Open Toda integrability: from Lax invariants to commuting integrals

Why do the spectral invariants of the open Toda chain establish Hamiltonian integrability? A Lax equation proves conservation, but Liouville integrability also needs enough functionally independent integrals that commute under the physical Poisson bracket. We prove all three statements for the finite open chain, then check them explicitly for three particles. This distinction is useful whenever a matrix representation is proposed as evidence of integrability.

Required background. Use Hamilton’s equations and the canonical Poisson bracket, as in derive the open Toda equations, and multiply matrices, as in build the Toda Lax pair. Helpful background. Extract spectral invariants introduces traces and characteristic polynomials; check involution and independence supplies guided practice with Jacobians.

The open Toda Hamiltonian and the claim

Use the dimensionless convention of the classical Toda model. For fixed $N\geq2$, the phase space is $\mathbb R^{2N}$ with coordinates $(q_1,\ldots,q_N,p_1,\ldots,p_N)$ and bracket

$$\{f,g\}=\sum_{i=1}^{N} \left( \frac{\partial f}{\partial q_i}\frac{\partial g}{\partial p_i} -\frac{\partial f}{\partial p_i}\frac{\partial g}{\partial q_i} \right).$$

The equal masses and positive interaction scale have been normalized to one. The Hamiltonian and equations are

$$\begin{aligned} H&=\frac12\sum_{i=1}^{N}p_i^2 +\sum_{i=1}^{N-1}e^{q_i-q_{i+1}},\\ \dot q_i&=p_i,\qquad \dot p_i=a_{i-1}^2-a_i^2,\\ a_i&=e^{(q_i-q_{i+1})/2}\gt0\quad(1\leq i\leq N-1), \qquad a_0=a_N=0. \end{aligned}$$

The zero endpoint bonds specify an open chain with freely moving endpoints, not periodic closure. The defining Hamiltonian agrees with Moser 1975, § 1, p. 467, equations (1.1)–(1.3), PDF.

Let $L$ be the real symmetric tridiagonal matrix with $L_{ii}=p_i$ and $L_{i,i+1}=L_{i+1,i}=a_i$. Define $I_k=\operatorname{tr}(L^k)/k$.

Finite open Toda integrability. The functions $I_1,\ldots,I_N$ are conserved, satisfy $\{I_j,I_k\}=0$ everywhere, and are functionally independent on an open dense subset of $\mathbb R^{2N}$. Since $I_2=H$, these are $N$ commuting integrals for a system with $N$ degrees of freedom. The argument below proves exactly this statement; it does not assume compact invariant levels or assert global action–angle coordinates.

The Lax equation, entry by entry

Set $b_i=p_i$. Differentiating the positive bond variables yields

$$\dot a_i=\frac12a_i(b_i-b_{i+1}),\qquad \dot b_i=a_{i-1}^2-a_i^2.$$

Define $B$ by

$$B_{i,i+1}=-\frac12a_i,\qquad B_{i+1,i}=\frac12a_i,$$

with all other entries zero. Direct matrix multiplication gives

$$\begin{aligned} [B,L]_{ii}&=a_{i-1}^2-a_i^2,\\ [B,L]_{i,i+1}&=\frac12a_i(b_i-b_{i+1}),\\ [B,L]_{i,i+2}&=-\frac12a_ia_{i+1} +\frac12a_ia_{i+1}=0. \end{aligned}$$

Entries outside this band vanish because both matrices are tridiagonal. The lower-triangular entries follow by symmetry. Therefore $\dot L=[B,L]$. Checking the second off-diagonal matters: a proposed matrix flow must preserve the tridiagonal form, not just reproduce its diagonal entries.

Conversion to Moser’s normalization

Moser uses $a_i^{\mathrm M}=\tfrac12e^{(q_i-q_{i+1})/2}$ and $b_i^{\mathrm M}=-\tfrac12p_i$. With $D=\operatorname{diag}(1,-1,1,-1,\ldots)$, the matrix conversion is

$$L^{\mathrm M}=-\frac12DLD,\qquad B^{\mathrm M}=DBD.$$

It preserves the same physical time and $\dot L^{\mathrm M}=[B^{\mathrm M},L^{\mathrm M}]$. This maps the signs, factors, and spectrum together; changing only a displayed off-diagonal sign would not be a valid conversion. See Moser 1975, § 2, p. 470, equation (2.1), and pp. 472–473, equation (2.7), PDF.

Conservation follows from the commutator

For each positive integer $k$, cyclicity of the finite matrix trace gives

$$\begin{aligned} \frac{dI_k}{dt} &=\operatorname{tr}\bigl(L^{k-1}[B,L]\bigr)\\ &=\operatorname{tr}(L^{k-1}BL)-\operatorname{tr}(L^kB)=0. \end{aligned}$$

In particular,

$$I_1=\sum_i p_i=P,\qquad I_2=\frac12\sum_i p_i^2+\sum_{i=1}^{N-1}a_i^2=H.$$

One can also solve $\dot U=BU$, $U(0)=\mathbf1$. Since $B^T=-B$, the matrix $U$ stays orthogonal, and differentiating $U^{-1}LU$ gives zero. Thus

$$L(t)=U(t)L(0)U(t)^{-1}.$$

The characteristic polynomial and eigenvalues are conserved as well. This is the finite isospectral argument in Moser 1975, § 2, p. 473, equations (2.7)–(2.8), PDF. It has so far proved conservation only; trace cyclicity does not, by itself, calculate a canonical Poisson bracket.

