The Kontsevich integral and universality

Form the coefficientwise formal expression $C_D(z)={\sum }_{j\ge 0}{c}_j(D)z^j$ from [F1]. It is a diagram invariant, has value $1$ on an unknot, and satisfies $C_{D_{+}}-C_{D_{-}}=z{C}_{D_0}$ coefficientwise. A local positive and negative curl on any component of the $r$-component unlink are both that unlink; the oriented smoothing is the $(r+1)$-component unlink. Thus $c_j(U_r)-c_j(U_r)=c_{j-1}(U_{r+1})$ for every $j\ge 1$. All coefficients of $U_{r+1}$ vanish. Therefore the unlink values of the formal expression are $1$ for one component and $0$ for more than one. This is also the forced unlink value for ANY polynomial-valued invariant satisfying the Definition.

Here is the finite argument that the expression is actually in $\mathbb{Z}[z]$. In a diagram choose finitely many component basepoints and an order. Call a crossing bad when its first encounter in this traversal is the underpass. If none is bad, the first component passes over every later component and has a descending self-diagram. Pull its successive arcs above everything they first encounter to isolate a round circle; repeat for the remaining components. Thus a descending diagram is an unlink. At a bad crossing the skein relation writes its value as the value of the switched diagram, plus or minus $z$ times its oriented smoothing. The switched diagram has one fewer bad crossing and the same number $N$ of crossings; the smoothing has $N-1$ crossings. Reset the finite component data on a smoothing if necessary. The nonnegative integer $N(N+1)\mathrm{/}2+b$, with $b$ the number of bad crossings, decreases in each branch. Hence this is a FINITE resolution tree. Its leaves have values $0$ or $1$ by step1.1. Working back through the tree uses only integer addition/subtraction and multiplication by $z$, so $C_D\in \mathbb{Z}[z]$. In fact every branch has at most $N$ smoothing operations, giving degree at most $N$. This proves finiteness and integrality without assuming an integer-valued formal sequence theorem or making an infinite selection.

Apply the SAME finite tree to any two polynomial-valued invariants satisfying the stated normalization and skein relation. Their unlink leaves agree by step1.1, and each internal skein relation then forces agreement. Their values agree on every diagram. Existence follows from [F1], step 2.1 and the ambient-link interpretation [F2], and this argument proves uniqueness. The separately fixed empty-link extension is unique as well, since its prescribed value is zero. Consequently the separate existence/uniqueness obligation mentioned in the Definition is now discharged: the resulting actual link invariant is exactly $\mathrm{\nabla }$, and its coefficient $[z^k]{\mathrm{\nabla }}_L$ is well-defined for every $k$. The finite local arguments use no choice axiom; the external constructive coefficient/diagram inputs are explicitly distinguished from them.

The critical points are isolated in the compact one-dimensional skeleton and absent from the tangle boundary, so there are finitely many critical points and critical values. On every open interval between consecutive critical values or boundary levels, the inverse image is a finite union of arcs on which $t$ is a local diffeomorphism. Thus the number of branches, and the finite set of allowed pairs at a level, are constant on each such interval.

At a regular level, each branch is a smooth graph $z=z_{\lambda }(t)$, so its evaluation extends smoothly to any regular face. Near a critical point, choose a local parameter $x$ with $t(0)=c$, $t^{\mathrm{\prime }}(0)=0$, and $t^{\mathrm{\prime }\mathrm{\prime }}(0)\mathrm{\ne }0$. The function $h(x)=(t(x)-c)\mathrm{/}{x}^2$ extends smoothly with $h(0)=t^{\mathrm{\prime }\mathrm{\prime }}(0)\mathrm{/}2\mathrm{\ne }0$; after shrinking the coordinate interval, $s=x\sqrt{\mathrm{\mid }h(x)\mathrm{\mid }}$ is a smooth coordinate with nonzero derivative and $t=c+\mathrm{sgn}(h(0))s^2$. Since $z=z(x(s))$ is smooth, the two local branches and their limiting common point extend to the critical face in this coordinate.

