Proof. The coefficient determinants of pairs in (2) belong, up to sign, to \(\{1,\lambda,1-\lambda^2\}\). The first two are units in \(\mathcal O\). Since the reciprocal of a nonzero algebraic integer \(v\) is \(\bar v/N(v)\), a common positive integer denominator gives the asserted \(D\). The group \(D^{-1}\mathcal O^2/\mathcal O^2\) is finite. Moreover, whether \(\ell_i(v)\) belongs to \(\mathcal O\) depends only on this residue class, so the directions through a vertex introduce no additional infinite choice. Every coset has dense ordinary image because \(\mathcal O^2\) does.

For the coordinate directions the tangent translation groups have bases \((1,0),(\tau,0)\) and \((0,1),(0,\tau)\). For the other directions they have bases \[(1,\lambda),\ \tau(1,\lambda) \quad\hbox{and}\quad (\lambda,1),\ \tau(\lambda,1).\] Multiplying both basis vectors by the unit \(\tau^{-b}\) leaves the group unchanged and makes their ordinary lengths as small as desired.

The two vectors \((1,1)\) and \((\tau,\bar\tau)\) generate the stated lattice, since their determinant is \(-\sqrt5\). Multiplication by \(\tau^k\) acts on the two factors with absolute scale factors \(\tau^k\) and \(\tau^{-k}\). A fixed lattice covering radius, followed by such a rescaling, therefore gives a constant \(C_0\) with this property: every axis-parallel rectangle of sufficiently large fixed area and side lengths comparable to \(L^{-1},L\), for \(L\ge1\), meets the lattice. More explicitly, start with a square containing a translate of a fundamental parallelogram around every point, and rescale by a unit with \(\tau^k\) within a factor \(\tau\) of \(L\); increasing one absolute constant absorbs this bounded factor.

Apply this observation with \(L\) a small fixed multiple of \(n\). For any chosen ordinary coordinate \(t\in[0,1]\), center the long conjugate interval so that \[i=\frac{z-\bar z}{\tau-\bar\tau}\] falls in the middle half of \(J\). A conjugate interval of length a fixed small multiple of \(n\), and an ordinary interval of length \(C_1/n\) about \(t\), then produce \(z=k+i\tau\) with \(i\in J\) and \(|z-t|\le C_1/n\). Constants may be enlarged to cover the bounded small values of \(n\). Thus these circular points form a \(C_1/n\)-net, which proves the gap bound with \(C=2C_1\).

For separation, choose the representative \(z=k+(i-j)\tau\) of a nonzero circular difference with \(|z|\le1/2\). It is a nonzero algebraic integer and \(|\bar z|=|z-(i-j)\sqrt5|\le\sqrt5 n+1/2\). Consequently \[1\le |N(z)|=|z\bar z| \le |z|(\sqrt5 n+1/2).\] Taking, for example, \(c=(\sqrt5+1)^{-1}\) proves the claim. ◻

Proof. We give separately the covariance, concentration, and signal estimates. The count following (6), now with \(t=N\eta\), shows that a row of \(M_r\) has \(O_\eta(r^2)\) nonzero entries, each \(O_\eta(r^{-2})\). Therefore \[ \|M_r(z,\cdot)\|_2\le C_\eta/r. \tag{12}\] The row sums are bounded by an absolute constant for all sufficiently large \(N\) at any fixed \(\eta\): omit \(P_r\) and \(\chi\), and use the uniform Riemann sum for \(K(u)K(v)\), whose integral is one. In particular this bound can be chosen independently of small \(\eta,\kappa\) once \(N\) is large enough.

Write \(\Delta_r=U_gM_rU_g^{-1}-M_r\). Compare its entries by comparing \((z,w)\) with \((gz,gw)\). If \(z,w\) lie in different continuity pieces on the same source track, some supporting line of the fixed continuity partition separates their flags. To see this, assign to a point the signs of all the finitely many supporting lines; each sign chamber lies in one continuity piece. Thus different pieces have different sign assignments. Its conjugate level is bounded independently of \(N\), so it is included for every threshold \(\kappa/s\) when \(N\) is large. The original separator factor is zero. Their images lie in different image pieces; the same reasoning makes the image factor zero if they are on the same track, and otherwise the image matrix entry is zero by definition. This also deals with points on different source tracks whose images share a track. Hence only pairs in a common continuity piece need be considered.

For such a pair, \(gz=z+u\) and \(gw=w+u\) in the chosen square charts, for a fixed \(u\in\mathcal O^2\). Both difference kernels in (7) are unchanged. In one direction let \(d\) be the least absolute conjugate level of a line separating \(z,w\), taking \(d=+\infty\) if there is none. The probability that the threshold is less than \(d\) is a function of \(d\) with Lipschitz constant \(s/\kappa\). Translation bijects the separating lines, changing their conjugate levels by \(\overline{\ell_j(u)}\). Either both minima are infinite, or they differ by a bounded amount. Taking the product over the four directions gives a change \(O_\kappa(s)\) in \(P_r\). Each cutoff factor changes by \(O(N^{-1})\). We have proved, on the union of the two entry supports, \[ |\Delta_r(z,w)|\le \frac{C_{\eta,\kappa}}{Nr}. \tag{13}\] This union contains \(O_\eta(N^2r^2)\) pairs, also after permutation. For \(r\le r'\), the intersection of the two entry supports contains at most the number of pairs in the smaller support. It follows that \[|\langle\Delta_r,\Delta_{r'}\rangle_{\mathrm{HS}}| \le C_{\eta,\kappa}\frac{r}{r'}.\] The scales are dyadic, so the sum of \(r/r'\) over \(r\le r'\) is \(O(|\mathcal R_N|)\). Consequently \[\|U_gB_NU_g^{-1}-B_N\|_{\mathrm{HS}}^2 \le C_{\eta,\kappa}(\log N)^{-3/2}|\mathcal R_N| =O_{\eta,\kappa}((\log N)^{-1/2}),\] which proves (9).

For concentration, (12) and the geometric sum over scales give \[\|B_N(z,\cdot)\|_2 \le C_\eta(\log N)^{-3/4}N^{-1/4}.\] The rowwise bounded-independent-summand estimate is Hoeffding’s inequality (Hoeffding 1963, Theorem 2), rederived here. If \(\xi\) is centered and \(|\xi|\le1\), convexity bounds its exponential moment by \(\cosh t\le e^{t^2/2}\). Multiplication of the independent exponential moments and optimization in \(t\) therefore give \[\Pr\{|(B_N\xi)_z|>1/2\} \le 2\exp\{-c_\eta N^{1/2}(\log N)^{3/2}\}.\] There are \(O(N^2)\) active rows; outside them \(B_N\xi=0\). The union bound proves (10).

It remains to explain why the matrices average the signal difference. On an image chart whose source translation is \(u\), the exact formula is \[h_N(z)=\theta\left(\frac{\bar z-\bar u}{N}\right) -\theta\left(\frac{\bar z}{N}\right).\] Thus \(|h_N|\le C/N\), it is supported on \(O(N^2)\) sites, and \(Nh_N\), viewed as a function of \(\bar z/N\), has a uniformly bounded Lipschitz constant on each fixed chart. In particular \(\|h_N\|_2\le C\). For large \(N\) its support, together with its \(2\eta\)-neighborhood in scaled conjugate coordinates, lies where \(\chi=1\) whenever \(\eta\) is sufficiently small. This guarantees that the cutoff introduces no error in averaging \(h_N\).

We bound the exceptional rows where a separator or a chart boundary can interfere with this averaging. For each direction \(\ell_j\) one may complete it to an \(\mathcal O\)-unimodular coordinate system, using respectively \[(x,y),\quad(y,x),\quad(y-\lambda x,x),\quad(x-\lambda y,y).\] In these coordinates the paired lattice is again a product of two one-dimensional quadratic lattices. In a bounded ordinary region and a conjugate ball of radius \(O(N)\), the number of sites within ordinary distance \(C s\) of a line \(\ell_j=c\) is, by (5), at most \[C(sN+1)(N+1)\le C sN^2,\] since \(sN=r\to\infty\). The number of lines meeting that ordinary region with \(|\bar c|\le2\kappa/s\) is \(O(\kappa/s+1)\). The rows within distance \(Cs\) of any of them consequently number at most \[ C(\kappa+s)N^2. \tag{14}\] The same estimate with only finitely many lines gives \(O(sN^2)\) rows near the fixed chart boundaries and square edges. These estimates include rows exactly on a line, independently of the flag convention.

Outside these exceptional rows, every contributing neighbor is in the same ordinary chart, and no line allowed in the definition of \(P_r\) separates it from the row. Hence \(P_r=1\). The normalized Riemann-sum argument in (6), together with the Lipschitz bound on \(Nh_N\), gives \[|(M_rh_N)(z)-h_N(z)| \le \frac{C}{N}\left(\eta+(r\eta)^{-1/2}\right).\] The assertion also holds for rows where \(h_N(z)=0\): a contributing nonzero value of \(h_N\) still lies in the cutoff-one neighborhood. On exceptional rows the bound \(C/N\) suffices, since the row sums are bounded. Outside a conjugate ball of radius \(RN\), with \(R\) fixed, both terms vanish. Squaring and summing yields, uniformly over \(r\in\mathcal R_N\), \[ \|M_rh_N-h_N\|_2 \le C\left(\eta+\sqrt\kappa+\sqrt{s}+(r\eta)^{-1/2}\right) = C(\eta+\sqrt\kappa)+o(1). \tag{15}\] The constant \(C\) in this estimate is independent of sufficiently small \(\eta,\kappa\); the \(o(1)\) is taken with them fixed.

Finally put \(e_N=B_Nh_N-W_Nh_N\). Equation (15) implies \(\|e_N\|_2\le W_N(C(\eta+\sqrt\kappa)+o(1))\). The identity \[(I+B_N)h_N=(1+W_N)h_N+e_N\] and the contraction bound for \((I+B_N)^{-1}\) give \[\|(I+B_N)^{-1}h_N\|_2 \le \frac{\|h_N\|_2}{1+W_N} +\frac{W_N}{1+W_N}\bigl(C(\eta+\sqrt\kappa)+o(1)\bigr).\] Since \(W_N\to\infty\), this proves (11). ◻

Proof. Set \(F(x)=\prod_{i=1}^d f(x_i)\), \(L=T_t^{-1}\dot T_t\) and \(v=T_t^{-1}\dot\mu_t\). Differentiate \((\det T_t)^{-1/2}F(T_t^{-1}(y-\mu_t))\) and pull back by \(y=\mu_t+T_tx\). The \(L^2\) norm of the derivative is the norm of \[(Lx+v)\cdot\nabla F+\tfrac12\operatorname{tr}(L)F.\] Under the product probability density \(F^2\), write \(a(x)=f'(x)/f(x)\) on the interior of the support. This notation is legitimate in \(L^2(f^2\,dx)\) because \(\int(f')^2<\infty\). Integration by parts gives \[\mathbb E a(\xi_i)=0,\quad \mathbb E[\xi_i a(\xi_i)]=-\tfrac12,\quad \mathbb E\xi_i=0.\] The diagonal summands are \(L_{ii}(\xi_i a(\xi_i)+1/2)\) and the mean summands are \(v_i a(\xi_i)\). They have mean zero, uniformly bounded second moments, and different coordinates are orthogonal. The two types on the same coordinate are orthogonal as well, by parity. An off-diagonal summand is \(L_{ij}\xi_j a(\xi_i)\); it has mean zero separately in each of its two coordinates. Therefore it is orthogonal to every single-coordinate summand and to off-diagonal summands with a different unordered coordinate pair. The two terms for \((i,j)\) and \((j,i)\) need not be orthogonal, but Cauchy–Schwarz bounds their combined second moment by \(C_f(L_{ij}^2+L_{ji}^2)\). Summing over unordered pairs proves the estimate. ◻

Proof. We first consider one signal and one flag type. The two noise matrices are \(T_0=I+B_N\) and \(T_1=I+U_gB_NU_g^{-1}\), and the two means are \(\mu_N\) and \(U_g\mu_N\). There is a finite set \(J_N\) of \(O(N^2)\) sites containing the row and column supports of both perturbation matrices and the supports of \(\mu_N-q\) and \(U_g\mu_N-q\), where \(q\) is the constant value of \(\theta\) outside a compact set. Outside \(J_N\) both laws have exactly the same independent background factor, namely the constant outside value of \(\theta\) plus independent noise with density \(f^2\). Thus their total variation distance on the whole product space is exactly the total variation distance of their restrictions to \(J_N\). This is the reason that the calculation below proves an assertion about all sites, rather than only about finitely many specified marginals.

