Proof. Define \[f\colon\mathcal F(\widetilde X)\longrightarrow\mathcal D^{\ell}, \qquad f(\alpha,T)=\alpha A_T.\] Formula (6) makes \(f\) order preserving: if \((\alpha\iota(u_0),R)\) is a face of \((\alpha,T)\), then \(\iota(u_0)\in A_T\) and \(R\subseteq T\), so \(\alpha\iota(u_0)A_R\subseteq\alpha A_T\).

Fix \(\beta A_T\in\mathcal D^{\ell}\). The inverse image of its lower ideal consists exactly of the cells \[ (\alpha,R),\qquad \alpha\in\beta A_T,\quad R\subseteq T. \tag{8}\] Indeed, the indicated conditions imply \(\alpha A_R\subseteq\beta A_T\), and the converse follows from the retained types and from \(\alpha\in\alpha A_R\).

By Theorem 2, write \(\alpha=\beta h\) uniquely with \(h\) an element of the Artin group on \(T\). Under this correspondence, every face formula becomes \[(h,R)\longmapsto(h\iota(u_0),Q), \qquad Q\subseteq R\subseteq T,\] with \(u_0\) the shortest representative of the specified \(W_Q\)-coset in \(W_R\). These are exactly the lifted face maps of \(X(W_T,T)\). Thus the poset in (8) is its universal-cover face poset. Its realization is contractible by Theorem 4. For \(T=\varnothing\) it is a single vertex.

Quillen’s Theorem A now gives \(|\mathcal F(\widetilde X)|\simeq|\mathcal D^{\ell}|\). Regularity identifies the left side with a subdivision of \(\widetilde X\), and inversion identifies the right side with \(|\mathcal D|\). This proves the lemma. ◻

Proof. Write \(C_T=2I-B_T\), where \(B_T\) is symmetric and entrywise nonnegative. Positive definiteness implies that the largest eigenvalue of \(B_T\) is less than \(2\). The Perron–Frobenius bound for a nonnegative symmetric matrix then gives spectral radius less than \(2\), so \[C_T^{-1}=\frac12\sum_{n\geq0}(B_T/2)^n.\] This is nonnegative, and a path between any two vertices in the same block gives a strictly positive entry in some power. Harmonic extension is explicitly \[p_T=C_T^{-1}B_{T,\widehat S\setminus T} p_{\widehat S\setminus T},\] which proves the first assertions, including uniqueness.

For the last assertion, every real vector \(x\) on \(G\) satisfies \[ x^{\mathsf t}C_Gx =\sum_{\substack{i<j\\i,j\in G}}b_{ij}v_iv_j \left(\frac{x_i}{v_i}-\frac{x_j}{v_j}\right)^2 +\sum_{i\in G}\frac{(C_Gv)_i}{v_i}x_i^2. \tag{11}\] If at least one coordinate of \(C_Gv\) is positive, vanishing of this quadratic form forces all \(x_i/v_i\) to be equal by connectedness, and then forces their common value to be zero. The form is therefore positive definite. If none is positive, the assumed inequalities say exactly that \(C_Gv=0\). ◻

Proof. For each individual finite edge of label \(m\geq4\), use an independent Temperley–Lieb factor with circle value \[d_m=2\cos(\pi/m),\qquad [j]=\frac{\sin(j\pi/m)}{\sin(\pi/m)}.\] Here is its construction. Begin with the pivotal diagram category whose objects are nonnegative tensor powers of a self-dual string and whose morphisms are \(\mathbb C\)-linear combinations of planar pairings. Composition is stacking, a closed circle is replaced by \(d_m\), and tensor product is juxtaposition. Closing a diagram gives its spherical trace. Take the additive idempotent completion, quotient by the ideal of negligible maps, and complete additively and idempotently again. A map \(f:X\to Y\) is negligible if \(\operatorname{tr}(gf)=0\) for every \(g:Y\to X\). Cyclicity proves stability under composition, and partial trace proves stability under tensoring; hence this is a tensor ideal.

For completeness, the idempotents and decompositions in this quotient can be obtained directly. Let \(f_0\) and \(f_1\) be the respective identity diagrams, and define, for \(1\leq j\leq m-2\), \[ f_{j+1}=p-\frac{[j]}{[j+1]}p e_jp, \qquad p=f_j\otimes\mathop{\mathrm{id}}, \tag{12}\] where \(e_j\) is the unnormalized cap–cup at the last two positions. All denominators in this recursion are nonzero. Inductively \(f_j\) is idempotent, kills every elementary turnback on either side, and its partial trace over the last string is \[ \operatorname{ptr}(f_j)=\frac{[j+1]}{[j]}f_{j-1} \qquad(1\leq j\leq m-1). \tag{13}\] Indeed the partial trace of the two terms defining \(f_j\) is \(d_m f_{j-1}-([j-1]/[j])f_{j-1}^2\), which is the right side of (13); the case \(j=1\) starts the induction. Sandwiched between the last cap–cups, \(p\) consequently gives \(([j+1]/[j])(f_{j-1}\otimes e)\), where \(e\) is the last cap–cup. Since \(p\) already factors through \(f_{j-1}\) on the first \(j-1\) strings, this identity shows that the subtracted term in (12) is a projection, proves idempotence of \(f_{j+1}\), and proves its turnback cancellation. The last cap and cup factor that projection through \(f_{j-1}\); their opposite composite is the nonzero scalar \([j+1]/[j]\). Thus its image is isomorphic to \(f_{j-1}\), giving \[ f_j\otimes f_1\cong f_{j+1}\oplus f_{j-1} \qquad(1\leq j<m-1). \tag{14}\] Closing (13) gives \(\operatorname{tr}(f_j)=[j+1]\).

A planar pairing between the corners of \(f_i\) and \(f_j\) has a turnback unless \(i=j\) and it is the straight pairing. Consequently these corners vanish for \(i\ne j\) and are one-dimensional for \(i=j\). The object \(f_{m-1}\), whose trace is \([m]=0\), is therefore negligible. The objects \(f_0,\ldots,f_{m-2}\) survive, are simple, and have positive dimensions \([1],\ldots,[m-1]\). The decompositions (14), with \(f_{m-1}\) now zero, generate every tensor power from these objects. Thus the quotient is finite semisimple. Reflecting the diagrams gives the same fusion rule on the other side: the turnback-killing idempotent with straight coefficient one is unique, as follows by multiplying any two such idempotents. Duality and the spherical trace descend to the quotient and its completions. This proves all the stated properties of a single factor.

