Tangential KZ regularization and the hump normalization

On the normalized three-strand slice, the three positions are $0,z,1$. The $t_{13}$ term of ${\mathrm{\Omega }}_3$ is constant in these coordinates, while $d\log (z_1-z_2)=d\log z$ and $d\log (z_2-z_3)=d\log (z-1)$. Thus the reduced equation is exactly the displayed equation with residues $a$ at $0$ and $b$ at $1$.

Near $z=0$, substitute $G_0(z)=f(z)z^a$ with $f(z)=1+{\sum }_{n\ge 1}{f}_n{z}^n$. Comparing coefficients in the differential equation gives $(n-{\mathrm{ad}}_a)f_n=-b{\sum }_{j=0}^{n-1}{f}_j$, where ${\mathrm{ad}}_a(x)=[a,x]$. On the degree-$N$ quotient, ${\mathrm{ad}}_a$ raises degree and is nilpotent, so $(n-{\mathrm{ad}}_a{)}^{-1}={\sum }_{s=0}^N({\mathrm{ad}}_a{)}^s\mathrm{/}{n}^{s+1}$ is defined. This determines every $f_n$ uniquely. If $S_n={\sum }_{j=0}^n\mathrm{\parallel }{f}_j\mathrm{\parallel }$ in any norm on the finite-dimensional quotient, the recurrence gives $S_n\le (1+C_N\mathrm{/}n)S_{n-1}$ for a constant $C_N$; hence $S_n$ grows at most polynomially, and $f$ converges for $\mathrm{\mid }z\mathrm{\mid }<1$. It solves the equation by termwise differentiation.

Replacing $z$ by $1-z$ and interchanging $a,b$ gives the same recurrence at $z=1$ and uniquely constructs $G_1(z)=g(1-z)(1-z{)}^b$ with $g(0)=1$. Any two fundamental solutions differ by a constant right multiplier; uniqueness of the analytic normalized factor follows from the recurrence in step 2.1. For a smooth tangential endpoint chart, write each within-cluster difference as $\rho u_{ij}(\rho )$ with $u_{ij}$ smooth and nonzero at $\rho =0$. Factoring these differences in the KZ form gives the radial equation $G^{\mathrm{\prime }}=(A\mathrm{/}\rho +B(\rho ))G$, where $A=(2\pi i{)}^{-1}{\sum }_{i<j\in S}{t}_{ij}$ and $B$ is smooth, not necessarily analytic. Construct $G=f(\rho ){\rho }^A$ degreewise with $f_0=1$ and, for $1\le d\le N$, $f_d(\rho )={\int }_0^{\rho }(\frac{[A,f_{d-1}(u)]}{u}+{\sum }_{j=1}^d{B}_j(u)f_{d-j}(u))du\mathrm{.}$ For $d=1$ the commutator with $f_0=1$ is zero. Inductively $f_{d-1}(0)=0$ for $d>1$, so $f_{d-1}(u)\mathrm{/}u={\int }_0^1{f}_{d-1}^{\mathrm{\prime }}(vu)dv$ is smooth. Thus each integrand is smooth, $f_d(0)=0$, and the integral uniquely solves the degree-$d$ equation. This proves existence and uniqueness of the smooth normalized factor with no Taylor expansion assumption. In the specified nested collision tree apply this construction successively in each radial variable, treating the remaining positive scales and the prescribed directions as parameters. Disjoint or nested cluster residues commute: for $S\subset T$, each $t_{ij}$ with $i,j\in S$ commutes with $t_{ik}+t_{jk}$ for $k\in T\setminus S$, and the internal terms commute with their total by the same relations in [F1]. Hence removing earlier residue factors preserves the next leading residue. Smooth dependence in the displayed finite recursion gives the successive tangential limits and their unique prescribed normalization. This concerns the fixed endpoint tree; it does not assert invariance under changes of parenthesization.

Both $G_0$ and $G_1$ solve the same left differential equation. Differentiating $G_1^{-1}{G}_0$ therefore gives zero, so their ratio is constant. For $G_{\epsilon }(z)=G_0(z)G_0(\epsilon {)}^{-1}$, the endpoint expansions and $G_0(\epsilon {)}^{-1}={\epsilon }^{-a}f(\epsilon {)}^{-1}$ give $\begin{array}{rl}{\displaystyle {\epsilon }^{-b}{G}_{\epsilon }(1-\epsilon ){\epsilon }^a}& {\displaystyle ={\epsilon }^{-b}f(1-\epsilon )(1-\epsilon {)}^a{\epsilon }^{-a}f(\epsilon {)}^{-1}{\epsilon }^a}\\ {\displaystyle }& {\displaystyle ={\epsilon }^{-b}g(\epsilon ){\epsilon }^b{\mathrm{\Phi }}_{\mathrm{K}\mathrm{Z}}{\epsilon }^{-a}f(\epsilon {)}^{-1}{\epsilon }^a,}\end{array}$ where the second equality follows from $G_1(1-\epsilon ){\mathrm{\Phi }}_{\mathrm{K}\mathrm{Z}}=G_0(1-\epsilon )$. In each degree, $f(\epsilon {)}^{-1}-1$ and $g(\epsilon )-1$ are $O(\epsilon )$. Since $a$ and $b$ have positive degree, conjugation by ${\epsilon }^{\pm a}$ or ${\epsilon }^{\pm b}$ is a finite polynomial in $\log \epsilon$ in the degree-$N$ quotient. Thus both outside factors in the last expression differ from $1$ by $O(\epsilon (1+\mathrm{\mid }\log \epsilon \mathrm{\mid }{)}^N)$ in each degree, which tends to zero; the limit is ${\mathrm{\Phi }}_{\mathrm{K}\mathrm{Z}}$.

The admissible standard one-in one-out hump tangle closes to the two-maxima unknot $H$ by [F3], identifying the normalization factor with the fixed hump in [F1]. For an arbitrary closed $K$, apply the closed-knot identity [F4] directly to its local insertion: $Z(K^{+})=Z(H)Z(K)$. A maximum-minimum birth adds exactly one maximum, so $r(K^{+})=r(K)+1$. Write $Z(H)=1+h$, where $h$ has positive degree. Through degree $N$ its inverse is ${\sum }_{j=0}^N(-h{)}^j$, and these inverses are compatible as $N$ increases. Connected-sum multiplication is commutative, so $I(K^{+})=Z(H)Z(K)Z(H{)}^{-r(K)-1}=Z(K)Z(H{)}^{-r(K)}=I(K)$. Deleting a standard hump is the inverse equality.