The spectral Hamiltonians commute

We now use the physical bracket to prove involution for arbitrary finite $N$. Introduce an auxiliary Hamiltonian flow generated by $I_m$, so $\partial_{t_m}f=\{f,I_m\}$. This is a proof device; physical Toda time is $t_2$.

Let $A=L^{m-1}$. The variation identity

$$dI_m=\operatorname{tr}(A\,dL)$$

implies

$$\frac{\partial I_m}{\partial b_i}=A_{ii},\qquad \frac{\partial I_m}{\partial a_i}=2A_{i,i+1}.$$

The factor two counts the two equal off-diagonal entries of $L$. Hamilton’s equations and the dependence of $a_i$ on position differences now give

$$\begin{aligned} \partial_{t_m}b_i &=a_{i-1}A_{i-1,i}-a_iA_{i,i+1},\\ \partial_{t_m}a_i &=\frac12a_i(A_{ii}-A_{i+1,i+1}). \end{aligned}$$

Terms whose indices lie outside the matrix are omitted. These equations were obtained from the canonical bracket, not inferred from conservation.

A Lax representation for each Hamiltonian flow

Denote the strictly upper- and lower-triangular parts of $A$ by $A_+$ and $A_-$. Define

$$B_m=\frac12(A_- - A_+).$$

Because $A$ is symmetric, $B_m$ is skew-symmetric. We claim that the canonical flow just computed satisfies

$$\partial_{t_m}L=[B_m,L].$$

Here are the necessary entry checks. The diagonal of the commutator is $a_{i-1}A_{i-1,i}-a_iA_{i,i+1}$. For the first upper diagonal, use $[A,L]=0$, which holds because $A$ is a power of $L$:

$$\begin{aligned} 0=[A,L]_{i,i+1} &=a_i(A_{ii}-A_{i+1,i+1})\\ &\quad +(b_{i+1}-b_i)A_{i,i+1}\\ &\quad +a_{i+1}A_{i,i+2}-a_{i-1}A_{i-1,i+1}. \end{aligned}$$

Multiplication of $B_m$ and $L$ gives minus one half of the last three terms, hence

$$[B_m,L]_{i,i+1}=\frac12a_i(A_{ii}-A_{i+1,i+1}).$$

For $j\gt i+1$, every contributing entry of $B_m$ is strictly upper-triangular; consequently

$$[B_m,L]_{ij}=-\frac12[A,L]_{ij}=0.$$

Symmetry gives the remaining entries. This proves the claimed Lax representation without importing a different Poisson structure.

It follows for all positive integers $k,m$ that

$$\{I_k,I_m\} =\partial_{t_m}I_k =\operatorname{tr}\bigl(L^{k-1}[B_m,L]\bigr)=0.$$

The new ingredient is a Hamiltonian Lax representation for every $I_m$, with respect to the same canonical bracket. A Lax representation for $H$ alone would not establish the result.

Why the matrix variables have one fewer coordinate

The induced brackets are

$$\{a_i,b_j\}=\frac12a_i(\delta_{ij}-\delta_{i+1,j}), \qquad \{a_i,a_j\}=\{b_i,b_j\}=0.$$

These $2N-1$ variables omit the common position translation. They are not canonical coordinates on the original $2N$-dimensional space. The function $P=\sum_i b_i$ is a Casimir of this induced bracket: its bracket with every $a_i,b_i$ vanishes. At fixed $P$, the induced bracket has rank $2N-2$, because the nearest-neighbor difference matrix has rank $N-1$.

There is no contradiction with $P$ generating a nonzero canonical flow. On the original phase space, $\{q_i,P\}=1$ translates all positions; its effect on $L$ is zero. Accordingly, $B_1=0$. We count $N$ integrals on the original phase space throughout this proof.

Generic independence requires a separate argument

Consider the $N\times N$ momentum minor of the full Jacobian of $(I_1,\ldots,I_N)$:

$$M_{kj}=\frac{\partial I_k}{\partial p_j} =(L^{k-1})_{jj},\qquad 1\leq k,j\leq N.$$

Choose distinct real momenta $p_1,\ldots,p_N$ and set every $a_i=\varepsilon\gt0$. This is an allowed configuration: for example,

$$q_i=2(N-i)\log\varepsilon$$

has precisely these bond variables. As $\varepsilon\to0^+$ at fixed momenta, $L$ tends to $\operatorname{diag}(p_1,\ldots,p_N)$. Therefore

$$\det M\longrightarrow \det[p_j^{k-1}]_{k,j=1}^{N} =\prod_{1\leq i\lt j\leq N}(p_j-p_i)\neq0.$$

For sufficiently small but strictly positive $\varepsilon$, the determinant is already nonzero at a finite phase-space point. The limiting configuration with zero bonds need not itself belong to the phase space.