Take the tensor product of these finitely many categories, with simple objects the tuples of simple objects in the factors and tensor product and Hom spaces given factorwise. If there are no such factors, take finite-dimensional vector spaces. For an edge of label \(m\geq4\), use the string object \(f_1\) in its factor and units in the other factors. Tensoring by it has the path-\(A_{m-1}\) multiplicities in that factor and fixes the others, proving the multiplicity bound. The choices of units and two units cover the remaining edges. All dimensions are products of positive dimensions. Choose dual objects for opposite orientations and use the opposite pivotal evaluations on the two orientations. ◻

Proof. The length-zero and length-two spaces pair perfectly. For length one, duality identifies maps \(L'\to D^1_{ts}\otimes L\) with maps \(D^1_{st}\otimes L'\to L\). Pairing these against \(L\to D^1_{st}\otimes L'\) by the semisimple trace is nondegenerate: on every simple summand the ordinary matrix trace is multiplied by its nonzero categorical dimension. This is exactly the backtrack pairing. Its symmetry is the pivotal partial-trace identity. Namely, the trace on \(L\) of the first backtrack equals the trace on \(D^1_{st}\otimes L'\) of the opposite composite; taking partial trace in \(D^1_{st}\) gives the second pivotal cap and then the trace on \(L'\). Hence \(\tau(ab)=\tau(ba)\). This is ordinary symmetry; it has no sign from internal degree.

The positive-length ideal is nilpotent and the quotient is the semisimple diagonal, so it is the radical. Tokens are annihilated by this radical. A nonzero length-one element has a nonzero product with a suitable opposite arrow by the perfect pairing, and a nonzero length-zero element has a nonzero product with its token. Therefore the token is the entire socle of each projective.

For (17), send \(f\) to \(\tau\circ f\). A homogeneous \(f\) of degree \(k\) gives a functional of degree \(k-2\), which explains the shift \(\langle2\rangle\). Nondegeneracy of the trace gives the usual Frobenius identification for every finite left module. It respects both bimodule actions: with \((a f b)(m)=f(ma)b\) one has \[\tau(f(ma)b)=\tau(bf(ma))=\tau(f(bma)),\] which is the action \((a\lambda b)(m)=\lambda(bma)\) on the vector dual. The shifted vector dual squares naturally to the identity, so this is a contravariant equivalence. Such an equivalence preserves the additive radical, characterized by invertibility of \(1-gf\) for all opposite maps \(g\).

The trace identifies \(D_\mathbb C(P_\alpha)=Q_\alpha\langle-2\rangle\) and \(D_\mathbb C(Q_\beta)=P_\beta\langle-2\rangle\). Vector duality reverses the factors of an external tensor product; combining its shift \(-4\) with the shift \(+2\) in (17) proves (18). The same trace identifies \(D_\mathbb C(Z)\langle2\rangle\) with \(Z\). ◻

Proof. The positive vector \(v=\phi(C_J^{-1}\mathbf1)\) satisfies \(\widetilde C_{\widetilde J}v=\phi\mathbf1>0\). Applying the scaled quadratic-form identity (11) to each connected component proves positive definiteness of the entire unfolded matrix. In particular this proves positivity on directions outside the image of \(\phi\). An infinite edge cannot lie in \(J\), since its two-by-two Cartan matrix is singular. The multiplicity assertion of Proposition 8 therefore makes every off-diagonal unfolded entry either \(0\) or \(-1\). The finite Coxeter criterion applies.

There are no edges between different vertices of a single color. Their reflections commute, and (20) shows that their product on a folded vector is the original reflection (10). Choose \(p\) with \(C_Jp>0\). Then \(\phi p\) lies in the interior of a full unfolded fundamental chamber, since every row of \(\widetilde C_{\widetilde J}\phi p\) is positive. For any word, intertwining identifies its image with \(\phi(wp)\). The row at an unfolded vertex \(\alpha\) on this image tests the sign of \(\widetilde w^{-1}\alpha\), while the corresponding original row tests the sign of \(w^{-1}\alpha_i\).

To see why this is a complete sign test, a root of either finite system has either all nonnegative or all nonpositive coefficients in the simple-root basis. Pairing it with one chamber-interior point is therefore strictly positive or strictly negative respectively. No assumption about the dimension of the folded subspace or about its orbit meeting every unfolded chamber is needed. The intertwining identity multiplies the original row by \(w_L>0\), proving the equality of left descent sets. Applying the same argument to reversed words proves the equality of right descent sets. The lifted product depends only on the original group element as well: a word for the identity fixes \(\phi p\), and the stabilizer of a chamber-interior point in the finite unfolded reflection group is trivial (Bourbaki 2002, V, §3, no. 3).

Suppose an original reduced expression is extended by an ascent letter. Every vertex of its color is an ascent of the lifted prefix by (21). After one reflection of that color, the others remain ascents, because they are orthogonal to it. Consequently the whole block is a sequence of successive ascents. Induction proves reducedness of the lift. Finally the longest original element has every original simple reflection as a descent. Its lift has every unfolded simple reflection as a descent, and hence is the longest unfolded element. ◻

Proof. The product has terms \(M_\alpha\) in degree \(-1\), \[Z\oplus P_\alpha\otimes_\mathbb C(e_\alpha Ze_\alpha) \otimes_\mathbb CQ_\alpha\langle-2\rangle\] in degree zero, and \(M_\alpha\langle-2\rangle\) in degree one. Split the middle factor as \(e_\alpha Ze_\alpha=\mathbb Ce_\alpha\oplus\mathbb C\theta_\alpha\). The incoming differential projects isomorphically, up to its cochain sign, onto the token summand. Explicitly, the token coefficient of \(q p_a\) is \(\tau(q p_a)\), and the identity \(\sum_a\tau(qp_a)q_a=q\) is the dual-basis identity. Cancel this degree-\(-1\)/degree-zero isomorphism. The outgoing differential from the remaining unit summand is the identity onto \(M_\alpha\langle-2\rangle\), by multiplication with \(e_\alpha\); cancel it as well. The remaining complex is the regular bimodule \(Z\) in degree zero. These are Gaussian cancellations of actual bimodule complexes. In the reverse tensor order the dual-basis identity and unit multiplication give the same two cancellations, with the opposite tensor order and the prescribed cochain signs. ◻

Proof. In the first case \(M_\alpha M_\beta=0=M_\beta M_\alpha\), so both products are the simultaneous evaluation cone with projective term \(M_\alpha\oplus M_\beta\).

