The linear Sklyanin bracket (9.11) of Lecture 9’s readings is an assumption about a Poisson bracket, so it must be consistent with the Jacobi identity. Working out what that requires produces the classical form of Lecture 13’s equation.

Put three legs in play, $V\otimes V\otimes V$, write $\leg{L}{i}$ for $L$ in leg $i$ at spectral parameter $\lambda_i$, and abbreviate $\leg{\rmat}{ij}=\leg{\rmat}{ij}(\lambda_i-\lambda_j)$. Two facts do the work: $\comm{\leg{\rmat}{ij}}{\leg{L}{k}}=0$ when $k\notin\{i,j\}$, because they act on disjoint legs; and $\leg{\rmat}{21}=-\leg{\rmat}{12}$, which Remark 9.13 showed to be sufficient for antisymmetry of the Poisson bracket but which the bracket does not by itself force. It is therefore a hypothesis of Theorem 14.7, not a consequence of Definition 9.12, and is assumed from here on.

One variant deserves its name in passing. Replacing the zero on the right of (14.12) by a specific $\rmat$-independent term gives the modified classical Yang–Baxter equation, whose solutions generate the same structures over a slightly larger class of Lie algebras. Its theory belongs to Lie-bialgebra and Poisson–Lie theory, which this course does not enter; it is named here so that meeting the phrase elsewhere is not a surprise.

The cancellation is a statement about the three spectral parameters and nothing else, which is why $c$ is free: the normalisation of $\rmat$ is fixed by the Lax matrix it is paired with, not by the CYBE. Lecture 9 fixes it for the periodic Toda chain, $c=-1$ in (9.10); the open chain of Lecture 9’s readings needs a different, constant $\rmat$.

Two further families solve (14.12): the trigonometric $\rmat$-matrices, in which $1/\lambda$ is replaced by $\cot\lambda$ or $\coth\lambda$ and the permutation is deformed, and the elliptic ones, built from Weierstrass functions. Belavin and Drinfeld classified the solutions under hypotheses this course has not set up: for $\rmat$ valued in $\glie\otimes\glie$ with $\glie$ a finite-dimensional simple Lie algebra, in difference form, and non-degenerate in their technical sense (the residue at the simple pole is proportional to the Casimir of an invariant form), every solution is equivalent to a rational, trigonometric or elliptic one cited, and the three degenerate into one another in limits. Those hypotheses are doing real work — dropped, the statement is false, and already fails for $d=1$, where every commutator in (14.12) vanishes and any function of $\lambda$ with a simple pole at the origin solves it. The same trichotomy reappears in the Calogero–Moser potentials (optional lecture B) and in the quantum chains XXX, XXZ and XYZ — it is the same list every time, which is not a coincidence.

The open chain’s constant $\rmat$ of (9.22) (Lecture 9’s readings) is skew, $\perm\rmat\perm=-\rmat$, which is (9.12) for a constant $\rmat$; so Theorem 14.7 applies to it. Does it solve the classical Yang–Baxter equation (14.12)? No. Its left-hand side is $$\comm{\leg{\rmat}{12}}{\leg{\rmat}{13}}+\comm{\leg{\rmat}{12}}{\leg{\rmat}{23}} +\comm{\leg{\rmat}{13}}{\leg{\rmat}{23}} =-\tfrac14\comm{\leg{\perm}{12}}{\leg{\perm}{13}}\;\ne0 .$$(14.13) This identity is cited, not proved here: Babelon–Bernard–Talon, eq. (4.32), state it in operator form, and scripts/check-l9-algebra.py checks it symbolically for $N=2,3$. An equation of this shape, with a fixed non-zero right-hand side in place of the zero, is the modified CYBE named above.

And yet the Jacobi identity holds, as it must: the bracket is the canonical one on $\R^{2N}$. Theorem 14.7 asks only that the left of (14.13) commute with $\leg{L}{1}+\leg{L}{2}+\leg{L}{3}$, and it does, for every matrix $L$: $\leg{\perm}{12}$ commutes with $\leg{L}{1}+\leg{L}{2}$ (it only swaps them) and with $\leg{L}{3}$ (another leg), so it commutes with the sum, and so does $\leg{\perm}{13}$, and so does their commutator. This is Remark 14.9 in a real model: the CYBE is sufficient for Jacobi, and the Toda chain satisfies Jacobi without it.