On \(J_N\) interpolate linearly between the matrices and means. Write \(\Delta=T_1-T_0\), \(T_t=T_0+t\Delta\), and \(h=h_N\). Each \(T_t\) is the identity plus a positive semidefinite matrix, so \(\|T_t^{-1}\|_{\mathrm{op}}\le1\). Also the resolvent identity gives \[ T_t^{-1}h=(I-tT_t^{-1}\Delta)T_0^{-1}h. \tag{17}\] Consequently \[\sup_{0\le t\le1}\|T_t^{-1}h\|_2 \le(1+\|\Delta\|_{\mathrm{op}})\|T_0^{-1}h\|_2, \qquad \|T_t^{-1}\Delta\|_{\mathrm{HS}}\le\|\Delta\|_{\mathrm{HS}}.\] Lemma 8, integrated over \(t\), bounds the \(L^2\) distance between the square-root densities by \[C_f\bigl(\|\Delta\|_{\mathrm{HS}} +(1+\|\Delta\|_{\mathrm{op}})\|T_0^{-1}h\|_2\bigr).\] For probability densities \(p,q\), Cauchy–Schwarz gives \(\frac12\|p-q\|_1\le\|\sqrt p-\sqrt q\|_2\). Lemma 7 therefore proves that the limiting total variation error is at most \(C(\eta+\sqrt\kappa)\). Choose these two parameters small and then \(N\) large. Thresholding the fields cannot increase total variation. For finitely many independent fields the joint error is bounded by the sum of their errors, using independent couplings.

On the event \(\|B_N\xi\|_\infty\le1/2\), the noise in each coordinate of \(Y_N\) has absolute value at most \(3/2\). Hence every signal value at least two gives bit one, and every value at most minus two gives bit zero, simultaneously. The event has probability tending to one by (10). Outside the finite matrix support the same statement holds deterministically because \(|\xi_z|\le1\). ◻

Proof. For the eight bit fields, Proposition 9 provides a coupling of an original sample \(b\) and a sample \(b'\) with the same law such that \[ b'(gz)=b(z) \tag{22}\] for both bits in every direction and every site, except on an event of arbitrarily small probability \(\varepsilon_N\). Such a coupling follows directly from the common part of the two probability measures: the mass of their pointwise minimum is one minus their total variation distance; on that mass make the samples identical, and couple the residual measures arbitrarily. This construction also applies to the countable product space, since the measures here differ only on a finite coordinate set. Intersect this coupling event with success of both samples. The resulting event has probability at least \(1-\varepsilon_N-2\alpha_N\). We prove the required equality deterministically on it.

First every fixed chart-edge supporting line meeting the square interior is a full barrier in both samples for large \(N\). Its conjugate level is bounded, so it is a candidate and satisfies the positive condition (20). Every subdivision start lies in the first-mark support and hence in the radius-\(R_1\) ball after scaling. Its second bit is therefore one. The incoming square endpoint is forced by the core, and the final endpoint is supplied geometrically, so all of this line’s segment in the square is covered by activated intervals. Square edges are barriers by definition. In particular the open cells of each sample stay in the corresponding source or image continuity charts.

Fix one open source chart \(U\), translated by \(u\) onto the open image chart \(V\). Consider a candidate line \(L\) that meets \(U\). Its image is the parallel allowable line \(L+u\), with conjugate level shifted by \(\overline{\ell_j(u)}\). First suppose that both lines are candidates. Interior marked points of \(L\cap U\) correspond exactly to interior marked points of \((L+u)\cap V\) by (22). At every transverse crossing of \(L\) with \(\partial U\), the subdivision point was already forced by (19); the corresponding intersection with \(\partial V\) is likewise forced independently. Thus the subdivision intervals in these open charts have exactly corresponding interiors and endpoints under translation.

The direction of the flag matters at the endpoints. If an interval starts at \(p\) and then enters \(U\), its positive tangent flag selects \(U\). Its initial second bit consequently maps to the bit at \(p+u\), which is the start of the corresponding interval in \(V\). Hence the two intervals have the same activation decision. If the terminal endpoint is \(q\), its positive tangent flag may enter a different source chart and use a translation different from \(u\). No identity between its second bit and the bit at \(q+u\) is needed: the preceding interval uses the bit at \(p\). The point \(q+u\) is instead an independently forced destination subdivision point when it lies on an interior chart edge, or a geometric truncation when it is a square exit. In particular, this comparison never inserts a new random subdivision into an already activated interval. Both source and destination were subdivided at all transverse chart crossings before the activation decisions were made. Figure 1 records the distinction.

At a vertex where several chart edges meet, an interval entering an open chart still has its positive tangent flag in that chart. If an interval runs along a chart boundary, that boundary is already a full barrier and requires no activation comparison. Parallel lines either do not cross an edge or coincide with its supporting line; these alternatives have just been covered. Intersections that only touch at a point make no difference to Boolean cells.

It remains to handle the finite line cutoff. If \(L\) is a candidate but \(L+u\) is not, then \[|\bar c|\ge4N-Q>3N\] for sufficiently large \(N\). Every second bit at a start on \(L\) is zero by (21), so the line contributes no active segment. The reverse mismatch is handled by the same inequality on the destination line. A line absent because it misses the square cannot produce a mismatch inside \(U,V\), since a line meeting one of these corresponding open charts translates to a line meeting the other. Square seam translations were included in the bound \(Q\) and in the endpoint core. Thus active barriers inside \(U\) translate exactly to the active barriers inside \(V\) in every case.

Since all chart edges are full barriers in both samples, this equality inside each pair of open charts identifies their connected complementary components. Each source cell therefore translates to precisely one destination cell, and all destination cells occur. It follows that \(\mathcal P'=g\mathcal P\). Finally choose \(\eta,\kappa\) small enough and then \(N\) large enough that \(\varepsilon_N+2\alpha_N<\delta\), simultaneously for the finite set \(\mathcal K\). ◻

Proof. Choose one representative \(t\) in every right coset \(Ht\) of \(H\) in \(J\), and write each \(x\in J\) uniquely as \(x=h t\). Define \(\pi(x)=h\). Then \(\pi(kx)=k\pi(x)\) for \(k\in H\). The pushforward \(p=\pi_*\nu\) is a probability measure on \(H\), and summing over fibers shows \[\|kp-p\|_1\le\|k\nu-\nu\|_1.\] Start with the sum on the right less than \(\varepsilon/2\). If \(K\) is empty any singleton suffices, so assume \(K\) is nonempty. Choose a finite \(E\subset H\) with \(p(E)>1-\varepsilon/(8|K|)\), decreasing this error further if necessary to ensure \(p(E)>0\), and put \(q=p\mathbf1_E/p(E)\). Direct calculation gives \(\|q-p\|_1=2(1-p(E))\). Therefore \[\sum_{k\in K}\|kq-q\|_1 \le \sum_{k\in K}\|kp-p\|_1 +4|K|(1-p(E))<\varepsilon.\] For \(t>0\) let \(A_t=\{h\in H:q(h)>t\}\). These sets are finite, and the identities for real numbers \(a,b\ge0\), \[\int_0^\infty\mathbf1_{\{a>t\}}\,dt=a, \qquad \int_0^\infty |\mathbf1_{\{a>t\}}-\mathbf1_{\{b>t\}}|\,dt=|a-b|,\] give, after finite summation, \[\int_0^\infty |A_t|\,dt=1, \qquad \int_0^\infty\sum_{k\in K}|kA_t\mathbin\triangle A_t|\,dt =\sum_{k\in K}\|kq-q\|_1<\varepsilon.\] If every nonempty \(A_t\) had its summed boundary at least \(\varepsilon|A_t|\), integration would contradict these two formulas. For some \(t\) the nonempty finite set \(A=A_t\) therefore satisfies the asserted strict inequality. ◻

Proof of Theorem 6. Fix a subgroup \(H\le F_m\), a finite \(K\subset H\), and \(\varepsilon>0\). For nonempty \(K\), apply the construction with \(\mathcal K=K\cup K^{-1}\) and with the total variation error in (23) less than \(\varepsilon/(4|K|)\). It supplies a probability measure on \(F_m\) whose summed \(\ell^1\) error for \(K\) is less than \(\varepsilon/2\). Lemma 11 supplies a nonempty finite \(A\subset H\) with summed boundary less than \(\varepsilon|A|\), and hence \(|kA\mathbin\triangle A|<\varepsilon|A|\) for every \(k\in K\). The empty \(K\) case is immediate. Taking \(H=F_m\) proves amenability of \(F_m\), and taking \(H=A_m\) gives the assertion needed for the simple group. All choices were made with \(m\) fixed; no passage to infinitely many tracks has occurred. ◻

Proof. Any \(b\) vertices have a common disjoint vertex if \(m>b+1\): the union of their images has area at most \(b\), so its complement has area greater than \(1\); Lemma 3 embeds \(X\) in that complement.

We construct a chain contraction in the asserted degrees. Choose a vertex \(v\) and put \(h_{-1}(1)=v\). Suppose \(h_{-1},\ldots,h_{p-1}\) have been constructed, satisfy \(\partial h+h\partial=\mathop{\mathrm{id}}\) in degrees below \(p\), and the vertices appearing in \(h_{p-1}(\rho)\) are bounded in number independently of the \((p-1)\)-simplex \(\rho\). For a \(p\)-simplex \(\sigma\), the chain \[z_\sigma=\sigma-h_{p-1}(\partial\sigma)\] is a cycle. Its vertices have a bound depending only on \(p\): if the bound for \(h_{p-1}\) is \(b_{p-1}\), a valid bound is \((p+1)+(p+1)b_{p-1}\). Choose a vertex \(w_\sigma\) disjoint from all these vertices and define \(h_p(\sigma)=w_\sigma*z_\sigma\), where the cone prepends \(w_\sigma\) to each ordered simplex. The identity \[\partial(w_\sigma*z)=z-w_\sigma*(\partial z)\] gives \(\partial h_p(\sigma)+h_{p-1}(\partial\sigma)=\sigma\). Moreover, \(b_p=1+(p+1)+(p+1)b_{p-1}\) is a uniform bound for the vertices appearing in \(h_p(\sigma)\). Starting with \(b_{-1}=1\), choose \(m\) larger than all the finitely many bounds needed for \(p\leq d\). Linear extension defines the required contraction. There is no requirement that the contraction be equivariant or that one vertex avoid an arbitrarily long cycle. ◻

Proof. Tensor the augmented configuration complex with a free right \(\mathbb Z[F_m]\)-resolution of \(\mathbb Z\) and take its total complex. One filtration, together with Lemma 13, shows that its homology vanishes in any prescribed bounded range when \(m\) is large. The other filtration has, in columns \(p\geq-1\), \[ E^1_{p,q}=H_q(F_{m-p-1}),\qquad E^1_{-1,q}=H_q(F_m). \tag{24}\] Here and below only boundedly many columns are needed. The identification follows directly from the permutation module on simplices: restricting the free resolution to a stabilizer is still a free resolution.

Every face inclusion of a standard stabilizer is a stabilization followed by conjugation in the target stabilizer. Inner conjugation induces the identity on homology. Thus \(d^1\) is stabilization when \(p\) is even and is zero when \(p\) is odd. In particular, in row \(q\) the entries at columns \(-1\) and \(0\) of \(E^2\) are respectively the cokernel and the kernel of \(H_q(F_{m-1})\to H_q(F_m)\).