Fix a chamber and let $k_r$ be the number of its ordered levels in the $r$th interval between critical or boundary levels. The chamber is a product of open simplices, one for each interval, with $k_r$ ordered coordinates. Its closure is the product of the corresponding closed simplices, hence compact. The facets of a closed simplex are exactly adjacent-coordinate equalities or a coordinate reaching one endpoint of its interval; across the finite family of chambers those endpoints are the knot extrema, tangle boundary levels, or critical values. Thus before resolving endpoint collisions the only faces are level collisions, outer endpoints, and critical-level events.

In a regular-level collision, endpoints on distinct branches remain distinct because $K$ is embedded; endpoints on one branch coalesce along a smooth arc. In the latter case, use local parameters $s_1,\dots ,s_k$ for the colliding endpoints, their center $p=k^{-1}\sum s_j$, the radius $r=({\sum }_j\mathrm{\mid }{s}_j-p{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}$, and the normalized shape $(s_j-p)\mathrm{/}r$. Replacing the diagonal $r=0$ by this shape data is the real oriented blow-up of that collision diagonal; its local coordinates are $r\ge 0$ and smooth shape coordinates. At a critical point, step 1.2 supplies the same form with the Morse coordinate $s$.

For a cluster of several endpoints, first blow up the full collision diagonal and then each proper collision diagonal in its limiting shape space, from larger clusters to smaller ones. Collision subsets in any limiting configuration are either nested or disjoint, so these blow-ups introduce finitely many nonnegative radii $r_C$ and smooth shape coordinates. Each resulting chart is a product of half-lines $r_C\ge 0$ and an open smooth coordinate set; transition maps are smooth because they are changes of center, scale, and unit direction. Hence the resolved cell is a manifold with corners. Each added boundary hypersurface is a collision cluster, and its boundary orientation is determined by the outward vector $-{\mathrm{\partial }}_{r_C}$. The original simplex orientation and the local coordinates $s\ge 0$ at critical levels give the stated orientations on all remaining faces.

Each chamber closure is compact by step 2.1, and real oriented blow-up replaces a closed collision diagonal by a compact sphere of limiting directions without losing compactness. Only finitely many chambers, branch pairings, and endpoint subsets occur at fixed $m$, so their resolved union is compact. The boundary labels from steps 2.1 and 3.1 exhaust its boundary faces; no additional face can occur because every boundary coordinate is either a level-gap, a distance to a critical/global endpoint, or one of the finitely many collision radii. The branch evaluations extend by steps 1.2 and 2.2.

The set of embeddings of the compact circle into ${\mathbb{R}}^3$ is open in the $C^1$ topology. To verify this for the given compact family, cover the compact parameter cylinder by finitely many coordinate rectangles on each of which one fixed component of the spatial tangent has constant sign and is bounded away from zero. Shrinking these rectangles gives a uniform neighborhood of the circle diagonal on which every sufficiently small $C^1$ perturbation is injective on each slice, by strict monotonicity of that component. On the compact complement of a neighborhood of the diagonal in $S^1\times S^1$, injectivity has a uniform positive separation because $(s,x,y)\mapsto \mathrm{\parallel }{K}_s(x)-K_s(y)\mathrm{\parallel }$ is continuous and positive. A sufficiently small $C^0$ perturbation preserves that separation. Thus a small perturbation of $h$ alone keeps every slice embedded and fixed at the two endpoint collars. The straight interpolation between $h$ and its perturbation is also $C^1$-small, hence stays embedded; after a smooth reparametrization stationary near both interpolation endpoints, [F6] extends this path to ambient isotopy.

Identify $S^1$ with $\mathbb{R}\mathrm{/}2\pi \mathbb{Z}$. Perturb $h$ by $\chi (s)(p(x)+sq(x))$, where $\chi$ vanishes on fixed endpoint collars and is positive on the remaining parameter interval, and $p,q$ range over real trigonometric polynomials of degree at most $16$. At up to four distinct circle points these polynomials prescribe all jets through order three independently: complexify, multiply by $e^{16ix}$, and use ordinary polynomial Hermite interpolation at the distinct nodes $e^{ix}$ (at most sixteen conditions in degree at most $32$); taking the real part recovers real data. The two coefficients $p,q$ independently prescribe these jets and their $s$-derivatives at any fixed interior $s$. Thus the universal maps for $h_x$, $(h_x,h_{xx})$, $(h_x,h_{xx},h_{xxx})$, and $(h_x,h_{xx},h_{xs})$ are transverse to zero. By [F4] choose arbitrarily small coefficients making all four slices transverse. Near the endpoint collars strict Morse behavior is already stable; shrink the coefficient ball to preserve it there. The critical locus is a one-manifold, the fold locus is zero-dimensional, and the two three-coordinate maps have empty zero sets by dimension. Every degeneracy therefore satisfies $h_x=h_{xx}=0$, $h_{xxx}\mathrm{\ne }0$, and $h_{xs}\mathrm{\ne }0$. The same finite family will impose the distinct-point conditions below simultaneously, rather than perturbing an already chosen slice.

