RTT gives commuting periodic transfer matrices

Put $s=u-v$. Since $P_{ab}^2=I$ and the mixed terms cancel, $R_{ab}(s)R_{ab}(-s)=-(s^2+1)I$; hence $R_{ab}(s)$ is invertible for $s\mathrm{\ne }i,-i$. By [F3], $T_b(v)T_a(u)=R_{ab}(s)T_a(u)T_b(v)R_{ab}(s{)}^{-1}$. The factor $R_{ab}$ acts only on the finite auxiliary space $V_a\otimes V_b$, so the ordinary finite trace over that space is cyclic even with operator-valued coefficients on $V_N$. Also ${\mathrm{tr}}_{ab}(T_a(u)T_b(v))=t(u)t(v)$ and ${\mathrm{tr}}_{ab}(T_b(v)T_a(u))=t(v)t(u)$, by expanding each auxiliary matrix in matrix units. Taking ${\mathrm{tr}}_{ab}$ therefore proves $[t(u),t(v)]=0$ whenever $u-v\mathrm{\ne }\pm i$.

Each entry of $t(u)$ is a polynomial in $u$, because it is the trace of a finite product of matrices whose entries are linear polynomials. Thus every entry of $[t(u),t(v)]$ is a polynomial in $(u,v)$. For each fixed $v$, it vanishes for every $u$ except at most $v+i$ and $v-i$; a one-variable polynomial with infinitely many zeros is zero. This holds for every $v$, so $[t(u),t(v)]=0$ for all complex $u,v$, including the two singular differences.

By [F2], $t(0)$ is invertible; hence $M(u)=t(0{)}^{-1}t(u)$ is analytic near $0$ and $M(0)=I$. Choose a submultiplicative matrix norm. Continuity of $M$ gives a neighborhood of $0$ where $\mathrm{\parallel }M(u)-I\mathrm{\parallel }<1$, and there define $\log M(u)={\sum }_{n\ge 1}(-1{)}^{n+1}(M(u)-I{)}^n\mathrm{/}n$, a convergent matrix power series. Step 2.1 implies $t(u)$ commutes with $t(w)$ and every $t(v)$; since $t(0)$ is invertible, $t(0{)}^{-1}$ commutes with them as well. Thus $M(u)$, $M(w)$, and each $t(v)$ commute. Their logarithmic power series preserve these commutations. Comparing the convergent two-variable Taylor coefficients proves the coefficient matrices commute pairwise and with every $t(v)$.