The row \(q=0\) is the alternating augmented row of copies of \(\mathbb Z\) and is exact. For \(q=1\), the only possible higher differentials into columns \(-1\) and \(0\) come from row zero, so both entries vanish, as the total homology vanishes. This proves stability in degree one. For \(q=2\), the possible sources for the entry \((-1,2)\) are \((1,1)\) and \((2,0)\), and those for \((0,2)\) are \((2,1)\) and \((3,0)\). The degree-one stability just proved kills the indicated row-one entries when \(m\) is increased by a fixed amount; row zero is exact. No higher differential has a nonnegative source row, and there are no outgoing higher differentials from columns \(-1\) or \(0\). Hence the kernel and cokernel in degree two are zero. Transitivity through column four and augmented acyclicity through degree two suffice for this argument; all the requirements are finite.

For \(\mathop{\mathrm{Sym}}(m)\) use ordered configurations of distinct points. The same bounded contraction works by choosing an unused point, the action is transitive, and its stabilizer is \(\mathop{\mathrm{Sym}}(m-p-1)\). The identical spectral sequence argument applies. ◻

Proof. We use the Serre homology spectral sequence in the form of (Hatcher 2004, Theorem 1.3, p. 8). For (i), apply it to the path-loop fibration, obtaining \[E^2_{a,b}=H_a(Z;H_b(\Omega Z))\Longrightarrow H_{a+b}(PZ), \qquad PZ\simeq *.\] All coefficients are constant because \(Z\) is simply connected. If the base homology is finitely generated through degree \(r+1\), induct on \(b\leq r\). The term \(E^2_{0,b}=H_b(\Omega Z)\) has no outgoing differential. Its incoming sources have strictly smaller fiber degree and base degree at most \(b+1\). They are finitely generated by induction and the universal coefficient theorem. The term at infinity is zero for \(b>0\), so finitely many finitely generated images exhaust \(H_b(\Omega Z)\). For the converse, induct on \(a\leq r+1\) in the row-zero term \(E^2_{a,0}=H_a(Z)\). It has no incoming differential, its outgoing targets have smaller base degree and fiber degree at most \(a-1\), and its limiting term is zero for \(a>0\). The successive quotients in this finite filtration are subgroups of finitely generated groups. This proves (i).

For (ii), the two products on based loops, concatenation and the \(H\)-space product, give the same operation on \(\pi_1(Y)\) and make it abelian. Thus \(\pi_1(Y)=H_1(Y)\). This fundamental group acts trivially on the homology of the universal cover. To see the latter claim, represent a deck transformation by a loop \(\alpha\) at the unit. Lift the family of maps \((t,y)\mapsto\alpha(t)y\) to the universal cover, starting with the identity (using the unit homotopy if necessary). Its terminal map is the deck transformation represented by \(\alpha\). Consequently that map is homotopic to the identity.

Put \(Q=\pi_1(Y)\). The spectral sequence for the universal cover is \[E^2_{a,b}=H_a(Q;H_b(\widetilde Y))\Longrightarrow H_{a+b}(Y),\] with constant coefficients. A finitely generated abelian group has finitely generated homology in each degree with finitely generated constant coefficients: tensor the resolutions for an infinite cyclic group and for a finite cyclic group. Induction on \(b\) in column zero now proves the forward implication, since all incoming sources have lower fiber degree and the limiting term is a subgroup of \(H_b(Y)\). The reverse implication follows directly from the finitely many terms on each total-degree diagonal. ◻

Proof. The monoid \(M=|N\mathcal B|\) is well based and has CW homotopy type. The symmetry supplies a homotopy between the two orders of multiplication, so \(\pi_0(M)\) is central in the integral Pontryagin ring. The integral group-completion theorem (McDuff and Segal 1976, Proposition 1) therefore identifies the homology of \(\Omega BM\) with the localization of \(H_*(M)\) at its components.

There is one cofinal object here. The object \(U=(U_1,\ldots,U_k)\) has complement \(U^c=(X\setminus U_1,\ldots,X\setminus U_k)\) and \(U\sqcup U^c\cong kX\). Hence inverting \(X\) inverts every component. Its localization is computed by the telescope \[T=\operatorname{tel}(M\xrightarrow{X\sqcup-}M \xrightarrow{X\sqcup-}\cdots).\] Equivalently, every right translation on this telescope is a homology equivalence: translation by \(X\) is invertible, tensor symmetry exchanges left and right translations, and a complement gives an inverse for translation by any object. This verifies the action hypothesis in the telescope form of group completion (McDuff and Segal 1976, Proposition 2 and p. 281); see also the corrected proof in (Miller and Palmer 2015, Theorem 4.1), applied with ordinary integral homology. In the nerve-realization model, \(M\) is a Hausdorff, locally contractible CW complex and its telescope is Hausdorff, as required there.

The bar two-skeleton identifies \(\pi_1(BM)\) with the group completion of the commutative monoid of object-isomorphism classes: an object supplies a loop, and a pair supplies the relation that its sum represents the product. The component of an object \(U\) at stage \(r\) of \(T\) is therefore \([U]-r[X]\). If this is zero, there is an object \(V\) with \(U\sqcup V\cong rX\sqcup V\). Choose \(V'\) with \(V\sqcup V'\cong kX\). Then \[U\sqcup kX\cong(r+k)X.\] Thus every component occurring in the neutral telescope component eventually becomes a standard component \(BF_{r+k}\). Choices of the identifying isomorphism differ by an inner automorphism and give the same map on homology. It follows that the neutral component has homology \(\operatorname{colim}_m H_j(F_m)\), with the usual stabilization maps. Group completion respects these component gradings, proving the first assertion. For \(\mathcal S\), use a singleton in place of \(X\); the same argument gives the second assertion. ◻

Proof. Finite groups have finitely generated integral homology in every degree, as their bar chain groups have finite rank in each degree. Lemma 14 therefore gives finite generation of the stable homology of \(\mathop{\mathrm{Sym}}(m)\) through degree two. By Lemma 16, this is the homology through degree two of \(\Omega_0 B^1\mathcal S\), or equivalently of \(\Omega\widetilde{B^1\mathcal S}\). Lemma 15(i) gives finite generation for the cover through degree three. The fundamental group of \(B^1\mathcal S\) is the group completion of \(\mathbb N\), hence \(\mathbb Z\). Part (ii) gives finite generation for \(B^1\mathcal S\) through degree three.

Ascend the bars using their skeletal filtrations and the product models (25). In the normalized filtration for the next bar, column zero is a point, so a positive-degree term of total degree at most \(d+2\) has bar degree at least one and internal degree at most \(d+1\). The induction hypothesis and the integral Künneth theorem make the homology of every finite product in this range finitely generated. There are finitely many bidegrees on each relevant diagonal. This proves the assertion successively for \(d=2\) and \(d=3\). ◻

Proof. Fix three bar indices \(\boldsymbol n=(n_1,n_2,n_3)\), and put \(\mathcal D=D_{n_1}D_{n_2}D_{n_3}\mathcal B\). Let \(\mathcal D^{\mathrm{pos}}\) have those objects of \(\mathcal D\) whose decomposition isomorphisms at every bar level preserve positions in \(X\), but retain all horizontal diagram isomorphisms between them. The inclusion \[ \mathcal D^{\mathrm{pos}}\longrightarrow\mathcal D \tag{29}\] is full and essentially surjective. Indeed, replace every multi-interval object by the lexicographically ordered sum of its elementary entries, indexed by products of consecutive intervals \([i_1-1,i_1]\times[i_2-1,i_2]\times[i_3-1,i_3]\) in the three bar-index sets. Its decomposition maps are canonical reorderings and preserve positions. The original coherent decomposition maps identify each ordered sum with the corresponding original multi-interval. They form a diagram isomorphism because changing the order of the cuts only introduces the specified symmetric reordering. The comparison is allowed to translate pieces: it is a horizontal arrow. For example, in one bar at degree two, the long edge \(U_{02}\) is replaced by \(U_{01}\sqcup U_{12}\), and the comparison on that edge is the original map \(\alpha_{012}\), regardless of whether it preserves positions.

The inclusions (29) commute with all bar operations. A face pulls back the existing interval diagram and its specified decomposition maps; it does not choose a new ordered-sum normalization. Composing position-preserving decompositions, inserting a unit, and reordering sums remain position-preserving. We use these inclusions as the natural maps; no strictly natural choice of an inverse normalization is required.

Now take the double nerve whose horizontal strings use arbitrary arrows of \(\mathcal D^{\mathrm{pos}}\) and whose vertical arrows preserve positions. Reading it horizontally first gives exactly the nerve of \(D_{n_1}D_{n_2}D_{n_3}\mathcal C_p\): its decomposition maps are position-preserving ladders, precisely the condition built into the morphisms of \(\mathcal C_p\).

Read the same double nerve vertically first, fixing a vertical string length \(q\). Its groupoid has objects strings \(V_0\xrightarrow{v_1}\cdots\xrightarrow{v_q}V_q\) of position-preserving arrows and has arbitrary horizontal commuting ladders. Evaluation at \(V_0\) is an equivalence with \(\mathcal D^{\mathrm{pos}}\). It is essentially surjective by constant strings. It is fully faithful because a horizontal arrow \(a_0:V_0\to W_0\) uniquely determines all its components: \[ a_i=w_{0i}\,a_0\,v_{0i}^{-1}, \qquad v_{0i}=v_i\cdots v_1, \quad w_{0i}=w_i\cdots w_1. \tag{30}\] Every map in (30) is an allowed horizontal diagram map. Constant-string inclusions are natural in \(q\) and in every bar index, and are levelwise equivalences. Thus the constant-string maps and (29) give the asserted zigzag after taking nerves and successive realizations. The levelwise-equivalence criterion applies to these simplicial-set models as explained before (26). ◻

Proof. Let \(V\) be the set of intersections of at least two nonparallel allowable lines. At \(z\in V\), let \(r(z)\in\{2,3,4\}\) be the number of allowable lines through \(z\). Their \(2r(z)\) sectors give a free abelian group of sector germs. Quotient it by the constant germ and the indicator germs of either half-plane for each incident line; call the quotient \(Q_z\).

We spell out this integral quotient. Number the sectors cyclically, put \(r=r(z)\), and write a germ as \((a_0,\ldots,a_{2r-1})\). With indices modulo \(2r\), let \[d_i=a_{i+1}-a_i,\qquad q_i=d_i+d_{i+r}\quad(0\leq i<r).\] Then \(\sum_iq_i=0\), and \[ Q_z\cong\{(q_0,\ldots,q_{r-1})\in\mathbb Z^r:\textstyle\sum_iq_i=0\} \cong\mathbb Z^{r-1}. \tag{35}\] Indeed, cyclic differences identify germs modulo constants with \(\{d\in\mathbb Z^{2r}:\sum_i d_i=0\}\). A half-plane germ has difference vector \(\pm(e_i-e_{i+r})\). These vectors generate exactly the kernel of the pair-sum map \(d\mapsto q\); the map is onto, since \(d=(q,0)\) has zero total sum. This proves (35) over \(\mathbb Z\).

For \(f\in L\), record its sector germ modulo this subgroup at every \(z\). The result is a translation-equivariant map \[c:L\longrightarrow\bigoplus_{z\in V}Q_z.\] This has finite support. To define the germ, use the finite line family defining \(f\) and take a small neighborhood excluding its nonincident lines; other allowable directions through \(z\) merely refine sectors with equal values. A nonzero symbol can occur only where two of the finitely many defining lines cross. The density of the collection of all allowable lines causes no difficulty.

The map \(c\) is injective. If \(c(f)=0\), the equation \(d_{i+r}=-d_i\) says that the jump across each incident oriented line is the same on both sides of the vertex. Fix a line in a finite defining family for \(f\). Its jump is constant between successive intersections with that family and, by the preceding identity, continues unchanged across every intersection. It is therefore constant along the whole line. Compact support makes the jump zero far away, hence everywhere. All the line jumps vanish. Any two chambers of the finite line arrangement can be joined by a path crossing its lines away from vertices; their function values consequently agree. The common value is zero outside the compact support, so \(f=0\).