In the second case, \(B_\alpha P_\beta\) has the nonzero arrow \(P_\alpha\langle1\rangle\to P_\beta\) in degrees \(-1,0\). The complex \(D_\beta P_\alpha\) has the nonzero arrow \(P_\alpha\to P_\beta\langle-1\rangle\) in degrees \(0,1\). Nondegeneracy of the edge pairing makes both arrows nonzero, and their common Hom space is one-dimensional. Therefore \[B_\alpha P_\beta\simeq (D_\beta P_\alpha)\langle1\rangle[1],\] after rescaling one term if needed. Their evaluation twists agree. Since \(T_{P_\gamma}=B_\gamma\), conjugation gives \[B_\alpha B_\beta D_\alpha \simeq D_\beta B_\alpha B_\beta.\] Multiplying on the left by \(B_\beta\) and on the right by \(B_\alpha\) gives the claimed braid relation. ◻

Proof. Distinct vertices in a color have no mutual Hom spaces. Their twists commute by Lemma 12, and their products are inverse by Lemma 11. Moreover \[ B_i\cong \mathop{\mathrm{Cone}}\left(\bigoplus_{\alpha\in\widetilde{\{i\}}} P_\alpha\otimes_\mathbb CQ_\alpha\longrightarrow Z\right): \tag{25}\] every mixed tensor term in the expansion vanishes by orthogonality.

If \(m_{ij}<\infty\), the subsystem on \(\{i,j\}\) is spherical. The two alternating words of length \(m_{ij}\) are reduced expressions for its longest element. By Lemma 10, both lift to reduced expressions for the same longest element of its finite simply-laced unfolding. Matsumoto’s theorem relates these expressions by commutations and three-term braid moves, all of which hold for the bimodules by Lemma 12. This proves the original Artin relation. When \(m_{ij}=\infty\) there is no relation to verify. Thus the assignment defines the asserted homotopy action for every Coxeter matrix under consideration.

On a morphism in (15), \(T_H\) is right tensoring by \(\mathop{\mathrm{id}}_H\), with the associativity identifications. All cap multiplications are to the left of the label, so this preserves composition and path length. In the evaluation term of (25) applied to \(P_{t,Y}\), semisimplicity gives the canonical identification \[\begin{align*} &\bigoplus_{a,L} P_{i,L}\langle a\rangle\otimes_\mathbb C \mathop{\mathrm{Hom}}_{\mathscr C}(L,D^a_{it}\otimes Y)\\ &\hspace{35mm}\cong \bigoplus_a P_{i,D^a_{it}\otimes Y}\langle a\rangle. \end{align*}\] Its evaluation is the tautological label map. Tensoring this description on the right by \(H\) is the same description with \(Y\) replaced by \(Y\otimes H\), naturally in \(Y\) and its projective morphisms. Hence the evaluation cones commute naturally with \(T_H\) on projectives and on their complexes. Whiskering this natural commutation by the inverse equivalence proves the same assertion for \(D_i\), without requiring \(T_H\) itself to be invertible.

Finally, tensoring an explicit chain homotopy and its equations with any module complex preserves those equations with the ordinary cochain tensor signs. The last assertion therefore follows from the homotopy-level inverse and braid calculations, without any flatness requirement on these additional module complexes. ◻

Proof. The level calculation shows that the indicated low terms form a subcomplex, with the indicated high quotient. An internal- and cochain-degree-zero map from a generator of level \(d\) to one of level \(e\) would require a path of length \(d-e\). Consequently every such map from levels at most \(0\) to levels at least \(1\) is zero. This gives the required orthogonality. The cochain shift \([1]\) decreases all levels by one, giving the two shift-closure axioms. The short exact sequence of low subcomplex, full complex, and high quotient gives the truncation triangle. There are finitely many occupied levels, so the \(t\)-structure is bounded.

Between pure objects of the same level, a degree-zero map has only length-zero entries. It is therefore exactly a graded map of their multiplicity arrays commuting with the coefficient arrows. A cochain homotopy would have cochain degree \(-1\) and would require path length \(-1\), so there are no homotopies to quotient out. Conversely every finite array satisfying the backtrack relation defines a pure complex of projectives. Kernels and cokernels of array maps, taken in each graded coefficient space, still satisfy the relation. This identifies the heart with the asserted abelian category. Applying the truncations twice identifies its cohomology layers with the complexes \(Y^d(E)\). ◻

Proof. In the positive case apply \(F\) to \[\tau_{\leq d-1}E\longrightarrow\tau_{\leq d}E \longrightarrow Y^d(E).\] The first term has levels at most \(d\), and the last term has levels \(d,d+1\). The flanking groups in the long exact sequence at \(d\) are \(H^{d-1}_{\mathrm{lay}}F(Y^d(E))\) and \(H^{d+1}_{\mathrm{lay}}F(\tau_{\leq d-1}E)\); both are zero. This gives a short exact sequence at that level. Applying \(F\) to the truncation below \(d-1\) identifies its left term with the one in (27), since levels at most \(d-2\) map to levels at most \(d-1\). Applying it to the truncation above \(d\) identifies the middle term with \(Y^d(FE)\), since levels at least \(d+1\) remain at least \(d+1\).

For the negative case use \(\tau_{\leq d}E\to\tau_{\leq d+1}E\to Y^{d+1}(E)\). After \(F\) the first term has levels at most \(d\), and the last has levels \(d,d+1\). The same exact-sequence argument, now discarding input levels below \(d\) and above \(d+1\), gives (28). ◻

Proof. After the first operation the quotient by its level-\(d\) truncation has levels at least \(d+1\). Every subsequent functor keeps that quotient in levels at least \(d+1\). The long exact sequence therefore identifies bottom cohomology before and after discarding it. Repeat this argument at each step. ◻

Proof. For a pure layer at \(d\), the paths in \(e_\alpha M\) have lengths \(0,1,2\), and their total degrees are respectively \(d,d+1,d+2\). At the length-zero end, being a cycle means being killed by every outgoing coefficient map. At the token end, the incoming maps span exactly the images of the incoming coefficient maps, because the opposite-arrow pairing is nondegenerate. This proves (29).

