Closure and connected sum for the Kontsevich integral of one-strand tangles

First consider a horizontal deformation of a closed strict Morse knot that fixes the height function and a neighborhood of every critical point. Cut at regular levels outside those neighborhoods. For the regular pieces of the Stokes argument, consider a compact tangle $T_s$ without critical points, with fixed boundary points and parameter $s\in [0,1]$. On a compactified branch chamber, write the full degree-$m$ form as the finite sum ${\omega }_m={\sum }_P(-1{)}^{\downarrow (P)}[D_P]{\bigwedge }_{j=1}^{m}d\log (z_j-z_j^{\mathrm{\prime }})$, with each differential taken in the level and deformation variables. Every branch difference is nonzero because the tangle is embedded and has no critical points; hence each logarithmic factor is closed and $d{\omega }_m=0$. Stokes on the chamber cylinder reduces equality of the two end integrals to vanishing on its codimension-one faces.

The upper and lower closure caps lie in height bands disjoint from $T$ and the return arc. Each cap has exactly two points at every regular level in its band. In a nonempty pairing on an upper cap, the chord at the highest selected level has an empty subarc on the cap side; on a lower cap use the chord at the lowest selected level. In either case that chord is isolated, so [F4] kills the diagram. The return arc has exactly one point at each level of $[a,b]$, so it has no internal pairing. Hence cap and return factors contribute only their degree-zero units.

On an outer-level face, the extreme endpoints are fixed, so the restricted form has no $ds$ component and vanishes. On a face where adjacent levels coincide, if the two chords use the same two branches, their logarithmic forms agree and their wedge is zero. If they use four distinct branches, exchanging the two chord choices preserves the diagram and reverses the wedge sign, so the terms cancel in pairs. If they use exactly three branches, the 4T relation [F4] groups the six choices into one common diagram coefficient times ${\alpha }_{12}\wedge {\alpha }_{23}+{\alpha }_{23}\wedge {\alpha }_{31}+{\alpha }_{31}\wedge {\alpha }_{12}$, where ${\alpha }_{ij}=d\log (z_i-z_j)$. Setting $f=z_1-z_2$, $g=z_2-z_3$, and $z_3-z_1=-(f+g)$, direct expansion gives the coefficient $1\mathrm{/}(fg)-1\mathrm{/}(g(f+g))-1\mathrm{/}(f(f+g))=0$. The real blow-ups in [F2] over collision centers of codimension at least two add no Stokes term: on this no-critical tangle the branch form extends smoothly to the unblown closed chamber, and the blowdown restricted to a front face times the deformation interval has rank at most $m-1$, so its pullback $m$-form vanishes there. A codimension-one center is an adjacent-level face already handled above. Thus the only remaining boundary integrals are the two ends $s=0,1$, which are equal. Summing the finitely many chambers proves invariance for the stated regular fixed-boundary tangle. The closed-knot conclusion across critical levels is the explicitly imported result [F6].

Put $M={\sup }_T\mathrm{\mid }z\mathrm{\mid }$ and take $R>M+1$. Any chord using the return arc over $[a,b]$ joins $R$ to $z\in T$ and has factor $dz\mathrm{/}(z-R)$. Choose finitely many small critical bands of the fixed $T$, with one critical point in each, and regular bands covering the rest. In a critical band write $t=c\pm s^2$. The colliding branches satisfy $z_{\pm }(s)=z_c\pm vs+O(s^2)$ with $v\mathrm{\ne }0$, so their short-chord factor is bounded in absolute value by $Cds\mathrm{/}s$. Every other pair of fixed branches has a nonzero limiting separation and hence a factor bounded by $Cds$, after shrinking the band. A return chord has factor bounded by $Cds\mathrm{/}(R-M)$ there, and by $Cdt\mathrm{/}(R-M)$ in a regular band. These constants depend on the fixed tangle, not $R$. For any nonzero pairing with short chords in one critical band, the short chord nearest the extremum must have a long endpoint between its ends through the extremum, or 1T kills it. If its short parameters are $0<s_k<\cdots <s_1<S$, retain that witness long factor and integrate its parameter first over $(0,s_k)$. This supplies $C{s}_k$, which cancels the innermost singularity: ${\int }_0^S\frac{d{s}_1}{s_1}{\int }_0^{s_1}\frac{d{s}_2}{s_2}\cdots {\int }_0^{s_{k-1}}\frac{d{s}_k}{s_k}{s}_k=S\mathrm{.}$ The height-reversed argument handles a minimum. Additional long factors have bounded integrals; dropping their ordering restrictions only enlarges this absolute bound. Apply the argument in each disjoint critical band and use bounded factors in regular bands. Retain one selected return-chord factor throughout: whether it is a witness or an additional long chord, its bound supplies the same factor $(R-M{)}^{-1}$. Thus every surviving pairing using the return arc has integral $O((R-M{)}^{-1})$, without deleting the factor that controls a short-chord singularity. The finitely many such pairings tend to zero; the remaining ones give $\mathrm{cl}(Z(T))$, with the cap units from step 1.2. This proves the degreewise limit. Varying $R$ preserves height and fixes every critical neighborhood of the closed knot, so [F6] makes $Z(K_R)$ constant; it equals that limit.

At the regular joining plane, smoothing the single matched upper and lower endpoints forms the neck in the connected-sum construction. Cutting that neck recovers the closures of $T_2$ and $T_1$, so the closure of their composite is their oriented connected sum as stated. On diagrams, the endpoints from $T_2$ occur before those from $T_1$ around the closure cut; applying $\mathrm{cl}$ to a stacked diagram therefore gives exactly the concatenated circle diagram defining $\mathrm{\#}$. Apply [F3] to the tangle composite and close the resulting degreewise identity to obtain the claimed series equality.