Lemma 2 places \(V\) inside \(D^{-1}\mathcal O^2\) for a fixed integer \(D>0\). Thus \(V\) has finitely many \(H\)-orbits; translation preserves the full set of incident directions, and a point has trivial translation stabilizer. After choosing one representative from each orbit, \[\bigoplus_{z\in V}Q_z \cong\bigoplus_{\alpha=1}^s R^{\,r(z_\alpha)-1}.\] This is a finite free \(R\)-module. Its submodule \(c(L)\) is finitely generated because \(R\) is noetherian, proving the Lemma. ◻

Proof. Put \(N=\mathbb Z^2\) and \(J=\tau\mathbb Z^2\). Then \(H=N\oplus J\) and \(J\cong\Gamma\). Cutting a compactly supported function along the integer square grid gives, as \(N\)-modules, \[ L\cong\mathbb Z[N]\otimes C(X,\mathbb Z). \tag{36}\] The second factor is represented by functions supported in one unit square, and \(N\) shifts the square index. Only finitely many squares meet any given support. The Boolean convention makes the half-open choice on square edges immaterial.

Tensor (36) with \(K\). Induction from the trivial subgroup, or the two-generator free abelian resolution, gives \[H_q(N;L\otimes K)= \begin{cases} C(X,\mathbb Z)\otimes K,&q=0,\\ 0,&q>0. \end{cases}\] On the coinvariants, a translation in \(J\) cuts and moves pieces back to the chosen square, so its action is exactly translation modulo one. Taking the two successive free abelian resolutions for \(N\) and \(J\) therefore identifies \[ H_i(H;L\otimes K) \cong H_i\bigl(\Gamma;C(X,\mathbb Z)\otimes K\bigr). \tag{37}\]

By Lemma 19, \(L\otimes K\) is a finitely generated \(R\)-module: tensor a finite set of \(R\)-module generators of \(L\) with a finite set of abelian generators of \(K\). The Koszul resolution on \(t_1-1,\ldots,t_4-1\) computes the left side of (37). Its chain groups are finite sums of \(L\otimes K\), and its homology is finitely generated over \(R\) by noetherianity. On this complex, exterior multiplication \(\varepsilon_a\) by the \(a\)th basis vector satisfies \[d\varepsilon_a+\varepsilon_a d=(t_a-1)\mathop{\mathrm{id}}.\] Thus each \(t_a-1\) acts trivially on homology. The homology is therefore a finitely generated module over \(R/(t_1-1,\ldots,t_4-1)=\mathbb Z\), as required. We have proved finite generation of \(L\) integrally before tensoring; no preservation of the corner-map injection under tensor product is being assumed. ◻

Proof. By Lemma 17, each \(K_j\) in (28) is finitely generated. Lemma 20 makes every relevant term of (34) finitely generated. There are finitely many terms of total degree at most five, apart from the rows already shown to vanish. Hence Lemma 18 gives finite generation of \(H_i(B^3\mathcal B)\) for \(i\leq5\).

Apply (26) and Lemma 15(i) twice. First \(H_i(B^2\mathcal B)\) is finitely generated for \(i\leq4\), and then \(H_i(B^1\mathcal B)\) is finitely generated for \(i\leq3\). Both bases in these applications are simply connected by (27). The connected \(H\)-space \(B^1\mathcal B\) has finitely generated abelian fundamental group, and part (ii) of the same Lemma gives finite generation of its universal-cover homology through degree three. Part (i) now gives finite generation of \(H_i(\Omega_0 B^1\mathcal B)\) for \(i\leq2\). Lemma 16 identifies these with the stable homology groups of \(F_m\), and Lemma 14 transfers the conclusion to every sufficiently large finite \(m\). ◻

Proof. Use the action of \(A_m\) on \(\mathcal E_m\). It is transitive in the bounded range of degrees needed here, by Lemma 4. For the standard \(p\)-simplex, its stabilizer is \[A_m\cap F_{m-p-1} =\ker\bigl(F_{m-p-1}\longrightarrow F_m/A_m\bigr).\] By Theorem 5, \(A_t=[F_t,F_t]\). The degree-one stability from Lemma 14 identifies \(H_1(F_{m-p-1})\to H_1(F_m)\) as an isomorphism when \(m\) is large and \(p\) is bounded. The displayed kernel is therefore exactly \(A_{m-p-1}\).

The augmented spectral sequence now has \(E^1_{p,q}=H_q(A_{m-p-1})\) and augmented column \(E^1_{-1,q}=H_q(A_m)\). Its \(E^2_{-1,2}\) is the cokernel of the stabilization map in the statement. Every \(A_t\) is perfect, so the row \(q=1\) vanishes; row zero is the exact alternating row of copies of \(\mathbb Z\). The only possible higher incoming differentials at \((-1,2)\) have sources \((1,1)\) for \(d^2\) and \((2,0)\) for \(d^3\). They are zero on their respective pages. There is no outgoing differential from column \(-1\). The limiting term vanishes by Lemma 13, so the cokernel is zero. ◻

Proof. Choose one finite integer \(b\) such that all the stabilization maps in Lemma 22 are surjective once the target index exceeds \(b\). Composition gives \[H_2(A_b)\twoheadrightarrow H_2(A_m)\qquad(m\geq b).\] Let \(Y\) be the first \(b\) tracks of \(mX\), and take \(m>10b\). For \(f\in F_m\), the restriction \(f|_Y:Y\to f(Y)\) is a piecewise translation between sets of area \(b\). Lemma 4 supplies \(a\in A_m\) that agrees with \(f\) on \(Y\). For every \(h\in A_b\), viewed as supported on \(Y\), one has the exact equality \[f h f^{-1}=a h a^{-1}.\] Conjugation by \(a\) is inner in \(A_m\) and acts trivially on its homology. The two induced maps agree on the image of \(H_2(A_b)\), which is all of \(H_2(A_m)\). This proves the claim. The bank \(Y\) is fixed independently of both \(f\) and the homology class; only the extending element \(a\) depends on \(f\). ◻

Proof of Theorem 12. Fix a finite \(m\) beyond the thresholds in Proposition 21 and Lemma 23. Put \(Q=F_m/A_m\). Theorem 5 identifies \(Q\) with \(H_1(F_m)\), so \(Q\) is finitely generated abelian. Apply the integral Lyndon–Hochschild–Serre spectral sequence (Weibel 1994, 6.8.2, pp. 195–196) to \(1\to A_m\to F_m\to Q\to1\): \[E^2_{p,q}=H_p(Q;H_q(A_m))\Longrightarrow H_{p+q}(F_m).\] By Lemma 23, \(E^2_{0,2}=H_2(A_m)\) itself. Perfection gives \(H_1(A_m)=0\), so the possible incoming \(d^2\) from \((2,1)\) vanishes. The only remaining possible incoming differential is \[d^3:E^3_{3,0}\longrightarrow E^3_{0,2}=H_2(A_m).\] Its source is a subquotient of \(H_3(Q)\) and is finitely generated. Later incoming differentials have negative source row, and none can leave column zero. Consequently \[E^\infty_{0,2}=H_2(A_m)/\mathop{\mathrm{im}}(d^3).\] This limiting term is the bottom filtration subgroup of \(H_2(F_m)\), which is finitely generated by Proposition 21. Both \(\mathop{\mathrm{im}}(d^3)\) and the displayed quotient are finitely generated, so \(H_2(A_m)\) is finitely generated. All thresholds used in choosing \(m\) are fixed finite integers, and the argument applies to every larger finite \(m\). ◻

Proof. First consider the Boolean algebra. Translate the coordinate interval test by \(i\alpha\), with \(i\) in an integer interval. The endpoints of these tests form a connected path in the orbit-index graph: each test joins \(i\alpha\) to \((i+1)\alpha\). Two different complementary arcs of the resulting endpoint set have different test assignments. Indeed, an arc between points with identical assignments contains either both endpoints of every such edge or neither; connectedness then says it contains all endpoints or none. The two points therefore belong to the same complementary arc. Enlarging the index interval consequently produces every circular interval whose endpoints lie in \(\mathcal O/\mathbb Z\). The two coordinate families generate the rectangular Boolean algebra.

Every allowed inclined line is a translate of its line through the origin. The allowable tangent translations are dense along the line by Lemma 2. Translated copies of \(Q\) therefore cover each relevant compact segment of that line. In each of finitely many such charts the translated basic sign test gives its side decision; rectangular tests clip the decision to the chart. A finite cover of the finitely many edges in a polygonal description proves that all polygonal tests are Boolean combinations of translates of \(\mathcal U_0\).

Fix an alphabet \(I\) with \(|I|\ge5\). The group \(\mathop{\mathrm{Alt}}(I)\) is perfect. For completeness, it is generated by \(3\)-cycles, and the commutator of two \(3\)-cycles meeting in one letter is a \(3\)-cycle. Conjugating this identity gives all the generators. If copies of this group on \(U\) and \(V\) are available in \(A_m\), then \[[g\text{ on }U,h\text{ on }V]=[g,h]\text{ on }U\cap V.\] Perfection gives every conditional permutation on \(U\cap V\). The product of the constant \(g\) and the inverse of \(g\) on \(U\) gives \(g\) on \(U^c\). This constructs all Boolean combinations in projection. A common translation on \(I\) can be balanced on one track outside \(I\). Hence it constructs all translated tests as well.

An alternating permutation of slots is a product of permutations of three slots. Subdivide their common parameter piece if necessary and add two spare slots, so that each such permutation lies in an alternating group on five slots. When their tracks are distinct, a balanced translation aligns the five offsets; the conditional group on the parameter piece is already available. If tracks repeat, move each slot to a private track using a conditional \(3\)-cycle with two private auxiliary tracks. These operations do not interfere on a repeated original track because the slots there are disjoint. After this routing the preceding distinct-track construction applies, and conjugating back gives the original slot permutation. Five slots require only finitely many private tracks. Processing each generator and each subdivision piece separately proves the assertion for one fixed sufficiently large \(m\). ◻

Proof. Take a free presentation \(E=P/R\) and put \[E^\ast=[P,P]/[P,R].\] Its map to \(E\) is surjective because \(E\) is perfect, and its kernel is central. In \(P/[P,R]\), the image of \(R\) is central and every element is a product of an element of the derived group and an element of that image. Taking commutators twice therefore shows that the derived group is perfect. This proves perfection of \(E^\ast\). Given a central lifting problem, choose free lifts of the generators of \(P\). They kill \([P,R]\), so their restriction to \([P,P]\) descends to the required homomorphism. Changing the free lifts multiplies them by central elements and hence does not change this restriction. Applying (a) to two lifts from the perfect group \(E^\ast\) proves uniqueness. This gives independence of presentation and functoriality.

For (a), the pointwise ratio of the two homomorphisms has central values and is consequently a homomorphism to an abelian group; it vanishes on a perfect group. For (b), when the relevant commutators are central, commutation with a fixed element is a homomorphism on the other subgroup. Perfection makes it trivial. Finally, if the image of \(h\in H\) is central in \(H/Z(H)\), then \(x\mapsto[h,x]\) has central values and is a homomorphism. It is trivial because \(H\) is perfect, so \(h\in Z(H)\), proving (c). ◻

Proof. Put \(C_0=C_{\Omega_0,I}\) and \(C_1=C_{\Omega_1,I}\). Restriction induces \(d:C_0\to C_1\): an old factor is repeated on all assignments extending it, and is killed if there are none. Write \(\Lambda:C_0^\ast\to H_0\) for the old canonical perfect-cover map, \(p:C_0^\ast\to C_0\) for its cover projection, and \(q:K\to K/Z(K)\). The new law gives \(\theta:C_1\to K/Z(K)\).