For the upper assertion, every term of \(e_\alpha E\) comes from a generator at \(\alpha\) or a neighbor. Its actual total degree is its generator level plus its path length, and is therefore at most \(d+2\). At degree \(d+2\) only the level-\(d\) token terms occur. Their incoming differential consists of the indicated linear maps: any additional contribution would need a higher generator at the same vertex or a neighbor. The lower assertion follows in the same way from the length-zero end. Finally, the coefficient arrows strictly raise cochain placement in a finite array. Iterating them cannot avoid a socle, and iterating incoming images cannot generate a nonzero array with zero top everywhere. ◻

Proof. For pure input at \(d\), the simultaneous evaluation cone inserts generators in color \(i\) at levels \(d+a-1\), for path lengths \(a=0,1,2\). Those at \(d-1\) are the length-zero evaluation copies. Their maps to the old generators of color \(i\) are identities modulo the additive radical. Cancelling these pairs removes all generators below \(d\). None were inserted above \(d+1\), proving the amplitude bound. Only the acted color can occur above \(d\).

For one vertex \(\alpha\) of this color put \(A=e_\alpha E\). All other vertices of the same color are orthogonal to it, so its simultaneous test is \[ \mathop{\mathrm{Cone}}\bigl(A\oplus A\langle2\rangle \xrightarrow{(\mathop{\mathrm{id}},\theta_\alpha)}A\bigr). \tag{32}\] Projection onto the token source summand \(A\langle2\rangle[1]\) is a chain map with kernel the cone of \(\mathop{\mathrm{id}}_A\). It is therefore a homotopy equivalence, proving (31) for arbitrary input. The same calculation works on the right. For pure input, the degree-\(d+3\) test on the output is the old degree-\(d+2\) test. Lemma 17 identifies it with the upper generator multiplicities and with the old top, respectively. There are no arrows among the vertices of the upper piece.

For completeness, signed generator counts are computed with \((-1)^d\). A contractible scalar pair has consecutive levels, so its two signed contributions cancel. In the cone for one vertex, the inserted signed contribution from a generator at \(\beta\) is minus the alternating path count \(2\delta_{\alpha\beta}-(2\mathop{\mathrm{id}}-\widetilde C)_{\alpha\beta}\). Thus the total signed vector transforms by \(r_i\). Since the output has only levels \(d,d+1\), subtracting its upper vector from its bottom vector gives \(r_i\dim M\). The upper vector is the old top. At the acted fiber this gives (30); outside it the multiplicities are unchanged.

At a vertex of color \(i\), the degree-\(d\) test of \(F_iM\) is zero by (31), so its bottom socle is zero. For a vertex \(\beta\) of another color the evaluation triangle for \(e_\beta F_iM\) has first term a sum of copies of \((e_\alpha M)\langle1\rangle\), with \(\alpha\) in the acted fiber adjacent to \(\beta\). Its cohomology at \(d\) is zero, and its cohomology at \(d+1\) is a sum of the old socles at those \(\alpha\). If the socles in \(J\) vanish, the long exact sequence at \(d\), together with the old socle vanishing at \(\beta\), gives the claimed vanishing at \(\beta\) after the move. ◻

Proof. It suffices to test the inverse on a heart object \(X\). If \(A\) has levels at most \(-2\), then \(FA\) has levels at most \(-1\), whence \[\mathop{\mathrm{Hom}}(A,F^{-1}X)=\mathop{\mathrm{Hom}}(FA,X)=0.\] Orthogonality implies that \(F^{-1}X\) has levels at least \(-1\). If \(B\) has levels at least \(1\), then \(FB\) does also, and \[\mathop{\mathrm{Hom}}(F^{-1}X,B)=\mathop{\mathrm{Hom}}(X,FB)=0.\] It follows that \(F^{-1}X\) has levels at most \(0\). This proves the inverse bound; shifting proves it for every pure level. Formula (33) follows by applying (31) to \(F_i^{-1}E\) and using invertibility. This argument is at the homotopy level also for nonprojective test inputs.

For \(\beta\) outside the acted color, the right simple \(U_\beta\) satisfies \[U_\beta B_{\sigma_i}^{\pm1}\simeq U_\beta,\] because all the inserted projective bimodule terms have left vertex in color \(i\). Tensoring a minimal left projective complex with \(U_\beta\) kills its radical differential and counts precisely its generators at \(\beta\), with both placements retained. The displayed identity therefore proves the last assertion in every layer. ◻

Proof. The test at \(\alpha\) shifts up by one in total degree. A nonneighbor’s test is unchanged. At a neighbor \(\beta\) in \(\widetilde J\) the evaluation triangle is \[ (e_\alpha X)\langle1\rangle\longrightarrow e_\beta X \longrightarrow e_\beta B_\alpha X. \tag{35}\] If the old maximum is \(t\), the neighboring new tests cannot exceed \(t\). Thus the maximum rises precisely when the test at \(\alpha\) was nonzero at \(t\), and only \(\alpha\) attains \(t+1\). If its old maximum was \(t-1\), its shifted test reaches \(t\). If it was below \(t-1\), any nonzero old degree-\(t\) test at another vertex survives, by the long exact sequence of (35). This proves monotonicity. The same triangles prove persistence of total vanishing. Reversing the module sides proves the right-action statements. ◻

Proof. We prove the left statement by induction on the positive word length, writing \(b=\alpha b'\) and \(Y=B_{b'}X\). If this last action raises height, its only new top vertex is \(\alpha\), and the assertion follows immediately. Otherwise let \(h_J(Y)=h_J(B_\alpha Y)=t>h_J(X)\). The top vertex \(\alpha\), if present, already left-divides \(b\). A top vertex \(\beta\ne \alpha\) not adjacent to \(\alpha\) was a top of \(Y\); by induction it divides \(b'\), and it commutes past \(\alpha\).