The fold set from step 1.2 is a compact zero-dimensional transverse preimage in the two-dimensional cylinder, hence finite. Near a fold $(s_0,x_0)$, the equation $h_x(s,x)=0$ has $h_x=h_{xx}=0$, $h_{xxx}\mathrm{\ne }0$, and $h_{xs}\mathrm{\ne }0$. By [F5], solve it locally as $s=\psi (x)$; differentiating gives ${\psi }^{\mathrm{\prime }}(x_0)=0$ and ${\psi }^{\mathrm{\prime }\mathrm{\prime }}(x_0)=-h_{xxx}\mathrm{/}{h}_{xs}\mathrm{\ne }0$. Thus on one side of $s_0$ there are exactly two nondegenerate critical points and on the other side none. Between these finitely many fold values, $h_x=0$ is a finite union of smooth branches by the implicit function theorem. At the birth, the two new critical values are distinct away from the fold: $h_x$ has constant nonzero sign between its two roots, so $h(s,x_{+})-h(s,x_{-})={\int }_{x_{-}}^{x_{+}}{h}_x(s,x)dx\mathrm{\ne }0$.

On the entire domain $(0,1)\times \{(x,y)\in (S^1{)}^2:x\mathrm{\ne }y\}$, including pairs involving a fold neighborhood, impose transversality to zero of $(h_x(s,x),h_x(s,y),h(s,x)-h(s,y))$. The independent distinct-point jets from step 1.2 make the universal map a submersion, so [F4] applies on this domain without a compactness assumption. At a zero the slice Jacobian has determinant, up to sign, $h_{xx}(s,x)h_{xx}(s,y)(h_s(s,x)-h_s(s,y))$. Thus both critical points are nondegenerate and their critical values cross transversely; in particular a fold cannot share a critical value with any other critical point. The zero set is discrete. It cannot accumulate at the removed diagonal: near a nondegenerate critical point there is only one critical branch, and near a fold the only two branches have unequal values by step 2.1. Compactness of the diagonal supplies finitely many such neighborhoods. The endpoint collars have distinct critical values and no events. Consequently the crossing set is closed in a compact subset away from the diagonal and endpoint collars, hence finite.

Use the same distinct-point jet family of step 1.2, so all genericity conditions hold for one arbitrarily small coefficient choice. For three distinct critical points, the event map with their three $x$-derivatives and two value differences has five scalar components on a four-dimensional parameter-and-point domain; [F4] makes its zero set empty. For two disjoint value crossings, four critical-point equations and two value differences give six components on a five-dimensional domain, so simultaneous crossings are absent. A birth/death coinciding in height with another critical point imposes $h_x=h_{xx}=0$ at the fold, $h_x=0$ at the other point, and equality of their values: four independent conditions on a three-dimensional domain, so this event is absent. A fold simultaneous with a value crossing at two other critical points imposes five independent conditions on a four-dimensional domain and is absent. Two folds at the same parameter impose four equations on a three-dimensional domain and are absent as well. The same finite-dimensional family imposes these conditions together with steps 1.2 and 3.1: the bad parameter set for each map is null by [F4], their finite union is null, and every coefficient ball contains a point outside that union.

Outside the finite fold and crossing parameters, every critical point is nondegenerate and the critical values are pairwise distinct, so each height slice is strict Morse. Steps 2.1–4.1 show that the only exceptional events are isolated max-min births or deaths and transverse crossings of two distinct critical values; the imposed three-point, four-point, and fold-pair conditions exclude simultaneous events and triple values. The total perturbation can be chosen arbitrarily small, so step 1.1 keeps every slice embedded and in the same oriented knot class. The perturbation vanishes near $s=0,1$, hence leaves the specified endpoint presentations fixed.

