Suzuki recursive cancellation of error terms

Put $A_j=-i{H}_j$ and $A={\sum }_j{A}_j$. The base word $S_2(h)$ is palindromic by [F1], so [F3] gives $S_2(-h)=S_2(h{)}^{-1}$; the same identity holds for $U(h)$ by [F3] and [F4]. Its derivative at zero is $A$, because the two half-time occurrences of every outer $A_j$ sum to one full occurrence, while the middle term occurs once. Also $U(0)=S_2(0)=I$. By [L2], $E(h)=S_2(h)-U(h)=C{h}^2+O(h^3)$ for some matrix $C$. The inverse identities imply $E(-h)=-S_2(h{)}^{-1}E(h)U(h{)}^{-1}$; comparing degree-two terms gives $C=-C$, hence $C=0$. Thus $E(h)=O(h^3)$, proving the base order-two claim as well as the symmetry base case.

Fix $j\ge 2$ and assume as the induction hypothesis [ih] that $S_{2j-2}(-h)=S_{2j-2}(h{)}^{-1}$ and that $S_{2j-2}$ agrees with $U$ through degree $2j-2$. Put $q=2j-1$, $p=p_j$, and $c=1-4p$. By [L2] and Taylor's theorem, $S_{2j-2}(h)-U(h)=E{h}^q+O(h^{q+1})$. By [F6], the unique nonnegative root ${\rho }_q$ with ${\rho }_q^q=4$ exists; it is positive because zero has zero $q$-th power. Since $q\ge 3$, [F7] gives ${\rho }_q<4$: otherwise ${\rho }_q^q\ge 4^q>4$, contradicting ${\rho }_q^q=4$. Thus $p=(4-{\rho }_q{)}^{-1}>1\mathrm{/}4$ and $c=1-4p=-{\rho }_{q}p<0$. The five scaled copies in [F2] have leading error coefficients $p^{q}E,p^{q}E,c^{q}E,p^{q}E,p^{q}E$. Their other factors equal $I+O(h)$, so multiplying an error term by those factors changes it only by $O(h^{q+1})$. The ideal factors $U(ph{)}^{2}U(ch)U(ph{)}^2$ equal $U((4p+c)h)=U(h)$ by [F3]. The coefficient $c$ is an expression in the rational and positive-root tokens of [F5], so the signed call has a finite exact code. Hence the coefficient of $h^q$ in $S_{2j}(h)-U(h)$ is $(4p^q+c^q)E$. Since $c=-{\rho }_{q}p$ and ${\rho }_q^q=4$, $c^q=-4p^q$, so this coefficient vanishes and the difference is $O(h^{q+1})=O(h^{2j})$.

Let $A(h)=S_{2j-2}(ph)$ and $B(h)=S_{2j-2}(ch)$. The recursive word is $A(h)A(h)B(h)A(h)A(h)$. The induction hypothesis in step 1.2 gives inverse symmetry for $A$ and $B$; reversing the five factors shows $S_{2j}(-h)=S_{2j}(h{)}^{-1}$. Also $U(-h)=U(h{)}^{-1}$. By step 1.2 and [L2], write $S_{2j}(h)-U(h)=D{h}^{2j}+O(h^{2j+1})$. The inverse identity gives $D{h}^{2j}=-D{h}^{2j}+O(h^{2j+1})$, hence $D=0$ and the difference is $O(h^{2j+1})$. Therefore $S_{2j}$ agrees with $U$ through degree $2j$ and has inverse symmetry. Together with the base case in step 1.1, induction proves both claims for every finite $k$.