It remains to treat an adjacent top vertex \(\beta\). Since height did not rise, \(H^t(e_\alpha Y)=0\). The triangle (35) gives \[ H^t(e_\beta B_\alpha Y) =\operatorname{coker}\bigl( H^{t-1}(e_\alpha Y)\longrightarrow H^t(e_\beta Y)\bigr). \tag{36}\] The nonzero cokernel implies that \(\beta\) was a top of \(Y\). Induction allows \(b'=\beta v\) using positive relations. Put \(Z_0=B_vX\), so \(Y\simeq B_\beta Z_0\).

We identify the arrow in (36) directly. Set \(A=e_\beta Z_0\), \(C=e_\alpha Z_0\), and choose arrows \[r\in e_\alpha Z e_\beta,\qquad s\in e_\beta Z e_\alpha,\qquad sr=\kappa\theta_\beta,\quad\kappa\ne0.\] The two test complexes are \[e_\beta Y=\mathop{\mathrm{Cone}}\bigl(A\oplus A\langle2\rangle \xrightarrow{(\mathop{\mathrm{id}},\theta_\beta)}A\bigr), \qquad e_\alpha Y=\mathop{\mathrm{Cone}}\bigl(A\langle1\rangle \xrightarrow{r}C\bigr).\] Let \(\pi:e_\beta Y\to A\langle2\rangle[1]\) be the token projection of (32). The arrow \(s:(e_\alpha Y)\langle1\rangle\to e_\beta Y\) maps its source summand to \(\kappa\) times the token source summand, and maps its target summand into the target \(A\), which \(\pi\) kills. Thus \(\pi s\) is exactly \(\kappa\) times the connecting projection of the second cone, up to a cochain sign. This is an identity of chain maps and needs no projectivity of \(Z_0\).

Since \(\pi\) is a homotopy equivalence, exactness of the second cone identifies the image of the map in (36) with \[ \ker\bigl(H^{t-1}(e_\beta Z_0) \longrightarrow H^t(e_\alpha Z_0)\bigr). \tag{37}\] The cokernel is nonzero, so the last map has nonzero image. Consequently \(H^t(e_\alpha Z_0)\ne0\). Height monotonicity gives \(h_J(Z_0)\leq h_J(Y)=t\), hence \(h_J(Z_0)=t>h_J(X)\). Induction applied to \(v\) makes \(\alpha\) a left divisor of \(v\). Writing \(v=\alpha v'\) and using the three-term braid gives \[b=\alpha\beta\alpha v'=\beta\alpha\beta v',\] which extracts \(\beta\) as required. All cases of the induction are proved. The right-sided cones and the opposite arrows give the right-divisor assertion by the same argument. ◻

Proof. First consider a reduced unfolded word. Its test height can increase at most once. Indeed, at a second increase the new letter was a highest-test vertex of the preceding result, whose height already exceeded the starting one. By Lemma 21 that letter could be extracted on the last-action side of the preceding positive word. Prepending it would then give two equal consecutive Coxeter generators, contrary to reducedness.

A pure input layer at \(d\) has \(h_J\leq d+2\). Positives never produce a layer below \(d\). Suppose that the output’s highest layer had level \(a>d+1\). Outside \(\widetilde J\) generator multiplicities in each layer are unchanged, so that highest layer is supported in \(\widetilde J\). It has a nonzero top there. Lemma 17 would give a test at degree \(a+2>d+3\), contrary to the preceding height bound. If the input tests all vanish, they remain zero and give the same contradiction. This proves the positive bound in the unfolding.

By Lemma 10, reduced expressions in \(W_J\) lift to reduced unfolded expressions, so the bound holds for their positive color actions. The inverse of a negative reduced lift is a positive reduced lift of the reversed expression. Lemma 19 therefore gives the negative bound. ◻

Proof. Proposition 13 supplies the additive functor \(P_{s,X}\mapsto P_{s,X\otimes H}\) and its natural commutation with twists. It extends termwise to complexes and commutes with cones and shifts. A positive-length differential entry remains of positive length after tensoring and decomposing the labels. Hence a minimal model stays minimal, with the same cochain and internal placements. Both halves of the layer \(t\)-structure are preserved, proving \(t\)-exactness.

The images under a \(t\)-exact functor of the two objects in a truncation triangle still have the required bounds. The uniqueness of \(t\)-truncations therefore gives natural commutation with \(\tau_{\leq d}\) and \(\tau_{\geq d}\), and with the induced layers. Commutation with inverse twists follows from the natural commutation with the positive equivalences. Finally a truncation-defined subobject is a composite of a canonical truncation map with the unit or counit of the inverse equivalences. The preceding natural isomorphisms commute with that composite. One may subsequently represent the resulting pure objects and subobjects by their coefficient arrays using Lemma 14. ◻

Proof. Let \(\mathbf1\) be the unit label. Natural label equivariance gives \[E_L(x)\simeq T_L E_{\mathbf1}(x),\qquad Y_L^d(x)\simeq T_LY_{\mathbf1}^d(x).\] For an object \(X\in\mathscr C\) and a simple label \(H\), rigidity and semisimplicity give \[ \sum_L w_L[X\otimes L:H] =w_H\dim_{\mathscr C}X. \tag{42}\] Indeed \([X\otimes L:H]\) is the multiplicity of \(L\) in \(X^*\otimes H\), by adjunction. Summing the dimensions of these simple summands gives \(\dim(X^*\otimes H)=\dim(X)\dim(H)\).

In a minimal representative of \(E_{\mathbf1}(x)\), collect the projective labels at each color and pair of placements into an object \(X\in\mathscr C\). Tensoring that summand by \(L\) gives \(P_{s,X\otimes L}\) with unchanged placements. Applying (42) and summing over the placements of a fixed level gives (39).

The same argument applies to tops and socles, as follows. For any such projective complex and any color \(s\), its test at \((s,H)\) is obtained by applying \(\mathop{\mathrm{Hom}}_{\mathscr C}(H,-)\) to a complex of objects of \(\mathscr C\): a term of type \(P_{t,X}\) contributes \(D^a_{st}\otimes X\) at path length \(a\), with the given placements. The maps are induced by the same \(D\)-multiplications as the projective differential. Tensoring labels by \(L\) tensors this entire label complex on the right by \(L\). Tensoring in the semisimple category is exact, so it commutes with its cohomology. Apply (42) to the cohomology in the two extremal total degrees from Lemma 17. This proves the proportionality for both tops and socles. Since every \(w_H\) and every family weight is positive, a nonzero proportional coordinate at one label is nonzero at all labels of that color, proving the occurrence assertion.