Put ${\alpha }_{ij}=d\log (z_i-z_j)$. Since $z_i-z_j$ is nonzero on ${\mathrm{Conf}}_q(\mathbb{C})$, in local coordinates ${\alpha }_{ij}=d(z_i-z_j)\mathrm{/}(z_i-z_j)$ and $d{\alpha }_{ij}=0$ by equality of mixed partial derivatives. Thus $d{\mathrm{\Omega }}_q=0$.

Write $a=t_{ij}$, $b=t_{kl}$ for two distinct chord generators. In the expansion of ${\mathrm{\Omega }}_q\wedge {\mathrm{\Omega }}_q$, the two ordered terms for these generators combine as $[a,b]{\alpha }_{ij}\wedge {\alpha }_{kl}$. If the pairs have four distinct indices, $[a,b]=0$ by [F1]; if they are the same pair, ${\alpha }_{ij}\wedge {\alpha }_{ij}=0$. Hence only pairs of generators supported on a common triple of strands can contribute.

Fix $i<j<k$ and put $a=t_{ij}$, $b=t_{ik}$, $c=t_{jk}$, $x={\alpha }_{ij}$, $y={\alpha }_{ik}$, and $z={\alpha }_{jk}$. The tangled 4T relation on these three strands is exactly the pair of infinitesimal braid relations $[a,b+c]=0$ and $[b,a+c]=0$. The contribution of this triple is $[a,b]x\wedge y+[a,c]x\wedge z+[b,c]y\wedge z$. The infinitesimal braid relations in [F1] give $[a,c]=-[a,b]$ and $[b,c]=[a,b]$, so this is $[a,b](x\wedge y-x\wedge z+y\wedge z)$. With $u=z_i-z_j$ and $v=z_j-z_k$, we have $x=du\mathrm{/}u$, $z=dv\mathrm{/}v$, and $y=d(u+v)\mathrm{/}(u+v)$. Direct expansion gives $x\wedge z+z\wedge y+y\wedge x=(1\mathrm{/}(uv)-1\mathrm{/}(v(u+v))-1\mathrm{/}(u(u+v)))du\wedge dv=0$, which is equivalent to $x\wedge y-x\wedge z+y\wedge z=0$. Thus every triple contribution vanishes.

The repeated-pair, disjoint-pair, and three-index cases exhaust the expansion of ${\mathrm{\Omega }}_q\wedge {\mathrm{\Omega }}_q$; steps 2.1–3.1 show that each contribution is zero. Together with step 1.1 this gives $d{\mathrm{\Omega }}_q=0$ and ${\mathrm{\Omega }}_q\wedge {\mathrm{\Omega }}_q=0$.

Fix a smooth path $\gamma$ and let $U(t)$ solve $\dot{U}=A_{t}U$, $U(0)=1$, where $A_t={\mathrm{\Omega }}_q(\dot{\gamma }(t))$. In each degree this is a finite linear ODE system, solved recursively by the iterated-integral formula. For a homotopy $\gamma (s,t)$ with fixed endpoints, set $A_s={\mathrm{\Omega }}_q({\mathrm{\partial }}_s\gamma )$ and $W=U^{-1}({\mathrm{\partial }}_{s}U-A_{s}U)$. Differentiating and using ${\mathrm{\partial }}_{t}U=A_{t}U$ gives ${\mathrm{\partial }}_{t}W=U^{-1}({\mathrm{\partial }}_s{A}_t-{\mathrm{\partial }}_t{A}_s+[A_t,A_s])U$. The coefficient in parentheses is the evaluation of the flatness expression from step 4.1, so it is zero. Since the initial endpoint is fixed, $W(s,0)=0$ and hence $W(s,t)=0$; at the fixed terminal endpoint $A_s=0$, so ${\mathrm{\partial }}_{s}U(s,1)=0$. The holonomy is therefore constant throughout the homotopy.