The homomorphisms \(q\Lambda\) and \(\theta dp\) agree on the canonical lifts of every defining old input copy: those inputs are the same elements of \(H_0\) and \(K\). These lifts generate \(C_0^\ast\). Indeed, their generated subgroup \(M\) surjects onto \(C_0\), because the defining conditional groups generate the assignment group. Thus \(C_0^\ast=M\ker p\); centrality of \(\ker p\) and perfection give \(C_0^\ast=[M,M]\le M\). Consequently \[q\Lambda=\theta dp.\] For an old assignment with no extension, this equality puts its original perfect image in \(Z(K)\), so that image is trivial. Otherwise it identifies the quotient maps of its old representation and the product of its extending new representations. The new factors commute, so their product is a homomorphism from \(\mathop{\mathrm{Alt}}(I)^\ast\). Uniqueness of lifts from a perfect source into \(K\to K/Z(K)\) identifies the two homomorphisms exactly. No inclusion \(Z(H_0)\subseteq Z(K)\) is needed. ◻

Proof. We give the support count, since it prevents reserves from growing under repeated refinement. Present \(\mathop{\mathrm{Sym}}(I)\) by transpositions \(t_{ab}\), imposing their involution relations, disjoint commutation, and the conjugation relations \[t_{ab}t_{bc}t_{ab}=t_{ac}\quad(a,b,c\text{ distinct}).\] These relations present the symmetric group: adjacent transpositions satisfy the usual braid and distant commutation relations, whose normal form moves the final letter to its prescribed position and then proceeds inductively; conversely the displayed relations express every transposition in adjacent ones. Each defining relation involves at most four letters.

Fix a five-letter reserve \(E\subseteq I\) and a transposition \(a\) on two of its letters. Reidemeister–Schreier rewriting for the parity subgroup, with coset representatives \(1,a\), gives generators \(a t_{bc}\) and their inverses. Each is an even permutation on at most four letters. To see the support bound directly, insert the running representative, \(1\) or \(a\), between successive letters of a symmetric relator. The rewritten relations use only the original at most four letters and the two letters of \(a\). Explicitly, using the equivalent symmetric relations \(x_tx_sx_t=x_{tst}\) indexed by transpositions, the rewritten presentation with \(u_t=at\) is \[u_a=1,\qquad u_t^{-1}u_su_t^{-1}u_{tst}=1,\qquad u_tu_s^{-1}u_tu_{tst}^{-1}=1 \quad(s,t\text{ transpositions}).\] The involution relations rewrite to cancellations. This also proves the subgroup presentation: insert the running representatives in a product of conjugates of symmetric relators representing a null word. No letter outside those in the relator and \(a\) is introduced.

For each \(\omega\in\Omega\), realize its conditional copy of \(a t_{bc}\) by a word in the test copies on \(E\cup\{b,c\}\), an alphabet of size at most seven. Such a word exists by the intersection-and-complement argument preceding (43); its length can depend on \(\Omega\). Use the rewritten alternating presentation on each assignment, and cross commutators between distinct assignments. Every relator so lifted involves at most the five reserve letters and four other letters, hence at most nine tracks.

An original generator on five tracks can be compared with the product of its assignment generators on the union of those tracks with \(E\), of size at most ten. Include all these comparisons and the internal relations of the input perfect copies. The resulting relations present the assignment group: after the comparisons every input is expressed by assignment generators, and the preceding product presentation then applies. Equivalently, the kernel of evaluation from the free group on the specified input symbols is normally generated by null words supported on at most ten tracks. This argument also applies to star inputs: their internal words evaluate through the underlying alternating permutations, and remain supported on the same five tracks.

Let \(r\) be any one of these small null words and let \(g\) belong to an original five-track perfect copy. Their combined alphabet has at most fifteen tracks. The central law there says \([r,g]=1\). Thus every normal generator of the evaluation kernel is central in the full subgroup \(H\). All its conjugates are therefore central, giving the global central law. For \(|I|<5\) one embeds in a five-letter alphabet; for \(5\le|I|\le15\) the conclusion is already among the hypotheses. The proof never requires more tracks for longer atom words or for more assignments. ◻

Proof. Two choices of \(d\) differ by a balanced translation zero on \(I\). The relations (ii)–(iii), first on a basis and then by additivity, make this difference commute with the basic copy. Since all translations commute, the same is true of any translate, then of every word on those translated copies. Canonical factors are represented by such words, so the assertion includes them.

For disjoint alphabets \(I,J\), choose one balanced vector that equals \(u\) on \(I\) and \(v\) on \(J\), compensating on a track outside their union. Both translated groups are conjugates by this single vector of their unshifted groups, which commute by the finite initial tables. Balance independence identifies them with any other chosen descriptions. Conjugating a word or a table proves (b); uniqueness of perfect lifts identifies the conjugated canonical factors with those for the translated family. The preceding commutation proves (c). The last statement in (c) follows from its imposed generating-shift instances and additivity. ◻

Proof. Part (a) follows from the commuting sector factors and (45). A formal predicate side over an actual rectangular cell satisfies this hypothesis for that cell and its known rectangular containers: those implications already hold in the retained rectangular coordinates of the assignment model. Geometric emptiness of the formal predicate side alone is not sufficient. For (b), test one perfect copy on an alphabet \(I\) of five tracks. Choose a fresh alphabet \(I'\) disjoint from the track support of \(u\) and an even permutation \(\sigma\) carrying \(I\) onto \(I'\). If necessary, two further unused letters correct its parity. Let \(v\) be a lift of this permutation conditional on \(J\), on the union of the relevant alphabets. By hypothesis \([u,v]=1\). Control gives \[v\rho_{P,I}(s)v^{-1} =\rho_{P,I'}(\sigma_*s),\] where \(\sigma_*\) is induced reindexing of the star group. The right side commutes with \(u\) by disjoint-alphabet commutation. Therefore \[ [u,\rho_{P,I}(s)] =v^{-1}[u,v\rho_{P,I}(s)v^{-1}]v=1. \tag{47}\] Only this one fragment is moved at a time.

Apply (b) to the ratio of two words to obtain the comparison assertion in (c). To obtain transitivity, suppose \(P\) is controlled inside \(J\) and \(J\) inside \(K\). The ratio of a \(K\)-conditional permutation and its constant counterpart centralizes the \(J\) copies; hence it centralizes \(P\) by (b).

For intersection control, write the four commuting factors of the joint \(J,K\) law as \[A=J\cap K,\quad B=J\cap K^c,\quad C=J^c\cap K,\quad D=J^c\cap K^c.\] Use the same star source on an alphabet large enough for the test under consideration, and abbreviate its representations by \(a,b,c,d\). Control inside \(J\) says that every \(c(s)d(s)\) centralizes the fragment; control inside \(K\) says that every \(b(t)d(t)\) does. Taking commutators gives \[[c(s)d(s),b(t)d(t)]=[d(s),d(t)].\] The image of \(d\) is perfect, so it centralizes the fragment. It follows that \(b\) and \(c\) do too. The \(a\) factor therefore has the same action as \(abcd\), the constant permutation. This proves control inside \(J\cap K\) without a joint law involving the fragment.

Finally, suppose one fragment is controlled on a side \(J\) and another on \(J^c\). Move the first to fresh tracks by a \(J\)-conditional permutation. This fixes the second by its own control and the \(J,J^c\) decomposition. The moved fragments commute by disjoint alphabets; conjugation back proves the assertion. Iterating handles disjoint unions of sectors. ◻

Proof. It suffices first to work on an alphabet with fixed spare tracks and then apply Lemma 27. Let \(H\) be generated by the conditional perfect groups. Choose a union \(J\) of atoms of one partition. Each input copy has a separate joint law with \(J,J^c\) and therefore an exact decomposition into its \(J\) and \(J^c\) factors. Let \(H_J\) and \(H_{J^c}\) be generated by all those respective factors. They are perfect and generate \(H\).

They commute. To check this, take a five-track factor from each side. A permutation conditional on \(J\) moves the first factor to fresh tracks, by its own pair law, and fixes the second, by the latter’s pair law. Their resulting disjoint alphabets commute. This calculation uses neither a triple law nor the desired simultaneous refinement. Thus \[H=H_JH_{J^c},\qquad [H_J,H_{J^c}]=1.\] The images in \(H/Z(H)\) form a direct product: an element in their intersection commutes with both images and hence is central in \(H/Z(H)\), whose center is trivial by Lemma 25(c).

We recall why different direct decompositions of a centerless group refine one another. Write \(Q=A\times B=C\times D\). For \(a\in A\), its \(C\) and \(D\) components separately commute with \(B\), since they do so after projecting the equality \([a,B]=1\) into the two factors. The centralizer of \(B\) in \(A\times B\) is \(A\), because \(B\) is centerless. Thus each component of \(a\) belongs to \(A\). The same applies to \(B\), so the two decompositions refine into their four intersections. Repeat this argument for the finitely many partition decompositions.

In every resulting factor \(Q_\omega\) of \(H/Z(H)\) an input conditional permutation is the constant permutation if its assigned predicate is true, and the identity otherwise. The factor is generated by these images, so it is a quotient of the constant \(\mathop{\mathrm{Alt}}(I)\). For star inputs their central kernels vanish in this factor: their pair law with each input group makes those kernels central in \(H\). The diagonal constant source is an exact alternating group by the initial tables. Consequently the simultaneous decomposition supplies the surjection \(C_{\Omega,I}\twoheadrightarrow H/Z(H)\), proving the formal central law.

We verify separately that earlier exclusions survive. Suppose the restriction of \(\omega\) to a subfamily \(\mathcal T\) is already forbidden. The usual commutator and complement words select that assignment in the formal model of \(\mathcal T\), with any prescribed alternating permutation as value there. In the already known law for \(\mathcal T\) these words evaluate to identity, so their images are central in \(H_{\mathcal T}\). In \(Q_\omega\), however, they give every image of the constant alternating group. The image of \(H_{\mathcal T}\) is the whole of \(Q_\omega\), since it includes that constant group. It follows that \(Q_\omega\) is abelian. It is also perfect, as a quotient of \(\mathop{\mathrm{Alt}}(I)\), and is therefore trivial. This removes the forbidden assignment without assuming that a center of a subfamily is central in all of \(H\).

Finally, each new assignment restricts to an allowed assignment of every old lawful subfamily, by the exclusions just proved. The input copies in these subfamilies are the same copies generating \(H\). Lemma 26 therefore identifies their canonical perfect factors with the corresponding products of new factors, including their unions. Applying Lemma 27 proves the asserted law on the full alphabet. ◻

Proof. For different cells, take one generating perfect copy in each. A permutation conditional on the first cell moves that copy to fresh letters and fixes the other one, using their separate laws with \(\mathcal P\). Disjoint-alphabet commutation and conjugation back show that \(H_P\) and \(H_{P'}\) commute. The exact splittings of the original copies show that the subgroup generated by the \(H_P\) contains \(H\).

Fix an original generator \(p(s)\) and one cell \(P\). Its own separate law with \(\mathcal P\) gives \[ p(s)=p_P(s)p_{P^c}(s),\qquad p_P(s)\in H_P. \tag{48}\] For a generating five-track fragment \(a_P\) of \(H_P\), move its letters off the support of \(p_{P^c}(s)\) by a permutation conditional on \(P\). This is the ordinary reindexing on \(a_P\), by \(a_P\)’s own \(P\)-law, and fixes \(p_{P^c}(s)\), by \(p\)’s own \(P\)-law. Disjoint-alphabet commutation gives \[ [p_{P^c}(s),a_P]=1. \tag{49}\] Thus \(p\) acts on \(H_P\) by its factor \(p_P(s)\), an inner automorphism of \(H_P\). This proves invariance using only two separate rectangular laws; no law relating the predicates of \(p\) and \(a_P\) was used.

All original conjugations can now be composed as automorphisms of the same \(H_P\). A replacement with the same action also preserves \(H_P\). Equality of their individual actions therefore implies equality of the actions of the two words. If the replacement word acts trivially, then \(w\) centralizes \(H_P\). Doing this for every cell proves that \(w\) centralizes the group they generate, and therefore \(H\).

For clarity, consider one container comparison. Let \(u\) be the ratio of two proposed replacements, and suppose their known joint law says that \(u\) centralizes every conditional perfect copy on \(J\). The fragment \(a_P\) is controlled inside \(P\), and actual rectangular containment and Lemma 29 make it controlled inside \(J\). Equation (47) gives \([u,a_P]=1\). Intersections of controlling rectangles are permitted by the intersection part of that lemma, so the argument includes clipped containers.