There are finitely many family members and finitely many generators in each of their complexes, proving finite level support. Fixing one simple label \(H\) in (39) writes \(m_s^d\) as a nonnegative integer combination of the finitely many positive numbers \(w_L/w_H\). Below a fixed bound, each integer coefficient is bounded. Thus there are only finitely many values; irrational label dimensions cause no accumulation through cancellation.

The outside-color statement follows from the right-simple test in Lemma 19. In particular \(U_{o,H}B_x\simeq U_{o,H}\), because the frame is never acted on. Its tensor product with \(P_{o,L}\) is one copy of \(\mathbb C\) in placements \((0,0)\) when \(H=L\), and is zero otherwise. The weighted count is therefore \(w_H\) in level zero and zero in the other levels. This proves (40).

Finally, the signed calculation in Lemma 18 gives the unfolded reflection on generator counts. Its inverse acts by the same reflection, which is an involution. The identity \(\widetilde C\phi=\phi C\) in (20) folds this to (41). ◻

Proof. Let \(R=\tau_{\geq q+1}E\). By the cap, every generator of \(R\) is outside \(J\cup\partial J\), so \(GR\simeq R\) by the preceding observation. Negative twists do not increase levels. Apply \(G\) to \[\tau_{\leq q}E\longrightarrow E\longrightarrow R.\] The first term has levels at most \(q\), and the third is the unchanged complex \(R\) in levels at least \(q+1\). The long exact sequence identifies level \(q\) of \(GE\) with level \(q\) of \(G\tau_{\leq q}E\), and all higher layers with the corresponding layers of \(R\). These higher layers remain outside the operated region, proving cap preservation. Now apply \(G\) to \[\tau_{\leq q-1}E\longrightarrow\tau_{\leq q}E \longrightarrow Y^q(E).\] Its first term still has levels at most \(q-1\), giving (44) by the long exact sequence. This argument does not require that \(G\) be a reduced word. ◻

Proof. Negative preservation follows by applying Lemma 26 to each family member. For the positive assertion use the same high quotient \(R=\tau_{\geq q+1}E_L\). It is unchanged by \(F_i\). On \(\tau_{\leq q}E_L\) the only possible level \(q+1\) contribution is the upper piece of \(F_iY_L^q\): input levels below \(q\) can rise by at most one. By Lemma 18, this upper piece is the old top at color \(i\), and vanishes under the hypothesis. The low part therefore stays at levels at most \(q\), while the high quotient stays outside \(J\cup\partial J\). This proves the positive cap assertion.

For accessibility there is nothing to prove if \(J\) is empty or already capped. Otherwise let \(t>q\) be the largest level with nonzero data on \(J\). The joint cap at \(t\) holds on \(J\cup\partial J\). In each \(Y_L^t\) the part supported on \(\widetilde J\) is a direct summand, because all boundary generators at that level vanish. If these parts are nonzero, some member has a nonzero top at a color \(i\in J\). Apply \(F_i^{-1}\) and repeat while level \(t\) persists. All these moves preserve the joint cap at \(t\).

We show that the chosen colors form a reduced expression in \(W_J\), read in their order of action. Suppose the current negative word represents a reduced Coxeter expression \(w\). Every left descent \(j\) of \(w\) can occur as its last-action letter, since reduced expressions are related by braid moves. The corresponding negative Artin words are related by the same braid moves. In that expression, let \(E'_L\) be the family immediately before its last \(F_j^{-1}\). All preceding negative moves preserve the joint \(t\)-cap. For \(\alpha\) in color \(j\), Formula (33) and Lemma 17 give \[H^{t+2}(e_\alpha F_j^{-1}E'_L) =H^{t+3}(e_\alpha E'_L)=0.\] Thus the current level-\(t\) top vanishes in every descent color. A chosen nonzero top must be at a left ascent, so appending its action increases Coxeter length. This proves the induction from the empty word.

The finite group \(W_J\) has a longest length. Therefore level \(t\) must disappear after at most that many chosen moves. Negative operations cannot recreate a larger level. Repeating at the next largest excess level terminates, since the initial largest level is finite and each stage decreases it toward \(q\). All operations lie in \(A_J\), and the exterior data stay fixed, proving the required accessibility. ◻

Proof. The exterior coordinates of \(m^q\) are fixed on \(A_Jx\), so the vector \(p\) in the hypothesis is the same for every representative under consideration. For each \(w\in W_J\), choose a reduced expression and let \(G_w\) be its negative lift acting on the left. Its action on signed folded counts is \(w\), because each negative letter acts by the same coordinate reflection as its positive inverse. Both \(G_w\) and its inverse have the reduced-word amplitude from Corollary 22.

Put \(M_L=Y_L^q(x)\) and \(N_L=G_wM_L\). The latter has levels \(q-1,q\). Its truncation triangle is \[ N_{L,\mathrm{low}}\longrightarrow N_L \longrightarrow N_{L,\mathrm{high}}, \tag{47}\] where the two end terms are pure at \(q-1\) and \(q\), respectively. The low term is supported on \(\widetilde J\): all outside-color multiplicities in each layer are unchanged under \(G_w\), and the input was pure at \(q\).

Apply \(G_w^{-1}\) to (47). Write its end terms as \(A_L\) and \(C_L\), so that the triangle is \[A_L\longrightarrow M_L\longrightarrow C_L.\] The amplitude bounds place \(A_L\) in \([q-1,q]\) and \(C_L\) in \([q,q+1]\). The long exact sequence at \(q-1\) forces \(H^{q-1}_{\mathrm{lay}}(A_L)=0\), since the adjacent middle and right groups there vanish. At \(q+1\) it forces \(H^{q+1}_{\mathrm{lay}}(C_L)=0\). Thus both end terms are pure at \(q\), and the sequence at \(q\) is short exact: \[ 0\longrightarrow A_L\longrightarrow M_L \longrightarrow C_L\longrightarrow0 \tag{48}\] in the shifted heart. By Lemma 14, \(A_L\) is an actual graded subrepresentation \(\mathcal S_L\) of \(M_L\). Its support is still in \(\widetilde J\), because a twist at a color in \(J\) sends a complex of projectives of colors in \(J\) to another such complex.