The KZ form is unchanged by relabeling the strands, since its sum runs over every unordered pair; endpoint strand permutations therefore act on holonomy by the same permutation of the $t_{ij}$. For $\mathrm{\mid }i-j\mathrm{\mid }\ge 2$, the paths for ${\sigma }_i$ and ${\sigma }_j$ lie in disjoint two-point disks, so exchanging their time intervals is a fixed-endpoint homotopy and step 5.1 gives ${\sigma }_i{\sigma }_j={\sigma }_j{\sigma }_i$. For three adjacent points, start at $(-1,0,1)$ and follow the configuration path $(-e^{\pi it},0,e^{\pi it})$, $0\le t\le 1$. Its endpoints agree as unordered configurations with the three points reversed, and its pairwise distances stay bounded below by $1$. At $t=1\mathrm{/}2$ the three real projections coincide. Small perturbations of the trajectories therefore remain collision-free and can separate the three projected crossing times in either of the two orders; both perturbed paths are homotopic to this rigid half-turn and give respectively ${\sigma }_i{\sigma }_{i+1}{\sigma }_i$ and ${\sigma }_{i+1}{\sigma }_i{\sigma }_{i+1}$. By step 5.1 their transports agree.

For any skein triple, the constant coefficients of ${\mathrm{\nabla }}_{L_{+}}$ and ${\mathrm{\nabla }}_{L_{-}}$ agree because their difference is divisible by $z$. Hence $c_0(L):=[z^0]{\mathrm{\nabla }}_L$ is unchanged by crossing switches. Choose a basepoint on each component of a link diagram and a finite order of the components; switch each crossing so its first encounter in that traversal is the overpass. The resulting diagram is descending: starting at each basepoint, every arc can be pulled successively over the portions it meets, in the chosen component order, until each component is a round circle in a separate ball. Thus every link can be changed by crossing switches to an unlink. For the $r$-component unlink $U_r$, the normalization gives $c_0(U_1)=1$. Add a local curl to one component of $U_r$; its positive and negative versions are both $U_r$, while its oriented smoothing is the split unlink $U_{r+1}$. The skein relation gives $z{\mathrm{\nabla }}_{U_{r+1}}=0$. Multiplication by the indeterminate $z$ is injective in $\mathbb{Z}[z]$: if $f={\sum }_{j=0}^N{a}_j{z}^j$ and $zf=0$, then each coefficient $a_j$ is zero. Hence ${\mathrm{\nabla }}_{U_{r+1}}=0$. Therefore $c_0(L)=1$ for knots and $c_0(L)=0$ for links with at least two components.

Comparing the coefficient of $z^1$ in the skein relation gives $c_1(L_{+})-c_1(L_{-})=c_0(L_0)$. In a two-component link, smoothing a mixed crossing yields a knot and therefore difference $1$; smoothing a self-crossing yields a three-component link and therefore difference $0$, by step 1.1. The value of this first difference depends only on whether the crossing is mixed, not on any other crossing resolution.

At one singular crossing of a knot, the skein relation gives $D_1{a}_2(K)=c_1(L_0)$ for its oriented two-component smoothing $L_0$. On a three-singular knot, resolve one marked point first and apply the two remaining differences to this $c_1(L_0)$. By step 2.1, taking the difference at either remaining crossing gives $c_0$ of its oriented smoothing; that smoothing has either one or three components, so its $c_0$ is constant across the resolution of the last marked crossing by step 1.1. The final alternating difference is therefore zero. Since the marked point chosen first was arbitrary, every third finite difference of $a_2$ vanishes.

For a two-singular knot, smooth the first marked point. Its two preimages split the circle into two arcs, which become the two components of $L_0$. The other marked crossing is mixed exactly when its two preimages lie on different arcs, equivalently when the two chord endpoints alternate with those of the first chord. By step 2.1 the resulting second difference is $1$ in that case and $0$ otherwise.

Define $w(D)=1$ on the intersecting two-chord diagram and $w(D)=0$ on the nonintersecting one. The latter is zero by 1T [F3]. If the cyclic order of the three fixed points is $p,q,r$, the four diagrams $D_1,D_2,D_3,D_4$ in the 4T generator have intersection indicators $(0,1,1,0)$; if the order is $p,r,q$, they have indicators $(1,0,0,1)$. These exhaust the cyclic orders up to rotation. In either case $w(D_1)-w(D_2)+w(D_3)-w(D_4)=0$. Thus $w$ factors through the unframed quotient. Since $w(D_{\times })=1$, that class is nonzero and the degree-two symbol in step 3.2 is nonzero.

