The Toda ring of Lecture 8 has $N$ particles on a circle, $$H=\frac12\sum_{n=1}^{N}p_n^{2}+\sum_{n=1}^{N}e^{q_n-q_{n+1}}, \qquad q_{N+1}=q_1 ,$$(9.1) and its Lax matrix $L$ has a spectrum that does not change in time (Theorem 8.2), so the $N$ numbers $\tr L^{k}$ are conserved. Lecture 8 also rewrote the eigenvalue problem of $L$ as a transport from site to site, $\Phi_{n+1}=L_n(\lambda)\Phi_n$, with the $2\times2$ site matrix $$L_n(\lambda)=\begin{pmatrix}\lambda-p_n&e^{-q_n}\\ -e^{q_n}&0\end{pmatrix}, \qquad \det L_n(\lambda)=1 ,$$(9.2) which contains the variables of site $n$ and nothing else.

Definition 5.5 asks for more than conservation: $N$ integrals that Poisson-commute with each other and are independent. Nothing in the Lax equation says anything about brackets between the $\tr L^{k}$. So the question of this lecture is

The plan has three steps. Build all the integrals at once, as the coefficients of one polynomial $\transfer(\lambda)$ (§9.2). Show that $\pb{\transfer(\lambda)}{\transfer(\mu)}=0$ follows from one identity satisfied by one site matrix (§§9.3 to 9.5). Then independence and Theorem 5.7 do the rest (§§9.6 and 9.7).

Transport all the way round the ring.

The Toda ring is ultralocal, since $L_n$ involves only $(q_n,p_n)$. By (9.2), $\det\mono=1$, and $\mono(\lambda)$ is what a solution of the recursion picks up in one circuit (Figure 9.1).

Expand the product (9.3) as a sum over paths through the index values $1,2$ of the $2\times2$ factors. The entry $(L_n)_{22}=0$ forbids a path to stay at $2$, so each visit to $2$ enters through $(L_n)_{21}=-e^{q_n}$ and leaves at the next site through $(L_{n+1})_{12}=e^{-q_{n+1}}$: a factor $-e^{q_n-q_{n+1}}$ in place of two factors $\lambda-p$. The trace closes the path, so the pair may straddle the join. Hence $$\transfer(\lambda)=\sum_{S}\ \prod_{n\in S}\big(-e^{q_n-q_{n+1}}\big) \prod_{m\notin\bar S}(\lambda-p_m),$$(9.4) the sum over sets $S$ of bonds, no two sharing a site, with $\bar S$ the sites they cover. Writing $\transfer(\lambda)=\sum_{k=0}^{N}c_k\lambda^{N-k}$, $$c_0=1,\qquad c_1=-\sum_np_n,\qquad c_2=\sum_{i<j}p_ip_j-\sum_ne^{q_n-q_{n+1}},\qquad H=\tfrac12c_1^{2}-c_2 .$$(9.5) So $\transfer(\lambda)$ is a polynomial of degree $N$ whose $N$ non-trivial coefficients $c_1,\dots,c_N$ are functions on phase space, and $H$ is one of their functions.

A number $\lambda$ is an eigenvalue of Lecture 8’s Lax matrix $L$ exactly when its eigenvalue recursion has a solution with $\psi_{n+N}=\psi_n$. The gauge that turned the recursion into $\Phi_{n+1}=L_n(\lambda)\Phi_n$ respects periodicity, since $q_{n+N}=q_n$, so this happens exactly when $\mono(\lambda)$ has an eigenvalue $1$: when $\det\big(\Id-\mono(\lambda)\big)=2-\transfer(\lambda)=0$, using $\det\mono=1$. For $N\ge3$, $\det(\lambda-L)$ and $\transfer(\lambda)-2$ are both monic of degree $N$. Where the spectrum of $L$ is simple they have the same $N$ roots, so they are equal there. Such points are dense: at fixed $q$, momenta far enough apart make the discs of Gershgorin’s theorem disjoint and the spectrum simple, so the discriminant of $\det(\lambda-L)$ is a non-zero polynomial in $p$, and its non-zero set is dense. Both sides are continuous in $(q,p)$, so sketched $$\det(\lambda-L)=\transfer(\lambda)-2 .$$(9.6) The $c_k$ are therefore, up to sign and the constant $2$, the elementary symmetric functions of the eigenvalues of $L$. Newton’s identities turn them into the $\tr L^{k}$ and back. They are Lecture 8’s integrals, reorganised, and they are the candidates for Definition 5.5. Everything now turns on their brackets.