The subobject morphism in (48) is \[G_w^{-1}\tau_{\leq q-1}(G_wM_L) \longrightarrow M_L.\] Lemma 23 shows that it is obtained from the unit-label morphism by right tensoring with \(L\), up to an isomorphism of inclusions. Applying the weighted fusion identity to its generator spaces therefore gives (46) for a vector \(s=s_w\). This argument uses canonical truncations; it makes no choice of elimination splittings simultaneously in the different labels.

The quotient family \(C_L\) has weighted dimension \(\phi(m^q(x)-s_w)\). Applying \(G_w\) to it recovers \(N_{L,\mathrm{high}}\), which is pure at \(q\). Signed generator counts consequently give the folded dimension of this high family as \[ w\bigl(m^q(x)-s_w\bigr). \tag{49}\] There is no contribution here from higher layers of the original \(E_L(x)\) outside the operated region: Lemma 26 identifies its level-\(q\) family after \(G_w\) with the level-\(q\) family of \(G_wM_L\), exactly the high family used in (49).

The representative obtained by \(G_w\) belongs to \(A_Jx\) and retains the joint cap, by Lemma 27. Applying the universal hypothesis to (49), and using \(wp=p\), gives \[ \bigl(w(e-s_w)\bigr)_j\geq0\qquad(j\in J). \tag{50}\] In the strict case every one of these coordinates is strictly positive.

We finish in the finite-dimensional real vector space of root coordinates on \(J\). The finite Coxeter chamber theorem says that every linear functional \(f\) on this space can be written \[f(v)=\sum_{j\in J}\lambda_j(wv)_j \quad\text{for some }w\in W_J\text{ and } \lambda_j\geq0.\] Indeed the nonnegative coordinate functionals form the closed fundamental chamber in the dual representation, whose \(W_J\)-translates cover the dual space. For the \(w\) chosen for \(f\), (50) implies \[ f(e)\geq f(s_w)\geq\min_{v\in\mathcal V} f(v), \qquad \mathcal V=\{s_u:u\in W_J\}\cup\{0\}. \tag{51}\] The set \(\mathcal V\) is finite. If \(e\) lay outside its convex hull, a separating linear functional would contradict (51). Thus \(e_J\) lies in that convex hull.

If the original inequalities were all strict, then for nonzero \(f\) at least one \(\lambda_j\) is positive, so \(f(e)>f(s_w)\geq\min_{v\in\mathcal V}f(v)\). A point in a finite convex hull which satisfies this strict inequality for every nonzero functional lies in its ordinary ambient interior: a boundary point, or a convex hull in a proper affine subspace, has a nonzero supporting functional with equality. This proves the strict assertion. For \(J=\varnothing\) the construction gives \(e=s_w=0\) and the stated zero-dimensional conclusion directly. ◻

Proof. Before minimizing, \(B_h\) has one regular bimodule \(Z\) in placements \((0,0)\); every other term is a shift of \(P_\alpha\otimes_{\mathbb C}Q_\beta\) with \(\alpha,\beta\in\widetilde T\). Tensor products of projective bimodules remain sums of such terms, since their middle tensor factor is a graded path space.

No indecomposable summand of the regular bimodule is isomorphic to one of these projective bimodules. To see this, decompose \(Z\) according to the connected components of its unfolded graph. The regular bimodule of one component is indecomposable: a decomposition would give a nontrivial central idempotent, whose degree-zero part would separate the vertices despite the nonzero arrows between them. Its top modulo the left and right radicals consists of the diagonal simple at every vertex of the component. The corresponding top of \(P_\alpha\otimes Q_\beta\) consists of just the single pair \((\alpha,\beta)\). Equality of these tops would force a singleton component and \(\alpha=\beta\). In that case the regular bimodule has dimension two, whereas \(P_\alpha\otimes Q_\alpha\) has dimension four. Isomorphism is again impossible. These observations are unaffected by shifts. Thus all regular summands remain in the minimal bimodule complex, and remain only in placements \((0,0)\).

A radical map between projective bimodules is a sum of path tensors with positive length on at least one side. The only possible term with both lengths zero is a scalar map between identical indecomposable shifted projectives; a nonzero such term would be an isomorphism modulo the radical and would have been canceled. Tensoring on the left by \(U_\ell\) kills terms with positive left length. Every surviving projective-to-projective differential therefore has positive right length and is radical. A map involving the regular term specializes to a map between \(U_\ell\) and a right projective, hence is also radical. There is no differential between two surviving simple terms, since the regular terms occur only in cochain degree zero. It follows that tensoring the minimal bimodule complex by \(U_\ell\) gives an already minimal right complex. Its projective counts are exactly the counts of the corresponding bimodule terms with first index \(\ell\).

By Lemma 9, adjoint duality on finite graded bimodules satisfies \[\mathop{\mathrm{Hom}}_{Z\text{-left}}(M,Z) \cong \mathop{\mathrm{Hom}}_\mathbb C(M,\mathbb C)\langle2\rangle.\] It is a contravariant equivalence, preserves additive radicals, and sends \[P_\ell\otimes Q_\beta \quad\longmapsto\quad (P_\beta\otimes Q_\ell)\langle-2\rangle.\] For complexes it also reverses cochain placements, with the ordinary cochain dual differential. Hence a generator at \((c,u)\) on the left corresponds to one at \((-c,-u-2)\) on the right. The total generator degree changes from \(r\) to \(-2-r\). The adjoint dual of \(B_h\) is homotopy equivalent to its inverse \(B_{h^{-1}}\), by Proposition 13. As the dual minimal complex is still minimal, the preceding specialization proves (53).

For \(b\in A_{J'}\) and \(\ell\notin\widetilde J'\), each generator or inverse generator occurring in \(b\) fixes \(U_\ell\): its projective evaluation or coevaluation term tensors to zero. Thus \[U_\ell B_{bh}\simeq U_\ell B_h,\] which proves row invariance. Apply (53) to \((ha)^{-1}=a^{-1}h^{-1}\), for \(a\in A_J\), to obtain column invariance. ◻

Proof. The inverse bimodule complex is \[D_\alpha= [Z\longrightarrow(P_\alpha\otimes Q_\alpha)\langle-2\rangle],\] with its second term in cochain degree one. An old projective \(Q_\beta\) with placements \((c,u)\), tested against a path of length \(a\) in \(e_\beta Ze_\alpha\), inserts a copy of \(Q_\alpha\) with placements \[ (c+1,u+a-2), \qquad\text{of level }c+u+a-1. \tag{56}\] The only insertions that can exceed the old level have \(a=2\), and then \(\beta=\alpha\). We cancel all of them simultaneously, keeping track of the old differential.