How unique is $\rmat$? Adding $x\Id+y\perm$ changes nothing in (9.21), since both commute with $\leg{L}{1}+\leg{L}{2}$ ($\perm$ only swaps the two legs). Babelon–Bernard–Talon, §4.7, derive (9.22) rather than check it, and show it is the only skew constant solution; scripts/check-rmatrix.py (V8) finds, at sampled points for $N=3$, no constant solutions beyond $\rmat+x\Id+y\perm$. Of these, the skew one is $\rmat$ itself, which fails the CYBE, while $\rmat\pm\tfrac12\perm$ do solve the CYBE (Babelon–Bernard–Talon, (4.33); checked for $N=2,3$) and are not skew, so Theorem 14.7 as proved does not cover them. So no constant $\rmat$ for the open chain is both skew and a solution of the CYBE: the route “skew plus CYBE, hence Jacobi” is closed here, and Jacobi holds by the weaker condition above. That is not a defect of the model; it is why Lie-bialgebra theory has a modified equation.

Three auxiliary spaces $1,2,3$ at parameters $\lambda_1,\lambda_2,\lambda_3$ give three relations (14.6), one for each pair, with $\Rmat_{ij}=\Rmat_{ij}(\lambda_i-\lambda_j)$.

For a model that satisfies $RTT$ this is automatic — the two routes rewrite one operator. It matters when $RTT$ is taken as the definition of an algebra generated by the entries of $\mono$, as a Sklyanin bracket can be posited (Lecture 9’s readings): then the two routes yield the commutation condition of the proposition among the generators, and $W_A=W_B$ makes it automatic. That is sufficient, not necessary: for $\Rmat=\sigma^{x}\otimes \sigma^{z}$, which fails the Yang–Baxter equation, $W_B=-W_A$ and the condition is again automatic (the check script). It is Remark 14.9 quantised: there the Jacobi identity required a combination of $\rmat$’s only to commute with $\leg{L}{1}+\leg{L}{2}+\leg{L}{3}$, here the associativity of the products requires $W_B^{-1}W_A$ only to commute with the triple products, and for a particular model that is a weaker demand. What makes the Yang–Baxter equation the right thing to impose is, again, that it mentions no model. (The recipe $\Lop{an}(\lambda)=-\Rmat_{an}(-\lambda)$ of (14.7) used $\Rmat_{12}=\Rmat_{21}$; for a solution without that symmetry it is $\Lop{an}(\lambda)=-\Rmat_{na}(-\lambda)$.) For the Heisenberg chain nothing is left to choose: the check script finds that (14.4) fixes $\Rmat$ up to a scalar for one and two sites, at generic $\lambda,\mu$.

YBE $\to$ CYBE. Give each line $i$ a parameter $\lambda_i$ and write $\Rmat_{ij}=\Rmat_{ij}(\lambda_i-\lambda_j)$, so (14.8) is $\Rmat_{12}\Rmat_{13}\Rmat_{23}=\Rmat_{23}\Rmat_{13}\Rmat_{12}$. Suppose $\Rmat=\Id+\ii\hbar\rmat+\hbar^{2}s+O(\hbar^{3})$, in difference form. The terms of order $\hbar$ are $\rmat_{12}+\rmat_{13}+\rmat_{23}$ on both sides. Those of order $\hbar^{2}$ containing $s$ are also the same on both sides; the rest are $(\ii\hbar)^{2}$ times the products of two $\rmat$’s in the order of the factors, so the difference of the two sides at order $\hbar^{2}$ is $$-\hbar^{2}\Big(\comm{\rmat_{12}}{\rmat_{13}}+\comm{\rmat_{12}}{\rmat_{23}} +\comm{\rmat_{13}}{\rmat_{23}}\Big):$$ a solution of the Yang–Baxter equation has a classical $\rmat$ that solves the CYBE (14.12), in the per-leg form of §14.10. For $\Rmat=\lambda-\ii\hbar P$, normalised by $1/\lambda$ (which does not change (14.8), the scalars being the same three on both sides), $\rmat=-\perm/\lambda$ is Proposition 14.10 at $c=-1$. This is the picture Figure 14.8 promised: the classical slide is the first correction to the quantum one proved; the check script repeats it with generic matrices.

The quantum determinant. At $\lambda-\mu=\ii$, $\Rmat(\ii)=\ii(1-P)$ is $2\ii$ times the projector onto the antisymmetric part of $\C^{2}\otimes\C^{2}$, and $RTT$ says that $\mono_a(\mu+\ii)\mono_b(\mu)$ maps the antisymmetric line to itself. The operator by which it multiplies that line (central; for XXX a number) is the quantum determinant, $$\operatorname{qdet}\mono(\mu)=A(\mu+\ii)D(\mu)-B(\mu+\ii)C(\mu),$$ the counterpart of $\det\mono=1$ for Toda (9.3). It commutes with every entry of $\mono$ (Korepin, Bogoliubov and Izergin, §VII.8, in their conventions). For the universal algebra defined by $RTT$ on a normalised formal series $\mono(u)=\Id+\mono^{(1)}u^{-1}+\cdots$, the Yangian $Y(\gllie_2)$, the coefficients of the quantum determinant generate the centre (Molev, ch. 1) cited; a spin chain gives a representation of it. For the Heisenberg chain it is the number $[\mu(\mu+2\ii)]^{N}$ (the check script, $N\le3$). Lecture 15 meets its two factors as the eigenvalues of $A$ and $D$ on the reference state.