$L_n(\lambda)$ is a $2\times2$ matrix whose entries are functions on phase space. What we need is the collection of all their brackets, $\pb{(L_n)_{ab}(\lambda)}{(L_n)_{cd}(\mu)}$, sixteen functions at once. The trick is to package them as a single $4\times4$ matrix acting on $V\otimes V$, with $V=\C^{2}$.

The subscript is a leg, not a matrix index: $\leg{A}{1}$ is the whole matrix $A$, placed in the first factor (Figure 9.2). The right-hand form of (9.7) is the one to memorise: the entry in row $(a,c)$ and column $(b,d)$ is the bracket of the $(a,b)$ entry with the $(c,d)$ entry.

One matrix on $V\otimes V$ appears everywhere below.

Here is the structural assumption that will force involution. It is a statement about the brackets of one site matrix at two values of the spectral parameter, and it involves one fixed numerical matrix.

The left-hand side is sixteen unknown brackets. The right-hand side says they are all determined by the matrices themselves and one numerical $\rmat$, which is slid past the product $\leg{L}{1}\leg{L}{2}$ from both sides (Figure 9.3). Why this form and not another is the content of the next theorem: it is the form that survives multiplication.

So the whole question of involution has been reduced to one identity for one $2\times2$ matrix. That is where the work of this lecture is.

Read the first displayed line as the tensor-leg drill: it is the sixteen brackets of (9.2), of which four are non-zero, written as one $4\times4$ matrix. If that line is not comfortable, nothing after it will be. As an exercise, find the four non-zero brackets and assemble the matrix.

The proof of the second theorem has four steps, and only the first two use the Toda chain at all. Write $x_n=e^{q_n-q_{n+1}}$ for the springs.

Step (i) is the only place the ring matters. On the open chain nothing ties the springs together, a single spring can be made as weak as one likes at fixed energy, and the level sets stay unbounded even after the drift is removed: the open chain scatters instead of winding.

The motion. On every regular level set the flow of $H$ is a straight line in coordinates on $T^{N-1}\times\R$: it winds round the torus while the centre of mass moves along $\R$ at the constant speed $\sum_np_n/N$, possibly zero. For $N=3$, with the centre of mass removed, the tori meet the Poincaré section of Figure 7.5 in its closed curves: the numerical plot of Lecture 7, beside the chaos of Hénon–Heiles at the same energy, illustrates this theorem at $E=1/6$.

Action–angle variables need one more step. Separate the centre of mass canonically, fix the total momentum, and $N-1$ relative degrees of freedom remain, whose regular level sets are the compact tori of Theorem 9.11. To those Theorem 6.5 applies. As an exercise, carry out the separation for $N=2$ with $Q=(q_1+q_2)/2$, $2\bar p=p_1+p_2$, $x=q_1-q_2$ and $p_x=(p_1-p_2)/2$; the two-particle ring at the end of the readings does it. Computing the actions needs the curve $z+1/z=\transfer(\lambda)$ of the reading below, with one action for each of its $N-1$ gaps cited: Flaschka and McLaughlin 1976.

The method. Three moves did everything, and none of them is special to Toda:

1. one identity, (9.9), for one site matrix;

2. composition along the chain, Theorem 9.7, which needs only ultralocality;

3. the trace, which kills a commutator.

Change the site matrix and the $\rmat$-matrix, and the same three moves give commuting integrals for another chain. Lecture 14 replaces the Poisson bracket by the commutator and $\rmat$ by an $\Rmat$-matrix with $\Rmat=\Id+\ii\hbar\,\rmat+O(\hbar^{2})$. The same three moves then give the commuting transfer matrices of the Heisenberg spin chain, and on the quantum Toda chain their limit $\hbar\to0$ is exactly (9.9) with (9.10). That is goal G5, and the one row of the classical-to-quantum dictionary that is a theorem rather than an analogy.

Three things a careful reader should know, collected here so that the arguments above can run without interruption. (The linear bracket, which this lecture does not need, is in the reading.)

1. Independence is generic, not universal. Theorem 9.10 proves independence on a dense open set, and that is all Definition 5.5 asks. It does fail somewhere: at the equilibrium $p_n=0$, $x_n=1$ every flow is a drift.

2. The ring drifts. The flow of $c_1=-\sum_np_n$ translates every $q_n$ and changes no spring, so no level set is compact. Tori appear only after the centre of mass is fixed, which is why Theorem 9.11 is stated for slices.

3. The sign of $\rmat$ belongs to the site matrix. Other books use the site matrix with $e^{\pm q}$ exchanged, which gives $\rmat=+\perm/\lambda$; mixing a sign with the wrong site matrix breaks (9.9).

