronger statement, expand the same minor:

$$\begin{aligned} D={}&(p_2-p_1)(p_3-p_1)(p_3-p_2)\\ &+u(p_1-p_2)+v(p_2-p_3). \end{aligned}$$

For every fixed $q$, this is a nonzero polynomial in $p$: its cubic Vandermonde term cannot be canceled by the terms of degree one. A nonzero polynomial cannot vanish on an open set. Thus every open phase-space neighborhood contains points with $D\ne0$, and continuity makes that set open as well. We have proved independence on an open dense set.

The test makes no assertion of dependence where $D=0$. Other minors can still be nonzero there. The standard initial state of this course is exactly such an example.

Exercises

Guided practice: the remaining Poisson bracket

Compute $\{H,J\}$ without substituting special numerical values. Begin with

$$\nabla_pH=(p_1,p_2,p_3),\qquad \nabla_qH=(u,v-u,-v).$$

Find $\nabla_pJ$ and use the coordinate derivatives of $J$ given above. Simplify each dot product in

$$\{H,J\}=\nabla_qH\mathbin{\cdot}\nabla_pJ -\nabla_pH\mathbin{\cdot}\nabla_qJ$$

before subtracting. Show explicitly why the terms containing products of $u$ and $v$ cancel. Explain how this result, the momentum brackets, and the rank calculation fit together.

Hint · Full solution

Independent practice: repair a misleading minor

At $q=(0,0,0)$ and $p=(1,0,-1)$, a learner finds that the momentum-only minor is zero and concludes that $P,H,J$ are dependent. Compute the full $3\times6$ Jacobian in the column order $(q_1,q_2,q_3,p_1,p_2,p_3)$. Then evaluate its minor with columns $(p_1,p_2,q_2)$ in that order. Assess the learner’s conclusion. Does $P=J=0$ at this point change the rank test?

Hint · Full solution

Changed setting: only two particles

Take the open chain with $N=2$ and $u=e^{q_1-q_2}\gt0$:

$$P=p_1+p_2,\qquad H=\frac12(p_1^2+p_2^2)+u.$$

How many independent commuting integrals are required on its canonical phase space? Check the bracket $\{P,H\}$ and the minor $\partial(P,H)/\partial(p_1,q_1)$. Does independence fail when $p_1=p_2$? Finally, calculate $J=\operatorname{tr}(L^3)/3$ for the $2\times2$ Toda matrix and show that

$$J=PH-\frac{P^3}{6}.$$

Explain why counting every conserved trace as a new independent integral would be wrong.

Hint · Full solution

Hints

Remaining bracket. $\nabla_pJ=(p_1^2+u,p_2^2+u+v,p_3^2+v)$. The terms with no momenta in the first dot product are $u^2+(v-u)(u+v)-v^2$.

Misleading minor. The coordinate derivatives of $J$ are $(1,-2,1)$ at this state, even though $J=0$. Use a coordinate column as well as momentum columns.

Two particles. The coordinate derivative $\partial_{q_1}H=u$ is never zero at a finite real $(q_1,q_2)$. For the cubic identity, expand $PH-P^3/6$ and cancel the mixed momentum terms.

Solutions and checks

Remaining bracket

The first dot product is

$$\begin{aligned} \nabla_qH\mathbin{\cdot}\nabla_pJ ={}&u(p_1^2+u)+(v-u)(p_2^2+u+v)\\ &-v(p_3^2+v)\\ ={}&u(p_1^2-p_2^2)+v(p_2^2-p_3^2), \end{aligned}$$

because $u^2+(v-u)(u+v)-v^2=0$. The second is

$$\begin{aligned} \nabla_pH\mathbin{\cdot}\nabla_qJ ={}&u(p_1-p_2)(p_1+p_2)\\ &+v(p_2-p_3)(p_2+p_3)\\ ={}&u(p_1^2-p_2^2)+v(p_2^2-p_3^2). \end{aligned}$$

Subtracting gives $\{H,J\}=0$ for every state, not just for the initial example. Together with $\{P,H\}=\{P,J\}=0$, all three functions commute pairwise. The rank calculation shows that they are independent on an open dense set of the six-dimensional canonical phase space. These are the required three integrals for Liouville integrability of this finite open three-particle system.

Misleading minor

The full Jacobian is

$$\frac{\partial(P,H,J)}{\partial(q_1,q_2,q_3,p_1,p_2,p_3)} =\begin{pmatrix} 0&0&0&1&1&1\\ 1&0&-1&1&0&-1\\ 1&-2&1&2&2&2 \end{pmatrix}.$$

The momentum-only minor vanishes because its third row is twice its first. However, the requested different minor is

$$\det\frac{\partial(P,H,J)}{\partial(p_1,p_2,q_2)} =\det\begin{pmatrix}1&1&0\\1&0&0\\2&2&-2\end{pmatrix}=2.$$

Therefore the full Jacobian has rank $3$. The initial state is regular for these three integrals. The values $P=J=0$ concern where the level sets sit; the differentials concern how those functions change under nearby variations. Equal function values at one point imply no functional dependence.

Two particles

The phase space is $\mathbb R^4$, so two independent commuting integrals are required. The momentum bracket is

$$\{P,H\}=-(\partial_{q_1}H+\partial_{q_2}H)=-(u-u)=0.$$

The requested minor is

$$\det\frac{\partial(P,H)}{\partial(p_1,q_1)} =\det\begin{pmatrix}1&0\\p_1&u\end{pmatrix}=u\gt0.$$

It is nonzero at every finite real point, including $p_1=p_2$. A momentum-only minor $p_2-p_1$ would have missed this fact. The limiting decoupled value $u=0$ is outside the finite-$q$ domain used in this argument.

The cubic trace is

$$J=\frac13(p_1^3+p_2^3)+u(p_1+p_2).$$

Expanding the proposed expression gives

$$\begin{aligned} PH-\frac{P^3}{6} ={}&\frac12(p_1+p_2)(p_1^2+p_2^2)\\ &-\frac16(p_1+p_2)^3+u(p_1+p_2)\\ ={}&\frac13(p_1^3+p_2^3)+u(p_1+p_2)=J. \end{aligned}$$

Thus $dJ=(H-P^2/2)dP+P\,dH$, so this third conserved trace adds no independent differential. Conservation does not override the functional relation.

What the three checks establish

For the stated three-particle open chain, you have conservation, pairwise involution, and independence on an open dense set. You also checked independence directly at the common initial state. The Library proof develops the corresponding finite-$N$ result.

These conclusions do not assert that the trajectories lie on compact invariant tori. A common translation $q\mapsto q+c\mathbf1$ leaves all three integrals unchanged and already supplies an unbounded direction in each level set. The distinction between the open-chain noncompact setting and compact-torus arguments is discussed in Moser 1975, § 1, pp. 468–469, PDF. Compactness and the other hypotheses of an action–angle theorem must be checked separately.

You are ready for the three-site computational project when you can explain why a conserved numerical spectrum would not, by itself, replace any of these analytic checks. The project tests trajectories and numerical error against the exact structure established here.

References

• Moser, Jürgen. “Finitely many mass points on the line under the influence of an exponential potential—an integrable system.” In Dynamical Systems, Theory and Applications, edited by Jürgen Moser, Lecture Notes in Physics 38, pp. 467–497. Springer, 1975. DOI. Open PDF.