Conservative second-order error

The claim is immediate on the zero-dimensional space, so assume $V\mathrm{\ne }\{0\}$. Write $A_j=-i{H}_j$, $A={\sum }_j{A}_j$, $P(h)=S_2(h)$, and $Q(h)=U_H(h)$. Every factor is unitary by [F1], and the palindromic word gives $P(-h)=P(h{)}^{-1}$; also $Q(-h)=Q(h{)}^{-1}$. Differentiating the factors at zero, the two occurrences with coefficient $1\mathrm{/}2$ contribute a total coefficient $1$ for each $A_j$, so $P(0)=Q(0)=I$ and $P^{\mathrm{\prime }}(0)=Q^{\mathrm{\prime }}(0)=A$.

Let $E(h)=P(h)-Q(h)$. Smoothness follows by repeatedly differentiating the constant-generator identity in [F1]. Thus $E(h)=C{h}^2+O(h^3)$ for some endomorphism $C$ by [L1]. The inverse identities in step 1.1 give $E(-h)=P(h{)}^{-1}-Q(h{)}^{-1}=-P(h{)}^{-1}E(h)Q(h{)}^{-1}$. Since both inverse factors equal $I+O(h)$, comparing degree-two coefficients yields $C=-C$, hence $C=0$. Therefore $P$ and $Q$ have matching Taylor coefficients through degree two.

Write $P(h)=F_1(h)\cdots F_{2m-1}(h)$ in its displayed order, with $F_{\mathrm{\ell }}(h)=e^{c_{\mathrm{\ell }}h{A}_{j_{\mathrm{\ell }}}}$ and the coefficients $c_{\mathrm{\ell }}$ equal to the half-step palindrome in [F2]. Repeated product differentiation and [F1] give $\mathrm{\parallel }{P}^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}(h)\mathrm{\parallel }\le ({\sum }_{\mathrm{\ell }}\mathrm{\mid }{c}_{\mathrm{\ell }}\mathrm{\mid }\mathrm{\parallel }{H}_{j_{\mathrm{\ell }}}\mathrm{\parallel }{)}^3=M^3$: each $j<m$ occurs twice with coefficient $1\mathrm{/}2$, and $j=m$ occurs once with coefficient $1$. Also $Q^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}(h)=A^{3}Q(h)$. For a unit vector $v$, the vector triangle inequality gives $\mathrm{\parallel }Av\mathrm{\parallel }\le {\sum }_j\mathrm{\parallel }{H}_{j}v\mathrm{\parallel }\le {\sum }_j\mathrm{\parallel }{H}_j\mathrm{\parallel }=M$, so [F4] yields $\mathrm{\parallel }A\mathrm{\parallel }\le M$. Then [F3], [F4], and unitarity give $\mathrm{\parallel }{Q}^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}(h)\mathrm{\parallel }\le M^3$. The integral Taylor remainder after degree two therefore gives $\mathrm{\parallel }P(h)-Q(h)\mathrm{\parallel }\le (2M^3)\mathrm{\mid }h{\mathrm{\mid }}^3\mathrm{/}3!=M^3\mathrm{\mid }h{\mathrm{\mid }}^3\mathrm{/}3$.

Expand $P(h{)}^r-Q(h{)}^r$ by telescoping. The definition [F4] and submultiplicativity [F3] show that unitary factors preserve the norm, so its norm is at most $r{M}^3\mathrm{\mid }h{\mathrm{\mid }}^3\mathrm{/}3$ by [step 3.1]. With $h=t\mathrm{/}r$, this is $\mathrm{\mid }t{\mathrm{\mid }}^3{M}^3\mathrm{/}(3r^2)$; [F1] gives $Q(h{)}^r=U_H(t)$.