Other spins. For a spin $s$ at each site, $\Lop{an}(\lambda)=\lambda+\tfrac{\ii}{2}+\ii\,\boldsymbol\sigma_a\cdot\mathbf S_n$ satisfies (14.4) with the same $\Rmat$, by a verification that uses only the commutation relations of $\mathbf S$ cited: Korepin, Bogoliubov and Izergin, §VIII.4, (4.6)–(4.8), in their conventions; the check script verifies $s=1$. The transfer matrix still commutes with itself; but the regular point is lost, since $\Lop{}(0)$ is no longer a permutation, and a local Hamiltonian needs a transfer matrix with a spin-$s$ auxiliary space (fusion: Kulish, Reshetikhin and Sklyanin 1981; not treated here).

Lecture 8’s and Lecture 9’s readings derived the conservation of $\transfer(\lambda)$ from the zero-curvature equation (8.23), $\dot L_n=V_{n+1}L_n-L_nV_n$, and Lecture 9 proved, separately, the Sklyanin bracket. This lecture quantised only the second. The zero-curvature equation is the equation of motion under one Hamiltonian, and in quantum mechanics equations of motion are not postulated: they follow from $\comm{\cdot}{H}$ once $H$ is known. What has to be quantised is the kinematics, the brackets, and from them the commuting family and its Hamiltonians follow. The site-by-site transport survives as the monodromy (14.5), and the bracket as $RTT$. The Heisenberg equation for $\mono(\lambda)$ under $H$ is then a consequence, and $\comm{H}{\transfer(\lambda)}=0$ is Theorem 14.3, not an input. That is the reason the classical limit of §14.6 was taken in the bracket, not in the equation of motion.

Lecture 12 set the Toda displacement beside Lecture 11’s $\theta'$ as a counterpart. For the Toda pair itself the link can be checked. In Proposition 12.1 the relative coordinate $x=q_1-q_2$ and its momentum $p=\tfrac12(p_1-p_2)$ have $H_{\mathrm{rel}}=p^{2}+e^{x}$, and the relative speed is $v=2p$. Quantised, $$-\hbar^{2}\psi''+e^{x}\psi=p^{2}\psi .$$ With $\hbar=1$ and $k=p$, the solution that decays under the wall is $\psi=K_{2\ii k}(2e^{x/2})$, the modified Bessel function: the chain rule turns this equation into Bessel’s. Far from the wall, $x\to-\infty$, the small-argument form of $K_\nu$ gives cited: DLMF §§10.25–10.27 $$\psi\approx\tfrac12\big[\Gamma(-2\ii k)\,e^{\ii kx}+\Gamma(2\ii k)\,e^{-\ii kx}\big]:$$ an incoming wave $e^{\ii kx}$, reflected with the phase $\Gamma(2\ii k)/\Gamma(-2\ii k)$. Since $\Gamma(-2\ii k)=\overline{\Gamma(2\ii k)}$, the stationary-phase argument of §11.3 puts the incoming packet on $x=2kt+x_w$ and the outgoing one on $x=-2kt+x_w$, with $x_w=2\operatorname{Re}\psi_0(2\ii k)$ and $\psi_0=\Gamma'/\Gamma$ the digamma function: the packet moves as if reflected by a hard wall at $x_w$. In the language of §11.3, with $k=p_1-p_2=2p$ the momentum difference, the reflected wave’s phase is $\theta(k)=2\arg\Gamma(\ii k)$ up to a constant, and $\theta'(k)=2\operatorname{Re}\psi_0(\ii k)=x_w$: the wall is $\theta'$. Restoring $\hbar$ shifts $x$ by $2\ln\hbar$ and replaces $k$ by $p/\hbar$, so $$x_w=2\ln\hbar+2\operatorname{Re}\psi_0(2\ii p/\hbar)\ \longrightarrow\ 2\ln(2p)=2\ln v \qquad(\hbar\to0),$$ by Stirling’s formula, $\operatorname{Re}\psi_0(\ii y)=\ln y+O(y^{-2})$. The classical orbit of Proposition 12.1, $x=\ln v^{2}-v\abs t+o(1)$, has its effective wall exactly there, and that wall is what displaces each momentum line by $\mp2\ln v$. So the classical displacement is the $\hbar\to0$ limit of the quantum one (scripts/check-l16-algebra.py checks the Bessel equation, the asymptotic expansion and the effective-wall limit; the stationary-phase step is argued, not checked; Sutherland §4.2 does the same for the whole hyperbolic family) sketched.