The unknot normalization gives ${\mathrm{\nabla }}_U=1$, hence $a_2(U)=0$. Let $H$ be the closure of a two-strand braid with two successive positive Artin half twists. Switching one crossing creates adjacent positive and negative half twists, which cancel by [F4] to the two-component unlink $U_2$; smoothing that crossing leaves a one-half-twist closure, an unknot with a removable type-I curl. Step 1.1 gives ${\mathrm{\nabla }}_{U_2}=0$, so the skein relation yields ${\mathrm{\nabla }}_H=z$. For the positive trefoil $T$, the closure of three successive positive half twists, switching one crossing leaves an adjacent positive-negative pair that cancels and hence an unknot; smoothing that crossing leaves $H$. Therefore ${\mathrm{\nabla }}_T-1=z{\mathrm{\nabla }}_H=z^2$, so $a_2(T)=1$.

In a Morse band the colliding pair has $\mathrm{\mid }{z}_{+}(s)-z_{-}(s)\mathrm{\mid }\ge as$, while its derivative is bounded, so its short-chord form is bounded in absolute value by $Cds\mathrm{/}s$. Every other pair has separation at least $d$ and bounded derivative, so its logarithmic form is bounded by $Cds$; in a regular band the bound is $Cdt$. These constants are uniform under the stated family hypotheses. For a single fixed tangle the same bounds hold with tangle-dependent constants: the Morse expansions are $z_{\pm }(s)=z_c\pm vs+O(s^2)$ with $v\mathrm{\ne }0$, and all other branch pairs have nonzero limiting separation; shrink finitely many disjoint critical bands and use compactness on the remaining regular bands. Minima are treated by reversing height.

Discard every pairing diagram killed by 1T before taking absolute values. In a remaining pairing, let $0<{\delta }_k<\cdots <{\delta }_1<S$ be all $k\ge 1$ short-chord parameters in one whole Morse band, including parameters outside a smaller cap. If no endpoint of another chord lies on the oriented critical arc between the ends of the innermost short chord at ${\delta }_k$, these ends are consecutive and 1T kills the diagram. Any such intervening endpoint must belong to a long chord in the same band, since no short chord lies closer to the critical point. Thus every surviving pairing has a long witness with parameter $0<r<{\delta }_k$. Retain that witness factor and drop other ordering constraints only when enlarging a nonnegative absolute majorant.

Integrating the retained witness first gives at most $C{\delta }_k$. The remaining short factors have ordered integral ${\int }_{0<{\delta }_k<\cdots <{\delta }_1<S}{\delta }_k{\prod }_{j=1}^k\frac{d{\delta }_j}{{\delta }_j}=S\mathrm{.}$ Indeed, the innermost integration gives ${\delta }_{k-1}$ and repeating gives $S$. Every additional long factor is integrated once over a bounded interval. Distinct critical bands use distinct level variables, and a witness chord in one band cannot be a witness in another band; their parameter intervals are disjoint. Apply this estimate in each band, and bounded-factor estimates in regular bands. There are finitely many branch chambers and pairing types at degree $m$, and every surviving term is absolutely integrable. The quotient norm is bounded by the sum of these absolute majorants times the norms of the finitely many diagram classes. This proves the fixed-tangle convergence assertion, with uniform bounds when all the family controls hold.

For the cap band put $S=S_c$ and, by enlarging the cap when necessary, use the cutoff $h=A\epsilon <S$. Define $J_0(h,S)=h$ and, for $k\ge 1$, $J_k(h,S):={\int }_{0<{\delta }_k<\cdots <{\delta }_1<S}\min \{h,{\delta }_k\}{\prod }_{j=1}^k\frac{d{\delta }_j}{{\delta }_j}\mathrm{.}$ For $L=\log (S\mathrm{/}h)\ge 0$, integrating the upper $k-1$ short parameters at fixed $u={\delta }_k$ gives $J_k(h,S)=\frac{1}{(k-1)!}({\int }_0^h\log (S\mathrm{/}u{)}^{k-1}du+h{\int }_h^S\log (S\mathrm{/}u{)}^{k-1}\frac{du}{u})=h{\sum }_{j=0}^k\frac{L^j}{j!}\mathrm{.}$ The last equality follows by integration by parts in the first integral and the substitution $v=\log (S\mathrm{/}u)$ in the second; the boundary term at $u=0$ is zero since $u\log (S\mathrm{/}u{)}^j\to 0$ for every fixed $j$. The same formula gives $J_0=h$ when the sum has only its $j=0$ term. In particular $J_k(S,S)=S$, consistently with step 3.1.