Conservation directly. (Lecture 8’s reading has the fuller version, with the plaquette picture.) Involution gave conservation for free, but $\transfer(\lambda)$ is also conserved for a more geometric reason. With $V_n(\lambda)=\big(\begin{smallmatrix}-\lambda/2&-e^{-q_n}\\ e^{q_{n-1}}&\lambda/2\end{smallmatrix}\big)$, the equations of motion of the ring are equivalent to $\dot L_n=V_{n+1}L_n-L_nV_n$ for every $n$ and every $\lambda$: transport in space and in time commute round every plaquette of the space-time lattice, a discrete zero-curvature condition. Differentiating the product (9.3), the sum telescopes to $\dot\mono=V_{N+1}\mono-\mono V_1 =\comm{V_1}{\mono}$, a Lax equation for the $2\times2$ matrix $\mono$, and $\transfer=\tr\mono$ is conserved. Periodicity is used once, to identify $V_{N+1}$ with $V_1$.

The route not taken. One can prove integrability of the open chain instead: its $N\times N$ Lax matrix satisfies the linear bracket with a constant $\rmat$, the involution theorem for that bracket gives $\pb{\tr L^{j}}{\tr L^{k}}=0$, and independence follows from a band lemma for tridiagonal matrices. That route reaches the same kind of theorem for a different chain, and §§9.11 and 9.12 take it. Its bracket does not compose, so it cannot be quantised site by site, and it is the ring’s quadratic bracket that Lecture 14 quantises.

The spectral curve. Setting $\det\big(z-\mono(\lambda)\big)=0$ gives $z+1/z=\transfer(\lambda)$, since $\det\mono=1$. Its $N-1$ finite gaps are the bounded intervals of real $\lambda$ where $\abs{\transfer(\lambda)}>2$, and when all are open they carry the $N-1$ action variables. §9.13 builds the curve from Bloch waves and proves these statements, or cites them.

The core proved the ring’s integrals commute through the quadratic bracket, because it composes. The older route goes through a bracket linear in $L$, for a single Lax matrix; the open-chain reading below applies it.

Read what (9.11) asserts. On the left is an arbitrary $d^{2}\times d^{2}$ matrix of Poisson brackets, $d^{4}$ functions on phase space with no reason to be related. On the right is a commutator of one fixed numerical matrix with something built linearly from $L$ itself. The assumption is enormously restrictive, and its whole force is that the same $\rmat$ works at every point of phase space and for every pair $\lambda,\mu$.

Two qualifications:

• Numerical. If $\rmat$ is allowed to depend on the dynamical variables the bracket is called dynamical, and optional lecture B meets the first real example — Calogero–Moser has no $\rmat$ of the constant form. Everything proved in the involution theorem below survives that generalisation; the classical Yang–Baxter equation of Lecture 14’s reading does not.

• Which $\rmat$? Part I teaches how to read and check an $\rmat$-matrix, never how to solve for one. Finding $\rmat$ from scratch is outside it.

The qualification is needed because the $\rmat$-matrices that occur have a pole at coincident arguments — the rational one is $c\perm/(\lambda-\mu)$ — so (9.11) is an assumption about $\lambda\ne\mu$ and says nothing at $\lambda=\mu$. Nothing is lost: the left of (9.13) is continuous in $\lambda$ and $\mu$ whenever $L$ is, and a continuous function vanishing off the diagonal vanishes on it too.

The proof is three steps and each is short.

This pays Lecture 8’s debt by the route not taken. Note how little was spent: the argument never opened $\rmat$, never used the Lax equation, and never mentioned a Hamiltonian. It also never used the particular combination $\leg{L}{1}+\leg{L}{2}$ — only that it commutes with $A$.

Theorem 9.14 is one-way: a Sklyanin bracket gives involution. The converse is true but weaker than one would like.

Read (9.15) carefully: it is not (9.11). Two independent matrices appear, and the single-commutator form is the special case $\leg{\rmat}{21}=-\leg{\rmat}{12}$. The converse does not deliver that special case, and the difference is not a technicality — it is the difference between a true statement and a false one.

So an $\rmat$-matrix always exists behind involution, but it need not be numerical and it need not put the bracket in the single-commutator form of Definition 9.12. The classification of solutions of the classical Yang–Baxter equation (Lecture 14’s reading) applies to neither. Optional lecture B meets a dynamical $\rmat$ in Calogero–Moser, so the caveat is not hypothetical.

The open Toda chain is the ring with its closing spring removed: $H=\frac12\sum_{n=1}^{N}p_n^{2}+\sum_{n=1}^{N-1}e^{q_n-q_{n+1}}$, the chain whose scattering Lecture 12 solves. This reading proves it Liouville integrable with the linear bracket, and shows what the torus theorem does and does not give when every fibre is non-compact.