Finally let \(r\) be a central relator in a template containing \(J\). The central-law assertion is applied on the union of the alphabet of \(r\), that of the particular fragment \(a_P\), and a fresh bank for the move. In that template group \(r\) commutes with the \(J\)-conditional mover. After the move, \(a_P\) has disjoint alphabet from \(r\), so they commute exactly; conjugating back proves \([r,a_P]=1\). This applies one generator of \(H_P\) at a time. It does not require \(H_P\) to belong to the template group. Nor does it promote centrality in a smaller subgroup without testing the additional letters. The proof therefore also applies to formal wrong-side fragments, whose actual rectangular control was available from the start. ◻

Proof. It suffices to transfer one coordinate at a time. Put \[L=\lfloor n/8\rfloor,\qquad S=\lfloor L/2\rfloor.\] For \(n\ge\max(160,40B)\), we have \(L\ge n/9\) and \(S\ge n/20\). Choose an \(L\)-window \(V_0\subset W\) and enclose \(I\) by the cells of \(\mathcal D(V_0)\) which meet it; call their union \(E_0\). Each end of \(I\) is enlarged by at most \(C/L\le9C/n\). The containment \(I\subset E_0\) is compared within \(W^{+B}\), whose cardinality is at most \(\lfloor0.9n\rfloor+2B<n\). Consequently it gives control by \(E_0\) using the old law. This first comparison discards the original endpoint labels: no later comparison needs to retain an extreme endpoint of \(W\).

Choose an \(L\)-window \(V_T\) containing the endpoint labels of the target interval and their prescribed margin. Its starting label can be reached from that of \(V_0\) in steps of at most \(S\), with \[T\le \lceil20(K+2)\rceil\] after increasing \(n\) to absorb the fixed \(B\) terms. Write the successive windows as \(V_0,\ldots,V_T\). The integer interval spanned by consecutive windows has length at most \(L+S\); even after adding the fixed margins its length is less than \(n\). Having obtained \(E_{j-1}\), enclose it by cells from \(\mathcal D(V_j)\) to obtain \(E_j\). Both enclosures occur in the old law on the interval spanned by \(V_{j-1}\cup V_j\). Thus they have their common canonical representations there, and \(E_{j-1}\subset E_j\) transfers control. Each endpoint grows by at most \(9C/n\) at this step.

It follows that \(E_T\) enlarges \(I\) at each end by at most \[ \frac{9C(T+1)}{n}. \tag{51}\] Choose \(n\) so large that this quantity is less than \(\delta\). Then \(E_T\subset J_\alpha\). The terminal comparison is in the old law on \(V_T\) with its margin. Transitivity of control proves the claim for that coordinate. A circular interval crossing the chosen seam is simply a union of cells; the same argument is made in a coordinate lift and then projected to the circle. The estimate and the available old laws are unchanged.

Apply this construction to the two coordinates. First change one coordinate window while keeping the other fixed, and then change the other. Every comparison uses an old window in each coordinate. Finally intersect the two coordinate controls using Lemma 29. This proves control in \(J\). ◻

Proof. For \(i<j\), conjugation by \(l_j(t)\) on \(l_i(E)\) is conjugation by \(f(t)\), because \(h_j(E)\) commutes with \(l_i(E)\). The latter conjugation is conjugation by \(l_i(t)\), because \(h_i(E)\) commutes with \(l_i(E)\). Therefore \(l_i(t)^{-1}l_j(t)\) centralizes \(l_i(E)\). Writing \(l_j=l_i(l_i^{-1}l_j)\) now shows directly from the homomorphism identities that \(s\mapsto l_i(s)^{-1}l_j(s)\) is a homomorphism. For successive indices its image centralizes every earlier prefix, since both relevant prefixes act there by the same conjugation \(\operatorname{Ad}f(t)\). It consequently centralizes every earlier difference. Set the first prefix equal to \(1\) and the last equal to \(f\) if they were not already listed. Multiplying the differences telescopes to \(f\), as required. ◻

Proof. We prove the implication \[\mathcal R(n)\ \Longrightarrow\ \mathcal R(\lfloor6n/5\rfloor)\] for every sufficiently large \(n\); iteration from \(n_0\) gives unbounded window lengths. Put \(n'=\lfloor6n/5\rfloor\). In a window of length \(n'\) on the \(x\) axis take the two end subwindows \(W_-\) and \(W_+\), each of length \(\lfloor0.9n\rfloor\). Their intersection has length at least \(0.59n\) for large \(n\). Its circular cells form an anchor partition of mesh at most \(2C/n\).

Fix an anchor cell \(U\). Each of the two old laws decomposes its canonical copy into prefixes and suffixes at that subwindow’s cuts. Both give the same copy on \(U\), since the common anchor partition already has its old law. Contributions on different anchor cells commute, including contributions obtained from different subwindows: move one small alphabet to unused letters conditionally on its anchor cell, as in Lemma 29. For a cut \(a\) in \(U\) write its prefix and suffix representations as \(l_a,h_a\); these have common star source \(\mathop{\mathrm{Alt}}(I)^\ast\). If the cut occurs in both subwindows it belongs to the anchor partition, and the two representations already agree. Lemma 33 shows that it remains to prove \[ [l_a(\mathop{\mathrm{Alt}}(I)^\ast),h_b(\mathop{\mathrm{Alt}}(I')^\ast)]=1\qquad(a<b) \tag{52}\] for arbitrary small alphabets, and then to apply Lemma 27. Pairs coming from one subwindow already commute by its old law.

Suppose, then, that \(a\) and \(b\) come from different subwindows. The two embeddings of \(\lambda\) have complementary roles in this comparison: its large ordinary value separates two thin rectangles, and its small conjugate value keeps the new transverse label in an old window. Use an ordinary lift of \(U\) to order them. The endpoint separation bound gives \[ b-a\ge c/n'. \tag{53}\] Split both copies by a \(y\)-grid from \(\lfloor n/4\rfloor\) consecutive labels. This is permitted separately for each copy by the old rectangular law. Contributions on different \(y\)-cells commute by the conditional move on those cells. Thus fix one lifted \(y\)-cell \([t,t+h]\), where \(h\le5C/n\). The two rectangles to be compared are \[R_-=(U\cap\{x<a\})\times[t,t+h],\qquad R_+=(U\cap\{x>b\})\times[t,t+h].\] They have diameter \(O(1/n)\). By (50) and (53), \[ h<\lambda(b-a). \tag{54}\] The allowable line \[ y-t=\lambda(x-a) \tag{55}\] therefore separates these rectangles: \(R_-\) is above it and \(R_+\) is below it.

This geometric separation must now be proved at the level of controlled copies. Put \[z_-=(a,t),\qquad z_+=(b,t+\lambda(b-a)).\] Both points lie in \(\mathcal O^2\) in the chosen planar lifts. In the chart at \(z_-\), the northwest quadrant lies above (55); in the chart at \(z_+\), the southeast quadrant lies below it. For large \(n\), both rectangles lie in the inner chart boxes. The actual quadrant tables, followed by intersection control, show that the copy on \(R_-\) is controlled in the first positive sign and the copy on \(R_+\) in the second negative sign. To justify using these tables we check the old coordinate windows explicitly. At \(z_-\) the cuts \(a,t\) are already available, and adding fixed chart sides to a window of length at most \(0.9n\) still leaves length less than \(n\). At \(z_+\) the new \(y\)-label is governed by \[ \nu(\lambda d)=\bar\lambda\nu(d)+\nu(\lambda)d, \qquad d=b-a. \tag{56}\] Here \(|\nu(d)|\le n'\) and \(|d|\le2C/n\). Consequently \(|\nu(\lambda d)|\le|\bar\lambda|n'+O(1)\). Adding this displacement and the fixed chart margins to the transverse window of length \(n/4\) gives a window of length less than \(n\). The \(x\) coordinates use the second subwindow with a fixed margin. Every quadrant and box containment just invoked is thus a consequence of \(\mathcal R(n)\).

It remains to identify the positive sign in the two charts on the copy on \(R_-\). Write the tangent displacement as an integral combination of the fixed short tangent basis. Since its ordinary coordinates are \(O(1/n)\) and its conjugate coordinates are \(O(n)\), the sum of the absolute values of these two coefficients is \(O(n)\). This follows by applying the fixed inverse of the two-embedding basis matrix. Order the signed increments so that the successive positions stay within one largest step of the segment from \(z_-\) to \(z_+\). Such an ordering exists: whenever a partial tangent coordinate is below the interval joining the endpoints, take a positive remaining increment; above the interval take a negative one. The required sign exists because the sum of the remaining increments leads to the final endpoint. Inside the interval take either available sign. Each crossing overshoots an endpoint by at most one increment.

By the choice of the small basis, the successive charts therefore have overlap rectangles containing both \(R_-\) and \(R_+\) with one fixed positive margin, independent of \(n\). The labels of all chart centers, relative to the starting labels, are bounded by \(Kn\) for a fixed \(K\): there are \(O(n)\) increments, each with fixed labels. Consecutive sign decisions and their overlap rectangle belong to one of the fixed tables after a common translation.

Apply Lemma 32 to the original \(R_-\) and each overlap rectangle separately. The source windows have lengths \(0.9n\) and \(n/4\) as required, and the overlap has the fixed margin just obtained. Thus the copy on \(R_-\) is controlled in that overlap. The two sign decisions agree there in the fixed table, so Lemma 29 says that their conjugations on the copy on \(R_-\) agree exactly. Composing these equalities identifies the first positive sign’s action with the last one’s action.

Although the chart chain has \(O(n)\) steps, its geometric error does not grow with that number. The only enlargement is (51), separately for each overlap and always starting from \(R_-\). The chart chain composes equalities of actions, rather than successively enlarged rectangles. Figure 2 summarizes the separation and the short-window transfer used for each overlap. The copy on \(R_-\) is now controlled in the last positive sign, whereas the copy on \(R_+\) is controlled in its negative sign. These two signs have a true split table. The disjoint-control part of Lemma 29 proves their commutation and hence (52).

Lemma 33 now splits all the new cuts of \(U\) into their actual consecutive intervals. Their perfect factor images commute. The central kernel of each source \(\mathop{\mathrm{Alt}}(I)^\ast\to\mathop{\mathrm{Alt}}(I)\) has central image in its own factor, and hence in the product of all factors. Every original input copy is the prescribed product of these interval copies. Quotienting their generated group by its center therefore gives a quotient of the product of ordinary alternating groups indexed by the actual intervals. Doing this in every anchor cell supplies exactly the assignment-group map for the new one-coordinate central law. The identical argument with \(x-\lambda y\) proves the \(y\) law. Every pair of one primitive \(x\) test and one primitive \(y\) test has a joint true table at arbitrary offsets: a translation in the two coordinates changes the offsets independently. Apply Lemma 30 to assemble the new coordinate laws. Their already established one-coordinate exclusions leave exactly the products of actual interval cells, so no spurious rectangular assignment remains. Lemma 27 extends this law simultaneously to the required alphabets, completing the induction. ◻

Proof. First suppose the two charts are primitive charts with \(\mathcal O^2\) vertices on their common line. At the first chart use one of the fixed recentering offsets to obtain an inner chart near \(z\). The first overlap is intersected with the original clipping box. On a cell on its interior side this is a rectangular container of \(P\). The fixed comparison table proves equality of the two decisions on that container. The same construction is made at the final chart. Exterior cells are not an assertion of equality of the unmasked signs: the original decision is zero by its actual box-complement law, and a replacement there is clipped by that same box.

The two recentered \(\mathcal O^2\) vertices differ by an element of the tangent group. Express this difference in the fixed short basis and order the signed increments as in the proof of Proposition 34. The centers remain in a small neighborhood of the segment joining the vertices and hence near \(z\). More precisely, choose a fixed \(\varepsilon>0\) much smaller than every inner chart margin. The finite recentering nets may be chosen to put both endpoints within \(\varepsilon/10\) of the tangential projection of \(z\), and each basis step may be chosen shorter than \(\varepsilon/10\). The greedy ordering keeps every intermediate center within \(\varepsilon/2\) of that projection. If the distance of \(z\) from the line and the diameter of \(P\cup\{z\}\) are less than \(\varepsilon/10\), every middle overlap therefore has a fixed positive margin around \(P\), independently of the number of steps and all labels. The first and last overlaps have the same margin in their uncut directions; they are simply clipped at the original box edges. This supplies a uniform smallness bound, without shrinking a cell once for each step of a long chain. All rectangular containers now have their laws by Proposition 34, irrespective of their labels. Every consecutive fixed table therefore gives equal actions on a controlled copy by Lemma 29. Their composition gives the desired equality.

