Exchange preserves total Lie spin

For any endomorphism $A$ of $V={\mathbb{C}}^2$, the two-factor flip obeys $P(A\otimes I)P=I\otimes A$ and $P(I\otimes A)P=A\otimes I$. Hence $P(A\otimes I+I\otimes A)=(A\otimes I+I\otimes A)P$. The same identity on factors $i,j$, while $P_{ij}$ commutes with every $A_k$ for $k\mathrm{\ne }i,j$, proves the three displayed commutators for $A=E,F,H\mathrm{/}2$. This is a direct matrix identity under the inherited assumption [A1] and the spin definition [F1].

For $N\ge 3$, every term in $H_N$ from [F3] is a real scalar multiple of a swap $P_{ij}$, including the periodic bond $P_{N,1}$, and the identity term commutes with every endomorphism. Step 1.1 therefore gives $[H_N,S_N^{+}]=[H_N,S_N^{-}]=[H_N,S_N^z]=0$. Each swap is a unitary involution on the standard tensor-product inner product, hence $P_{ij}^{\ast }=P_{ij}$; the real linear combination $H_N$ is Hermitian.

On $V_a\otimes {\mathcal{H}}_N$, the flip $P_{aj}$ commutes with $A_a+{\sum }_{k=1}^N{A}_k$ for each $A\in \{E,F,H\mathrm{/}2\}$: the two-factor flip identity from step 1.1 applies to $A_a+A_j$, and $P_{aj}$ commutes with $A_k$ for $k\mathrm{\ne }j$. Thus $R_{aj}(u)=uI+i{P}_{aj}$ commutes with this total generator. A product of operators commuting with a fixed operator also commutes with it, since $[XY,G]=X[Y,G]+[X,G]Y$. Hence $T_a(u)$ commutes with $A_a+{\sum }_k{A}_k$.

The partial trace defined in the Statement is cyclic for an endomorphism acting only on $V_a$. In an auxiliary basis, write $X=(X_{rs})$ in chain-operator blocks and $A=(A_{rs})$ in scalar entries. Then ${\mathrm{tr}}_a((A_a\otimes I)X)={\sum }_{r,s}{A}_{rs}{X}_{sr}$ and ${\mathrm{tr}}_a(X(A_a\otimes I))={\sum }_{r,s}{X}_{rs}{A}_{sr}$; relabelling $r,s$ in the second sum and commuting scalar coefficients shows equality. Taking ${\mathrm{tr}}_a$ of $[A_a+{\sum }_k{A}_k,T_a(u)]=0$ therefore gives $[{\sum }_k{A}_k,{\mathrm{tr}}_a{T}_a(u)]=0$: the chain generators act only on the untraced tensor factors, so their commutator passes through ${\mathrm{tr}}_a$. With $t(u)$ as in [F2], this is each asserted transfer-matrix commutator.

By [F1], $S_N^z$ is diagonal in the tensor basis with eigenvalues $-N\mathrm{/}2,-N\mathrm{/}2+1,\dots ,N\mathrm{/}2$, so its weight spaces form an orthogonal direct sum of ${\mathcal{H}}_N$. Since $[H_N,S_N^z]=0$ by step 2.1, each ${\mathcal{H}}_{N,m}$ is invariant under $H_N$ and under every power $H_N^r$. The series defining $\exp (-\beta H_N)$ converges in operator norm by [F4]. Its partial sums preserve every ${\mathcal{H}}_{N,m}$, and their limits do as well because these subspaces are closed. The exponential is therefore block diagonal for the orthogonal decomposition ${\mathcal{H}}_N={⨁}_m{\mathcal{H}}_{N,m}$. In a basis formed by concatenating orthonormal bases of the blocks, its matrix is block diagonal, so its trace is the sum of the traces of the restrictions. Basis independence [F5] gives the stated identity.