The phase space is $\R^{2N}$ with coordinates $(q,p)$, so $n=N$ degrees of freedom. §8.4 put the equations of motion in Lax form, with Flaschka’s variables (8.4), $a_n=e^{(q_n-q_{n+1})/2}$ and $b_n=p_n$, and $$L=\begin{pmatrix} b_1&a_1&&\\ a_1&b_2&\ddots&\\ &\ddots&\ddots&a_{N-1}\\ &&a_{N-1}&b_N \end{pmatrix}, \qquad M_{n,n+1}=-\tfrac12a_n=-M_{n+1,n},\qquad \dot L=\comm{M}{L}$$(9.16) (Proposition 8.3). By Theorem 8.2 the eigenvalues of $L$ are conserved, and so are the $N$ functions $$F_k=\tr L^{k},\qquad k=1,\dots,N,$$(9.17) which begin $F_1=\sum_np_n$, $F_2=2H$ (8.8). They are our candidates for Definition 5.5, with $H=\tfrac12F_2$.

Three facts about $L$ will be used over and over.

1. (L1)

   Simple spectrum. Every $a_n>0$, so $L$ is a real symmetric (Jacobi) matrix with $N$ distinct real eigenvalues $\lambda_1<\dots<\lambda_N$ (proof of Proposition 8.3). Here $\lambda_k$ are eigenvalues, not a spectral parameter: the open chain has none.

2. (L2)

   Bounded entries. For a real symmetric matrix $\abs{L_{ij}}\le\max_k\abs{\lambda_k}$, since $\abs{L_{ij}}=\abs{e_i\cdot Le_j}$ is at most the largest stretching factor of $L$. The eigenvalues are conserved, so along any motion every entry of $L$ stays bounded by a constant fixed at $t=0$.

3. (L3)

   Brackets. Since $\pb{f}{p_m}=\partial f/\partial q_m$, $$\pb{a_n}{b_m}=\tfrac12a_n\,(\delta_{nm}-\delta_{m,n+1}),\qquad \pb{a_n}{a_m}=\pb{b_n}{b_m}=0 .$$(9.18) The bond variable $a_n$ is shifted only by the momenta at its two ends, with opposite signs.

For $N=3$ the traces are $F_1=\sum p_n$, $F_2=\sum p_n^{2}+2(a_1^{2}+a_2^{2})$ and $F_3=\sum p_n^{3}+3a_1^{2}(p_1+p_2)+3a_2^{2}(p_2+p_3)$. Their derivatives with respect to the momenta alone form a $3\times3$ matrix with determinant $$\det\frac{\partial(F_1,F_2,F_3)}{\partial(p_1,p_2,p_3)} =6\Big[(p_2-p_1)(p_3-p_1)(p_3-p_2)-a_1^{2}(p_2-p_1)-a_2^{2}(p_3-p_2)\Big].$$(9.19) When the particles are far apart, or their momentum differences large compared with the $a_n$ (a common boost changes nothing), the $a_n^{2}$ terms are negligible, the $F_k$ are nearly the power sums $\sum p_n^{k}$, and (9.19) is six times the Vandermonde determinant of the momenta: non-zero as soon as the momenta are distinct. That is why the far-apart (or large momentum-separation) regime is the right place to look first: there the chain is three free particles, and the three integrals are visibly three different functions.

But (9.19) is not a proof of anything everywhere. It vanishes when all three momenta are equal, and also at $p=(0,1,2)$, $a_1=a_2=1$, where the momenta are distinct. At such points independence has to come from the $q$-derivatives. The general proof uses both, and is organised by $L$.

Both conditions on $X$ are needed: $X=\Id$ commutes with $L$ and has a zero off-diagonal; and for $N=2$ with $b_1=b_2$, $X=E_{12}+E_{21}$ commutes with $L$ and has a zero diagonal.

The Vandermonde determinant of the $N=3$ example reappears at the end, with the momenta replaced by the eigenvalues, which are distinct at every point. That upgrades “independent where the momenta are distinct” to “independent everywhere”. Definition 5.1 asks only for a dense open set; §9.12 needs the stronger statement.

Theorem 9.14 gives $\pb{\tr L^{j}}{\tr L^{k}}=0$ as soon as $L$ satisfies a Sklyanin bracket (9.11). The open chain’s $L$ has no spectral parameter, so we look for a constant matrix $\rmat$ on $\R^{N}\otimes\R^{N}$ with $$\pb{\leg{L}{1}}{\leg{L}{2}}=\comm{\rmat}{\leg{L}{1}+\leg{L}{2}} .$$(9.21)