For a template with vertex \(r+u\), where \(u\in\mathcal O^2\), use the chosen \(\mathcal O^2\) anchor on each line of its representative at \(r\), translated by \(u\). The initial and terminal primitive anchors lie on the same labeled line; their difference is exactly in its tangent group. The fixed terminal comparison identifies the last primitive chart with the template chart. This accounts explicitly for vertices in \(D^{-1}\mathcal O^2\) without pretending that the vertex itself is an allowable translation. A single line through an arbitrary ordinary point needs only an \(\mathcal O^2\) anchor on that line sufficiently close to the point.

At either clipped endpoint every overlap is intersected with the appropriate inside coordinate half-planes of its box. These are rectangular conditions with their established law. Thus the argument applies separately to each actual rectangular side of the clipping boundary. It does not require the controlled copy to split by any intermediate sloping sign. Its sole use of the copy is its control in the rectangular overlap. ◻

Proof. We first compare one sloping predicate \(p\) with one translated primitive coordinate predicate \(v\). Around \(p\) choose the fixed fine rectangular grid included in its initial table. Include its clipping edges and its planar chart seams in the grid. The mesh is chosen smaller than the gaps between the two boundaries of a primitive coordinate predicate, and sufficiently small compared with the fixed chart margins and the finitely many angles. Translate the whole grid with \(p\).

This partition has an actual joint law separately with \(p\), by the fixed table, and with \(v\), by the rectangular law. On one of its cells \(P\), form \(H_P\) from the restricted copies of these two predicates and their complements, as in Lemma 31. If either predicate is constant on \(P\), its action on \(H_P\) is the indicated constant action. This assertion follows from its individual actual split with \(P\), together with the conditional move on \(P\); it does not use the pair law being proved. Only the other individual split is then needed to check a formal two-predicate relation.

If both predicates vary, \(P\) is inside the clipping box of \(p\), one sloping line crosses it, and exactly one boundary line of \(v\) crosses it. Their intersection is within a fixed multiple of the diameter of \(P\): solve the two line equations, whose angle is one of a fixed finite list. It lies in \(D^{-1}\mathcal O^2\). The grid was chosen fine enough that the appropriate translated concurrent template has a margin box containing \(P\). The intersection itself can lie just outside the clipping box of \(p\). In applying the line comparison choose instead a point \(z_0\) of its sloping line in \(\overline P\). The template intersection is within \(C_\lambda\operatorname{diam}P\) of \(z_0\), and its chosen \(\mathcal O^2\) anchor is within a fixed arbitrarily small distance of that intersection. Choose the mesh once so that this distance and the recentering errors fit the uniform margin established in Lemma 35. The tangent walk then stays near \(P\) regardless of its number of steps. No refinement depending on the relative label of \(v\) is needed. Replace \(p\) by its corresponding line decision in that template, using Lemma 35, and replace \(v\) by the matching axis decision, using only rectangular control. The replacements have exactly the same actions on \(H_P\) as the original generators.

The true template has the four ordinary sectors of these two crossing lines. Extend it outside its margin box by any one of those four assignments. Therefore a word trivial on all formal assignments of \((p,v)\) is trivial in this pointwise template model. Its lifted value is central in the template subgroup. Its rectangular margin box controls every generator of \(H_P\), since it contains \(P\) and all rectangular laws are available. Take the template law on the enlarged alphabets required by Lemma 31. The template-relator conclusion of that lemma now makes the replacement word act trivially on \(H_P\).

Thus every formal two-predicate relator acts trivially on each \(H_P\). Lemma 31 and then Lemma 27 prove the pair law. Apply this to the coordinate predicates generating any desired rectangular partition, and apply Lemma 30. Retain all exclusions already supplied by the rectangular law. The resulting sectors have an actual rectangular cell coordinate, although their sloping side may still be formal. Refining by another rectangle and using the same pair laws proves that each sector is controlled in every rectangular container of its cell. No compatibility of two different sloping predicates has been used. ◻

Proof. Outside \(B\) the actual law of \(p\) with its own clipping box gives the assertion, by exterior rectangular control. Consider the interior side, allowing \(z\) itself to be on an edge or corner of \(B\). We give the argument for \(L=\{y-\lambda x=c_0\}\); interchange coordinates for the other slope. Put \(d=z_y-\lambda z_x-c_0\ne0\).

Suppose first that \(d>0\). The point on \(L\) with coordinates \[(z_x+d/(2\lambda),\ z_y-d/2)\] has \(z\) strictly northwest of it. By density choose \(q_x\in\mathcal O\) within \(d/(4\lambda)\) of its first coordinate, and put \(q_y=\lambda q_x+c_0\in\mathcal O\). Then \[ q_x-z_x>d/(4\lambda),\qquad z_y-q_y>d/4. \tag{57}\] The northwest quadrant based at \(q=(q_x,q_y)\) lies wholly on the positive side of \(L\). For \(d<0\), the same formula with \(|d|/(4\lambda)\) as the allowed perturbation puts \(z\) strictly southeast of \(q\); that quadrant lies wholly on the negative side. Thus the coordinate inequalities have positive room even when the inactive line is arbitrarily close to \(z\).

If \(|d|\) is below a fixed small threshold, choose that threshold so that \(q\) and \(z\) fit in the inner margins of a line chart. Lemma 35 compares the original predicate with that chart near \(z\), clipping the first overlap to \(B\). By (57), all sufficiently small cells on the interior side of \(B\) are controlled in the indicated quadrant and in the chart box. The genuine line–quadrant table says that its sign decision there is respectively the constant permutation or the identity. Intersection and transitivity of control give the claimed action on the controlled copy. No sign attached to a formal sector was used to infer its location.

For \(|d|\) above the fixed threshold, use the finite rectangular grid included with the primitive chart. Choose its mesh small enough that every cell meeting the part with this lower bound on \(|d|\) lies strictly on the same side of the line. This is possible because the defining linear form has fixed norm. The individual actual table of \(p\) with the grid gives its constant action there. Rectangular control transfers the assertion to any sufficiently small \(P\) near \(z\). Cells meeting a grid boundary may be further split by that fixed grid; this changes neither the conclusion nor the initial data.

If a clipping edge is active at \(z\), treat interior cells by the quadrant or grid just described and exterior cells by the first paragraph. At a corner there are four rectangular sides and the same procedure applies to each. Therefore the local action is exactly the Boolean combination of the constant sloping sign with the actual clipping decisions on all sides of the boundary. ◻

Proof. First take a finite family of translated primitive predicates, including coordinate predicates if needed. By Lemma 27, we may fix an alphabet of its prescribed bounded size and a word \(w\) entirely on that alphabet whose pointwise permutation is the identity on every actual assignment. It suffices to prove that \(w\) acts trivially on the subgroups needed to test centrality. Include those additional small-support copies and their spare alphabets throughout; the geometric family is still finite.

Fix \(z\in T\). Call a supporting boundary line active at \(z\) if it contains \(z\); include the coordinate lines which form clipping edges. Inactive supporting lines have positive distance from \(z\), and the family is finite. In a sufficiently small ordinary neighborhood of \(z\) the only varying signs are therefore the active ones. There is at most one distinct active line of each of the four directions.

Choose the corresponding translated concurrent template. If at least two directions are active, \(z\in D^{-1}\mathcal O^2\) and the template is provided by its coset representative and direction subset. With one active direction use a nearby \(\mathcal O^2\) anchor on that line, and with none use a constant model. Each original predicate becomes, in this model, the Boolean combination of its active sign and clipping decisions, with the other decisions replaced by their actual local constants. The allowed assignments are precisely the ordinary open sectors of the active lines, with these combinations evaluated on them. Each such sector occurs arbitrarily near \(z\) in the original Boolean space. Hence the word \(w\) is the identity on every assignment of this template model. Extend the model outside its margin box by any one allowed sector, so that it remains a finite pointwise test model on the whole square.

We verify the equality of its actions with the original actions on \(H_P\) for every sufficiently small cell near \(z\). All cells will be cut by every original clipping edge and by any additional finite rectangular subdivisions needed in these comparisons.

1. Coordinate decisions compare with the template by the established rectangular law. An inactive coordinate decision is its actual constant on the relevant cells.

2. An active sloping line inside its clipping box compares with the same oriented line in the template by Lemma 35. If \(z\) lies on a clipping edge, make that comparison only on the interior rectangular side. The template includes that active axis decision, and its true table identifies the resulting clipped sign there. On exterior cells the original predicate is trivial by control in the exterior of its own box. This treats all sides of a corner as well.

3. An inactive sloping line is replaced by its actual constant-side decision by Lemma 37. If some of its clipping edges are active, keep those coordinate decisions in the replacement. The interior and exterior arguments in that lemma establish precisely this clipped Boolean combination, even for a formal wrong-side copy over \(P\).

Every equality in this list follows from equal actions on a known rectangular container of \(P\) and the conditional move of Lemma 29. Thus it holds on every generator of \(H_P\), not just in projection. The original generators preserve \(H_P\) by Lemma 31. Their actions can therefore be composed, and the action of \(w\) on \(H_P\) equals the action of its template replacement.

The replacement is central in the true template table. Its margin box is a rectangular container of \(P\), so it controls every generator of \(H_P\). Apply the table on the enlarged alphabets required by Lemma 31. The template-relator conclusion of that lemma shows that the replacement, and hence \(w\), centralizes \(H_P\), including its possibly formal sloping factors.

We finally choose one finite partition on which these local checks all apply. For every \(z\) the preceding construction provides an ordinary neighborhood and an upper bound on the permitted cell diameter. A finite subcover exists by compactness of \(T\). A sufficiently fine \(\mathcal O\) rectangular grid, refined by the finitely many clipping edges and auxiliary rectangular cuts of that subcover, has every cell in one of those neighborhoods and below its required diameter. This follows, for example, by applying the Lebesgue number property to the finite subcover and then using the coordinate mesh bound. Make the separate splits of every original predicate with this partition by Lemma 36. We have proved that \(w\) centralizes every resulting \(H_P\). The group generated by these subgroups contains the original test subgroup, so Lemma 31 proves its actual central law. Lemma 27 gives the assertion on all the required alphabets simultaneously.

Apply this established law to a larger finite family containing translated primitive coordinate tests that generate the partition just used. Apply Lemma 26 first to the coordinate subfamily: its original partition copies agree with the corresponding coordinate factors of the enlarged actual law. We may therefore include those same copies among the inputs of that law. For each old formal law, the enlarged family now contains exactly its specified input copies, including the constant copy. Its subgroup \(H_0\) is thus contained in the enlarged defining-copy subgroup \(K\). Every new actual assignment restricts to an allowed old assignment, since the old law retained all actual assignments. Lemma 26 now identifies each original formal factor with the product of its actual refinements and makes it trivial when it has no actual extension.

Every polygon is a Boolean combination of finitely many translated primitive tests. Apply the law just proved to a common finite family realizing the polygons. The canonical perfect-cover construction supplies their representations. To compare two Boolean expressions, pass to their common finite family and apply Lemma 26; the resulting representations agree there. Within this lawful family, uniqueness of lifts from a perfect source gives exact multiplicative splitting over a finite disjoint refinement. Finally a common allowable translation on a proper alphabet is implemented by the balanced translations of Lemma 28; apply uniqueness to the translated test family. This proves the claimed translation covariance and alphabet compatibility. ◻

Proof. For (i), choose \(h\in B_m\) equal to \(v\) on \(I\). The common-translation identity for aligned copies is \(t(h)\rho_{U,I}t(h)^{-1}=\rho_{U+v,I}\). A balancing vector for the new offsets, added to \(h\), agrees with \(f\) on \(I\). Independence of the balance then proves the assertion. Part (ii) is the aligned refinement identity conjugated by \(t(f)\).

