Logarithmic derivative is periodic exchange

Fix $u\in \mathbb{C}$. Differentiate $[t(u),t(v)]=0$ from [F3] with respect to $v$ at $v=0$; the entries are polynomials, so this gives $[t(u),t^{\mathrm{\prime }}(0)]=0$. Also $[t(u),t(0)]=0$. Since $t(0)$ is invertible by [F2], multiplying $t(u)t(0)=t(0)t(u)$ on the left and right by $t(0{)}^{-1}$ shows $[t(u),t(0{)}^{-1}]=0$. Therefore $t(u)$ commutes with $t^{\mathrm{\prime }}(0)t(0{)}^{-1}$.

Substitute [F2] into step 1.1 and multiply by the scalar $iJ$. The sum ${\sum }_j{P}_{j,j+1}$ therefore commutes with every $t(u)$, and so does $H_N=J{\sum }_j{P}_{j,j+1}+cI$, since scalar multiples of the identity commute with all endomorphisms.

Each $P_{j,j+1}$ is a real permutation matrix with $P_{j,j+1}^2=I$; therefore $P_{j,j+1}^{\ast }=P_{j,j+1}^{-1}=P_{j,j+1}$, so the real linear combination $H_N$ is Hermitian. On the ordered basis $\mathrm{\mid }\uparrow \uparrow ⟩,\mathrm{\mid }\uparrow \downarrow ⟩,\mathrm{\mid }\downarrow \uparrow ⟩,\mathrm{\mid }\downarrow \downarrow ⟩$ of $V\otimes V$, the swap $P$ fixes the first and fourth vectors and exchanges the middle two. Direct multiplication of the displayed Pauli matrices gives $\begin{array}{rl}{\displaystyle \frac{1}{2}(I+X\otimes X+Y\otimes Y+Z\otimes Z)\mathrm{\mid }\uparrow \uparrow ⟩}& {\displaystyle =\mathrm{\mid }\uparrow \uparrow ⟩,}\\ {\displaystyle \frac{1}{2}(I+X\otimes X+Y\otimes Y+Z\otimes Z)\mathrm{\mid }\downarrow \downarrow ⟩}& {\displaystyle =\mathrm{\mid }\downarrow \downarrow ⟩,}\\ {\displaystyle \frac{1}{2}(I+X\otimes X+Y\otimes Y+Z\otimes Z)\mathrm{\mid }\uparrow \downarrow ⟩}& {\displaystyle =\mathrm{\mid }\downarrow \uparrow ⟩,}\\ {\displaystyle \frac{1}{2}(I+X\otimes X+Y\otimes Y+Z\otimes Z)\mathrm{\mid }\downarrow \uparrow ⟩}& {\displaystyle =\mathrm{\mid }\uparrow \downarrow ⟩\mathrm{.}}\end{array}$ These four vectors form a basis, so the displayed Pauli operator equals $P$.