Quotient receipts for planar factorizations

The quotient picture

What is forgotten, and what must be kept?

Begin with a disk containing marked points and labeled interior boundary circles. Fill each interior circle by a punctured disk. The resulting marked disk retains the visible braid motion, but a full twist around a filled circle disappears. The integer vector of those lost twists is the receipt for capping.

A branched double cover asks a second question: can a base mapping class be lifted while fixing every lifted boundary point? Here one parity bit is not enough. The answer depends on which boundary components lie across an odd curve. Finally, two positive factorizations can have the same product and matching per-factor types while occupying different Hurwitz orbits. A finite group quotient can sometimes certify that separation.

QA-BK-D01 · definition

The capping quotient and its integral receipt

Let $D_n^b$ be a disk with$n$ marked points,$b$ labeled interior boundary components $\delta_1,\ldots,\delta_b$, and an exterior boundary $\delta_0$. Mapping classes fix every boundary pointwise. Capping the interior boundaries by once-marked disks gives

$$\pi:\operatorname{Mod}(D_n^b) \longrightarrow \operatorname{Mod}(D_{n+b},Q).$$

Its kernel is the free abelian group generated by the interior boundary twists. Buckman's Dehn-twist vector is normalized by

$$D=(D_1,\ldots,D_b),\qquad D(t_{\delta_j})=e_j.$$

Labels matter. If the boundary components may permute, the coordinates are permuted too and the direct-product theorem below must be replaced by a semidirect statement.

QA-BK-T01 · exact reconstruction theorem

$D$ is the complete capping receipt

Proof

If $(\pi,D)(g)=(1,0)$, then$g\in\ker\pi$, so$g=\prod_j t_{\delta_j}^{a_j}$. The normalization gives $D(g)=a$, hence $a=0$ and$g=1$. Thus the map is injective. Given $(h,z)$, choose any lift $g$ of$h$ and multiply it by the boundary multitwist with exponent vector$z-D(g)$. This produces the required preimage and proves surjectivity.

No lower-rank free-abelian receipt can separate all capping fibres: its restriction would have to inject$\mathbb Z^b$ into a smaller free abelian group. In this doctrine, all$b$ coordinates are necessary.

The capping kernel and framed-braid splitting are classical. The identification of Buckman's displayed vector as a complete quotient receipt is the FPRD deduction. Compare Bellingeri–Gervais.

QA-BK-T02 · cohomological receipt

The same vector is an explicit cocycle primitive

Choose any normalized set-section$\sigma$ of$\pi$. Its failure to preserve multiplication is a factor set

$$\sigma(h_1)\sigma(h_2)\sigma(h_1h_2)^{-1} =k(c_\sigma(h_1,h_2)),$$

valued in the forgotten lattice. Applying$D$ and writing$r=D\circ\sigma$ gives the exact identity

$$c_\sigma(h_1,h_2) =r(h_1)+r(h_2)-r(h_1h_2)=\delta r(h_1,h_2).$$

Thus $D$ does not merely detect a sewing failure: it supplies the cochain that removes it. The corrected section$s_D(h)=k(-r(h))\sigma(h)$ is homomorphic and has receipt zero.

QA-BK-N02 · failed obstruction

Where the expected obstruction does not exist

A search for a nonzero$H^2$ obstruction in this exact capping problem is doomed: the previous formula writes every section cocycle as a coboundary. This negative result is scoped. Permuting boundary labels changes the coefficient action, and analogous extensions over closed surfaces need not split. Those are different doctrines, not counterexamples to the labeled-disk theorem.

QA-BK-T03 · positive resource theorem

Group solvability does not erase positive cost

Suppose two lifted expressions $U,V$have the same capped image. Put$\Delta=D(V)-D(U)$. If positive boundary multitwists with exponent vectors$x,y\in\mathbb N^b$ may be added to the two sides, balance requires$x-y=\Delta$. Every solution is

$$x=\Delta_++c,\qquad y=\Delta_-+c, \qquad c\in\mathbb N^b.$$

Hence the unique minimum total number of added boundary twists is$\|\Delta\|_1$, attained at$(\Delta_+,\Delta_-)$. In the lantern comparison used by Buckman, the discrepancy is$(1,1,1)$; the three added boundary twists are therefore forced and resource-minimal for this correction model.