For (iii), the initial relations give \(c\,t(f)c^{-1}=t(c\cdot f)\) and exact reindexing of the aligned representations. These are precisely the data defining the new frame. For (iv), additivity gives \[t(d)\rho_S(s)t(d)^{-1} =t(d+f)\rho_{U,I}(\iota_{\boldsymbol i}^\ast(s))t(d+f)^{-1};\] \(d+f\) has the required offsets on \(I\). Finally, compatibility under restriction follows from the corresponding compatibility of aligned canonical representations and from independence of balances. This also proves the assertion about two frames. ◻

Proof. For each leg \(j\), choose four private tracks \(p_{j,1},\ldots,p_{j,4}\), all distinct and outside \(J\cup I\), and put \[V_j=U+u_j, \qquad L_j=\{i_j,p_{j,1},\ldots,p_{j,4}\}, \qquad K_j=\rho_{V_j,L_j}(\mathop{\mathrm{Alt}}(L_j)^\ast).\] The projected element \(\pi(b)\) fixes the source track pointwise on \(V_j\) and fixes every private track. It therefore commutes with the projection of \(K_j\). Take the finite family consisting of all tests occurring in a chosen aligned expression for \(b\), the five tests \(V_j\), and constants, on the union of the track sets involved. Theorem 38 gives a central extension of its pointwise group. Within that subgroup, \([b,k]\) is central for \(k\in K_j\). Since \(K_j\) is perfect, the central-commutator argument above proves \[ [b,K_j]=1\qquad (1\leq j\leq5). \tag{62}\]

Choose \(a_j\in K_j\) projecting to the cycle \((i_j\ p_{j,1}\ p_{j,2})\) on \(V_j\), and set \(a=a_5\cdots a_1\). There is one common offset frame on the union of the \(L_j\): assign offset \(u_j\) to every track in \(L_j\). In this frame, \(K_j\) is the aligned conditional copy on \(U\) with track set \(L_j\). Thus \(a\) and \(\rho_S\) lie in one conjugate of a subgroup with an aligned central law. In its pointwise model, \(a\) sends the \(j\)th slot to \((p_{j,1},U+u_j)\). Uniqueness for homomorphisms from \(E^\ast\) in this central extension gives the exact identity \[ a\rho_S(s)a^{-1}=\rho_T(s), \qquad T=(U;(p_{j,1},u_j)_{j=1}^5). \tag{63}\] Here compatibility of subframes in Lemma 39 identifies the two maps with their definitions in (59).

The tracks occupied by \(T\) are outside \(J\). Write its representation using a balancing vector \(f_T\) which is zero on \(J\): prescribe \(u_j\) on \(p_{j,1}\) and balance on a further track outside both sets. The aligned base copy for \(T\) commutes with \(b\) by disjoint-track commutation, and \(t(f_T)\) commutes with \(b\) because \(f_T\) vanishes on \(J\). Consequently \([b,\rho_T(E^\ast)]=1\). Equation (62) also gives \([b,a]=1\). Conjugating (63) back proves \([b,\rho_S(E^\ast)]=1\). ◻

Proof. Extend the bijection \(i_j\mapsto k_j\) to a permutation of the union of the two five-element sets. If it is odd, multiply by a transposition on two additional tracks. The resulting even constant permutation \(c\) has support at most twelve and sends \(S\) to \(T\) parameterwise. The aligned word \(c^{-1}b\) fixes \(S\) pointwise. Apply Lemma 40, and then the constant-reindexing identity of Lemma 39. ◻

Proof. Let \(r_1:S\to T_1\) and \(r_2:S\to T_2\) be two admissible routings. The aligned word \(b=r_2r_1^{-1}\) sends \(T_1\) to \(T_2\) parameterwise. Both terminal tuples have distinct tracks, so Lemma 41 applies and gives \[r_2r_1^{-1}\rho_{T_1}(s)r_1r_2^{-1}=\rho_{T_2}(s).\] After multiplying by \(r_2^{-1}\) and \(r_2\), this is precisely \(\rho_S^{\,r_1}(s)=\rho_S^{\,r_2}(s)\). A distinct-track tuple admits the identity routing, proving agreement with its earlier definition.

Fix one routing \(r:S\to T\). It is also an admissible routing of every \(S|U_\alpha\) to \(T|U_\alpha\): its support bound has not changed. Apply Lemma 39(ii) to \(T\), and conjugate the identity and its commutation assertions by \(r^{-1}\). This proves (66). For common reparameterization, use the same routing, which depends only on the actual parameterized slot maps after their domain identification, and apply Lemma 39(i) at its distinct-track target. ◻

Proof. Choose admissible routings \(r_S:S\to S'\) and \(r_T:T\to T'\). The word \(r_Tbr_S^{-1}\) is aligned and sends the distinct-track tuple \(S'\) to \(T'\). Lemma 41 gives \[(r_Tbr_S^{-1})\rho_{S'}(s)(r_Tbr_S^{-1})^{-1}=\rho_{T'}(s).\] Conjugate by \(r_T^{-1}\) and use the definitions of \(\rho_S\) and \(\rho_T\). The support of the transition word is at most \(25+5+25=55\), so this application uses a fixed reserve.

For the last assertion, the finitely many polygon tests in \(b\) induce a finite partition on each map \(x\mapsto x+u_j\). Their common refinement makes all five terminal track labels constant. Apply the proved identity on every part and multiply, using (66). ◻

Proof. It suffices to treat the fixed generators of \(B_m\), each supported on two tracks, and their inverses. The general statement then follows by composition and additivity of \(t\).

Choose the private-block routing \(r:S\to T\) used in the existence proof above. Its \(j\)th factor lies in \(\rho_{V_j,L_j}(\mathop{\mathrm{Alt}}(L_j)^\ast)\), where \(V_j=U+u_j\). Choose \(d'\in B_m\) as follows: \[ d'_i=d_i\quad\text{on every occupied source track }i, \qquad d'_{p_{j,a}}=d_{i_j}\quad(1\leq j\leq5,\ 1\leq a\leq4). \tag{69}\] Put \(d'=0\) on the other tracks except for one unused track, on which we balance the sum. Repeated source labels give consistent prescriptions because they prescribe the same \(d_i\). Thus \(d'\) is constant on each \(L_j\), and has support at most \(26\). Set \[\delta=d-d'.\] The translation \(\delta\) vanishes on all occupied source tracks and has bounded support (at most \(28\) is sufficient).

We first show \[ [t(\delta),\rho_S(E^\ast)]=1. \tag{70}\] Choose another private-block routing \(q:S\to T_0\), with all its private tracks outside the support of \(\delta\). Every track in each of its five blocks then has \(\delta_i=0\). Outside-support translation commutation, from Lemma 28, gives \([t(\delta),q]=1\) factor by factor. The tuple \(T_0\) has distinct tracks, all outside the support of \(\delta\). Write \[\rho_{T_0}(s)=t(f)\rho_{U,I_0} (\iota_{\boldsymbol k}^\ast(s))t(f)^{-1}.\] The element \(t(\delta)\) commutes with \(t(f)\) by additivity and with the aligned representation on \(I_0\) because it is zero on \(I_0\). It therefore centralizes \(\rho_{T_0}(E^\ast)\). By the already proved routing independence, \(\rho_S=q^{-1}\rho_{T_0}q\), proving (70). This argument uses only the distinct-track frame formula and outside-support commutation.

Next conjugate the original routing by \(t(d')\). Since \(d'\) is constant on \(L_j\), the common-translation law for aligned polygon representations replaces its \(j\)th conditional factor on \(V_j\) by the same factor on \(V_j+d_{i_j}\). Hence \[r'=t(d')rt(d')^{-1}\] is an aligned word on the same at-most-\(25\) tracks. It routes \(d'S\) to \(d'T\), preserving base coordinates at this new configuration. It is therefore an admissible routing. The distinct-track formula (61) gives \[\begin{split} t(d')\rho_S(s)t(d')^{-1} &=r'^{-1}\bigl(t(d')\rho_T(s)t(d')^{-1}\bigr)r'\\ &=r'^{-1}\rho_{d'T}(s)r' =\rho_{d'S}(s). \end{split}\] Finally, \(d'S=dS\) because \(d'=d\) on the occupied source tracks, while \(t(d)=t(d')t(\delta)\). Equation (70) now proves (68). ◻

Proof. Each of the initial conditional alternating track groups is generated by its subgroups on five tracks. The canonical representations on those five tracks agree exactly with their initially specified lifts, by the finite initial tables and uniqueness for perfect sources. Thus \(N\) contains all conditional track-permutation generators, including the constant ones.

It remains to include the balanced translation generators. Let \(v\) be one of the two fixed generators of \(\Gamma\), and choose four distinct tracks \(i,j,k,h\). Put \[c=(i\ j\ k),\qquad a=v e_i-v e_h\in B_m,\] where \(e_i\) denotes placement in the \(i\)th track coordinate. We use the same symbol \(c\) for the specified constant lift. The additive and constant-reindexing relations give \[ [c,t(a)]=t(c\cdot a-a)=t(v e_j-v e_i). \tag{71}\] The commutator convention is \([x,y]=xyx^{-1}y^{-1}\). On the other hand, \[[c,t(a)]=c\,\bigl(t(a)c^{-1}t(a)^{-1}\bigr).\] The first factor belongs to a constant five-track representation, after adjoining two track labels to the three moved by \(c\). The second belongs to its offset-frame conjugate and hence, by (59), to another five-slot representation. Both factors are in \(N\), so \(t(v e_j-v e_i)\in N\).

Such differences generate \(B_m\). Only the fixed basis differences are needed for the finite generator list, and the finitely many instances of (71) are among the initial relations of Section 6 (they also follow from its additive and reindexing relations). This proves \(N=\widehat G\) without assuming that any conjugating translation already belongs to \(N\). ◻

Proof. Choose \(m\) beyond the fixed track requirements of the lifting and propagation arguments and \(M_{\mathrm{tr}}\). Choose the finite initial relation menu in the order specified in Sections 6 and 7. This produces the finitely presented group \(\widehat G\) and the surjection \(\pi\); the transport lemmas have all been proved inside this group.

Take \(w\in\ker\pi\), and write it as a finite word in the chosen generators and their inverses. Each conditional permutation letter can be written, using its initial finite group table, as a product of letters supported on five tracks. Thus we may regard every letter as either a five-track aligned element or a fixed balanced translation generator or its inverse. Write the resulting word as \(w=g_n\cdots g_1\) and put \(w_k=g_k\cdots g_1\), so that the letters act in the order \(g_1,\ldots,g_n\). Fix a configuration \(S=(U;(i_j,u_j)_{j=1}^5)\).

There is a finite polygonal partition \(U=\bigsqcup_\alpha U_\alpha\) such that, along every \(w_k\) and for each of the five legs, its image on \(U_\alpha\) has one fixed track label and one fixed translation offset. To construct the partition, pull back the finitely many continuity pieces of each next letter along the five current partial translations and intersect with the pieces already obtained. The Boolean algebra is closed under these operations. Induction on the finite word length therefore gives a finite partition; there is no claimed bound on its number of pieces.

On a nonempty \(U_\alpha\), follow the resulting five-slot tuple through successive letters. The images remain disjoint because every projected letter is a bijection. Lemma 43 applies to aligned letters and Lemma 44 to translation letters. Composing their exact conjugation identities gives \[ w\rho_{S|U_\alpha}(s)w^{-1} =\rho_{\pi(w)(S|U_\alpha)}(s). \tag{72}\] Since \(\pi(w)=1\), the terminal tuple equals the initial tuple parameterwise. In particular the terminal track labels are the same; the terminal offsets are the same in \(\Gamma\), since the translation action on \(X\) is free. The right side of (72) is consequently \(\rho_{S|U_\alpha}(s)\).

Multiply these identities over \(\alpha\) and use the exact refinement formula (66). We obtain \([w,\rho_S(E^\ast)]=1\) for every configuration \(S\). Lemma 45 says that these images generate \(\widehat G\), so \(w\in Z(\widehat G)\). Since \(w\) was arbitrary, \(\ker\pi\subseteq Z(\widehat G)\), as required. ◻