Finally, $\det M$ is real analytic on the connected space $\mathbb R^{2N}$. A real-analytic function that is nonzero somewhere cannot vanish on an open subset. Its nonzero set is consequently open and dense. On that set, the full Jacobian has rank $N$, proving generic functional independence.

This proof supplies the integrability claim stated above. It does not classify every possible zero of this particular minor. If a chosen minor vanishes, another minor can still demonstrate full rank, as the next example shows.

Three integrals for three particles

For $N=3$, abbreviate $u=e^{q_1-q_2}$ and $v=e^{q_2-q_3}$. These are positive bond strengths squared. Expanding the traces gives

$$\begin{aligned} P&=p_1+p_2+p_3,\\ H&=\frac12(p_1^2+p_2^2+p_3^2)+u+v,\\ J:=I_3&=\frac13(p_1^3+p_2^3+p_3^3) +u(p_1+p_2)+v(p_2+p_3). \end{aligned}$$

The terms proportional to $u,v$ in $J$ are essential. The sum of momentum cubes alone is not conserved.

Direct Poisson-bracket check

Simultaneously translating all $q_i$ leaves $u,v$ unchanged. Since $\{P,f\}=-\sum_i\partial_{q_i}f$, we have $\{P,H\}=\{P,J\}=0$.

For the remaining pair, the gradients are

$$\begin{aligned} \nabla_q H&=(u,v-u,-v),\\ \nabla_p H&=(p_1,p_2,p_3),\\ \nabla_p J&=(p_1^2+u,\ p_2^2+u+v,\ p_3^2+v),\\ \partial_{q_1}J&=u(p_1+p_2),\\ \partial_{q_2}J&=-u(p_1+p_2)+v(p_2+p_3),\\ \partial_{q_3}J&=-v(p_2+p_3). \end{aligned}$$

Their two contractions agree:

$$\begin{aligned} \nabla_q H\cdot\nabla_p J &=u(p_1^2-p_2^2)+v(p_2^2-p_3^2),\\ \nabla_p H\cdot\nabla_q J &=u(p_1^2-p_2^2)+v(p_2^2-p_3^2). \end{aligned}$$

The quadratic bond terms in the first line cancel because $u^2+(v-u)(u+v)-v^2=0$. Thus $\{H,J\}=0$ exactly for arbitrary real positions and momenta, not merely at one initial condition. In particular $\dot J=\{J,H\}=0$.

The shared initial condition and a rank pitfall

At $q=(0,0,0)$ and $p=(1,0,-1)$, one finds

$$P=0,\quad H=3,\quad J=0,\qquad L=\begin{pmatrix}1&1&0\\1&0&1\\0&1&-1\end{pmatrix}.$$

Its characteristic polynomial is $\lambda^3-3\lambda$, with eigenvalues $-\sqrt3,0,\sqrt3$. To check independence at this point, use columns $(p_1,p_2,q_2)$ of the full Jacobian:

$$\left.\frac{\partial(P,H,J)}{\partial(p_1,p_2,q_2)}\right|_{(q,p)} =\begin{pmatrix} 1&1&0\\ 1&0&0\\ 2&2&-2 \end{pmatrix},\qquad \det=2.$$

The three differentials are therefore independent here. In contrast, the momentum-only minor has rows $(1,1,1)$, $(1,0,-1)$, and $(2,2,2)$, so its determinant vanishes. Independence is a property of the full differentials, not of one preferred choice of columns. Nor does the value $J=0$ mean that the differential $dJ$ vanishes.

What has been proved, and what comes next

The chain has $N$ canonical degrees of freedom, $N$ conserved traces including its Hamiltonian, pairwise zero Poisson brackets, and generic functional independence. Those are the ingredients of the finite-dimensional Liouville-integrability claim made here.

Compactness does not follow. Every joint level contains the line $q_i\mapsto q_i+c$, $c\in\mathbb R$, and the endpoint momentum satisfies $\dot p_1=-a_1^2\lt0$. We cannot apply the compact-torus conclusion of Liouville–Arnold to these full open-chain levels. A reconstruction of arbitrary trajectories from spectral data, including the non-spectral center coordinate, requires further work; Moser gives a finite inverse spectral construction in § 3, pp. 475–480, PDF.

Closing the endpoint bond or passing to an infinite chain changes the problem and invalidates the present endpoint and finite-trace arguments as written. Likewise, numerical conservation of several quantities would be a useful implementation check, but it would not replace the identities and independence proof above.

For practice, return to check involution and independence. For a reproducible trajectory and numerical error checks, continue to the three-particle Toda project. The open Toda sequence puts those steps in order, and the Lax reference collects the matrix convention.

References

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