QA-BK-T04 · double-cover lift theorem

Parity needs a spatial address

Let $p:\widetilde S\to S$ be a connected double cover, classified away from its branch set by$\omega:\pi_1(S\setminus B)\to\mathbb F_2$. Choose arcs $a_j$ from the exterior boundary to each interior boundary. For a mapping class preserving $\omega$, define

$$\varepsilon_j(f)=\omega\!\left(f(a_j)a_j^{-1}\right).$$

The vector is independent of the arc choices modulo a common phase, so$\varepsilon(f)\in \widetilde H^0(\partial S;\mathbb F_2)$. Path lifting gives the criterion

For a planar Dehn twist the calculation becomes

$$\boxed{\varepsilon(t_c^m) =(m\bmod2)\,\omega(c)\,\chi_c},$$

where $\chi_c$ records which boundary components the curve $c$separates from the exterior boundary. The bit$\omega(c)$ says whether a twist changes sheets; the cut vector says where that change is visible. Scalar parity alone loses the second datum.

QA-BK-N01 · presentation boundary

A liftable product can have unliftable displayed factors

If $\varepsilon(t_c)\ne0$, then

$$\varepsilon(t_c^2)=0, \qquad t_c^2=(t_c)(t_c).$$

The squared twist is liftable as one factor, while the two displayed elementary factors are individually unliftable. A Hurwitz sequence may bring identical twists next to one another, but recognizing the pair as the atom$t_c^2$ is a packetization step, not another Hurwitz move.

This is already covered by the thesis's squared-twist and marked-point-to-boundary conventions. Its role here is algorithmic: any comparison procedure must declare whether packetization is part of its grammar. Strict Hurwitz equivalence, factorwise liftability, and substitution equivalence are distinct questions.

QA-BK-T05 · finite-quotient Hurwitz obstruction

Matching unary receipts do not determine the Hurwitz orbit

In $B_3=\langle x,y\mid xyx=yxy\rangle$, put

$$\alpha=x^2yx^{-2},\qquad \beta=(y^{-1}x)y(y^{-1}x)^{-1},$$

and consider$A=(\alpha,y,y)$ and$B=(x,\beta,y)$. Every entry is a positive half-twist—a conjugate of a standard generator. Using$xyx^{-1}=y^{-1}xy$,

$$\alpha y^2=x^2yx^{-2}y^2 =xy^{-1}xyx^{-1}y^2=x\beta y.$$

Thus the two tuples have the same product, length, factor type, conjugacy multiset, and factorwise abelian receipt.

The separating receipt

Map $B_3\to S_3$ by$x\mapsto s=(12)$ and$y\mapsto t=(23)$. Then

$$\rho(A)=(t,t,t),\qquad \rho(B)=(s,s,t).$$

The first tuple generates a subgroup of order two; the second generates all of $S_3$, of order six. A Hurwitz move$(a,b)\mapsto(b,b^{-1}ab)$preserves the subgroup generated jointly by the factors. Simultaneous conjugation changes it only by conjugacy. Therefore$A$ and$B$ are inequivalent under either doctrine.

Complete breadth-first search in the finite quotient gives Hurwitz-orbit sizes one and eight, respectively. The verifier and its certificate replay the product, subgroup, and orbit calculations without searching for new examples.

This witness does not solve the general Hurwitz-equivalence problem. Orevkov gives algorithms for relevant$B_3$ classes; unrestricted factorization problems become undecidable in larger braid groups. See Orevkov and Liberman–Teicher.

What remains open

The actual decision problem starts after these receipts

The cheap tower now distinguishes three layers: integral capping data, relative double-cover phase, and relational subgroup data in finite quotients. None is claimed complete for positive planar Hurwitz equivalence. The next bounded test is stricter: fix a boundary multitwist, a length, and the positive factor types used in Buckman's Chapter 4, then enumerate the finite class and ask where this receipt tower first merges distinct Hurwitz orbits.

A quotient separation is a sound negative certificate. A path in a finite quotient is not yet a positive certificate upstairs; it carries a kernel-valued lifting obligation. That is the remaining local-to-global problem.