Suppose a selected cap-meeting chord is long and $k\ge 1$. If it is the witness retained in step 2.1, its parameter is restricted to $0<r<\min \{h,{\delta }_k\}$; integrating this one long factor supplies $C\min \{h,{\delta }_k\}$ and hence the majorant $C{J}_k(h,S)$. If it is distinct from the chosen witness, retain both factors: integrate the selected cap long chord over a parameter interval of length at most $h$, and the witness over $(0,{\delta }_k)$. Dropping their mutual order enlarges the absolute integral and supplies $Ch{\delta }_k$, so step 3.1 gives the bound $ChS$. This is at most a fixed constant times $J_k(h,S)$, since $J_k(h,S)\ge h$ and $S$ is fixed. Neither long factor has been omitted or used twice. If $k=0$, a cap-meeting long factor directly supplies $Ch=C{J}_0(h,S)$.

Suppose instead that a selected cap-meeting chord is short. Some short parameter is at most $h$, hence the innermost one has ${\delta }_k\le h$. The witness from step 2.1 lies closer to the critical point, so it also lies inside the cap cutoff. Its integration supplies $C{\delta }_k$, while all upper short parameters still range through the whole band up to $S$. Therefore the ordered bound is ${\int }_{\begin{array}{c}0<{\delta }_k<\cdots <{\delta }_1<S\\ {\delta }_k\le h\end{array}}{\delta }_k{\prod }_{j=1}^k\frac{d{\delta }_j}{{\delta }_j}=h{\sum }_{j=0}^{k-1}\frac{\log (S\mathrm{/}h{)}^j}{j!}\le J_k(h,S)\mathrm{.}$ This follows from the first integral in step 4.1. In particular no parameter of an outer short chord has been artificially restricted to $h$.

The event that a pairing meets $N_{\epsilon }$ is covered by the finite union over its at most $m$ selected chords and the long/short cases in steps 5.1 and 5.2. Keep the associated cap and witness factors in that band; in every other critical band keep its separate witness as in step 2.1. Integrating all remaining bounded factors once and summing over the finitely many diagrams gives $C_m{J}_k(A\epsilon ,S)$ for some $k\le m-1$, or $C_m\epsilon$ in the case of distinct retained long factors. The inequality $k\le m-1$ holds because any surviving short-chord contribution has at least one long witness. Formula 4.1 gives the stated $C_m\epsilon (1+\mathrm{\mid }\log \epsilon \mathrm{\mid }{)}^{m-1}$ bound. For fixed $m$ it tends to zero, since with $x=-\log \epsilon$ it is bounded by a constant times $e^{-x}(1+x{)}^{m-1}$. If $m=0$, there is no chord endpoint and the cap-meeting contribution is exactly zero.

The final special-critical-movement assertion is exactly the separately labelled external input [F4] under its stated disk-region and single-critical-point hypotheses. It makes no assertion that individual pairing integrals, or arbitrary degenerating families, obey a uniform linear bound. The local convergence and controlled-cap estimates above were established independently.

In the degree-$m$ ordered simplex, the set where some level $t_j=c$ has measure zero. Absolute convergence from [F2] allows removal of this set without changing the integral. The remaining domain is the disjoint union, for $k=0,\dots ,m$, of regions with exactly $k$ levels above $c$ and $m-k$ below it; each such region is the product of the corresponding ordered simplex for $T_2$ and that for $T_1$.

On the region indexed by $k$, every selected pair at a level above $c$ lies on $T_2$, and every pair below $c$ lies on $T_1$. Thus a pairing on the composite decomposes uniquely into a pairing on each piece. The product differential form separates into the upper and lower factors in height order, and the resulting chord diagram is their vertical stack.

Fubini's theorem applies to each product region because both factors are absolutely integrable by [F2]. Therefore its integral is the product of the degree-$k$ coefficient for $T_2$ and the degree-$(m-k)$ coefficient for $T_1$. Summing over the finite set $k=0,\dots ,m$ is exactly the degree-$m$ coefficient convolution of $Z(T_2)Z(T_1)$. Since $m$ was arbitrary, the displayed completed-series identity follows.