Write \(\mathcal E=\operatorname{Tot}(R D_\alpha)\). In its underlying bigraded module, let \(A\) be the sum of all old \(Q_\alpha\) terms in the \(Z\)-column, and let \(B\) be the sum of their inserted length-two copies. The copy paired with an old term at \((c,u)\) lies at \((c+1,u)\). The token–unit summand of the Frobenius coevaluation gives an isomorphism \[\Phi=\pi_B d|_A:A\longrightarrow B\] of cochain degree one, with signed nonzero scalar identity entries. The old differential stays within its tensor column and contributes nothing to this particular block.

Consider \[\mathcal K=A\oplus d(A)\subseteq\mathcal E.\] Projection onto \(B\) identifies \(d(A)\) with \(B\), so \(A\cap d(A)=0\). The displayed sum is a subcomplex: its differential maps \(A\) isomorphically to \(d(A)\) and kills \(d(A)\). It is contractible, and is a direct summand of the underlying bigraded module. Thus the quotient \(\mathcal E\longrightarrow\mathcal E/\mathcal K\) is a homotopy equivalence. More explicitly, if \(\mathcal E=A\oplus B\oplus W\) as a bigraded module, the quotient has underlying terms \(W\). Representing a quotient class by its \(W\)-component gives the differential \[ d'_W=d_{WW}-d_{WA}\Phi^{-1}d_{BW}. \tag{57}\] This formula includes any scalar entries that the old differential induces among the inserted projectives; their vanishing is not needed.

The remaining projective terms are the old \(Q_\beta\) with \(\beta\ne\alpha\), and the length-zero and length-one insertions in (56). Their levels are, respectively, the old level, the old level minus one, and the old level. The original simple remains at normalized level \(-1\). If \(\alpha=\ell\), the simple also inserts a projective with placements \((1,-2)\), of level \(-1\); if \(\alpha\ne\ell\), it inserts nothing. All surviving terms have level at most \(r\). Further minimization removes summands without introducing new ones, so preserves this bound. Applying this argument successively to the orthogonal vertices of a color proves the color statement.

Finally, the inverse of the right-hand test-shift formula for the positive twist in Lemma 18 is (55). Equivalently, it follows by canceling the token–unit component of \(D_\alpha e_\alpha\). Its total shift is \(-2+1=-1\), so in particular \[H^t((R D_\alpha)e_\alpha) \cong H^{t+1}(Re_\alpha).\] The shift formula holds for the nonprojective simple input as well, since the cancellations are bimodule homotopies before tensoring. ◻

Proof. Filter the entire vector complex \(R_\ell\otimes_Z E_{L'}(z)\) by middle path lengths zero, one, and two, assigning length one to a simple tensor. Minimal projective differentials have positive path length. The map from a projective to a simple sends a length-zero tensor to the assigned length-one tensor and kills positive middle paths; the map from the simple to a projective sends its assigned length-one tensor to a token tensor. A left radical differential tensored with a right simple is zero. Thus every differential strictly increases this filtration index.

For a right projective generator at \((c,u)\), a left generator at \((c',u')\), and a middle path of length \(a\), the total degree is \[(c+u)+(c'+u')+a.\] For the simple tensor it is \(c'+u'=(-1)+(c'+u')+1\), the same rule with the normalized right level. The first differential of the filtration increases length by one and preserves each pair of levels. Its cohomology at the pair \((r,q)\) is precisely \(H_j(N,M)\). With only three length columns, the only possible remaining differential goes from length zero to length two. Accordingly, every middle class survives to the associated graded of actual cohomology.

Every right projective has its vertex in \(\widetilde T\). The tensor \(Q_\beta\otimes_Z P_\gamma=e_\beta Ze_\gamma\) is nonzero only if \(\gamma=\beta\) or \(\gamma\) is a neighbor of \(\beta\). A simple tensor is nonzero only at its own vertex in \(\widetilde T\). Hence every nonzero term of the entire tensor complex uses a left color in \(T\cup\partial T\), where the cap is \(q\). Layers above \(q\) outside this set give zero tensor terms.

A secondary differential into length two at \((r,q)\) would start in length zero at a pair whose level sum is \(r+q+1\): the differential raises actual total degree by one while raising length by two. Such a pair is absent, since the right level is at most \(r\) and every relevant left level is at most \(q\). Thus \(H_2(N,M)\) survives as well. Its total degree is \(r+q+2>q\), so (58) forces it to be zero. The middle degree is \(r+q+1\), proving (59).

For the last assertion, project the token tensors at \(\beta\) to the tensor product of the two coefficient spaces modulo their incoming images, and send all other token tensors to zero. Every incoming length-one boundary has, at \(\beta\), an incoming coefficient on at least one side. The projection therefore annihilates boundaries and is surjective onto the tensor product of the two top quotients. If a right top at a vertex of color \(j\) is nonzero, \(H_2(N,M)=0\) forces the left top at that same vertex to be zero for every family member. By Lemma 24, a top occurring anywhere in a color of the left family occurs at every unfolded label of that color in some member. Hence no left-family top in color \(j\) can occur. ◻

Proof. Suppose a right top at the current highest level \(r\) occurs in color \(j\in J\). Lemma 33 shows that the left-family top in this color at level \(q\) vanishes. The positive operation \(F_j\) therefore preserves the cap on \(T\cup\partial T\), by Lemma 27 applied with spherical set \(T\). Replace \[z\longmapsto\sigma_jz, \qquad h\longmapsto h\sigma_j^{-1}, \qquad R_\ell\longmapsto R_\ell B_{\sigma_j}^{-1}.\] The candidate, its harmonic vector \(p\), and the exterior cap are unchanged. The tensor products still compute the same complexes \(U_\ell E_{L'}(z')\). By Lemma 32, the new maximum normalized level is at most \(r\). Thus the vanishing of the left top needed to license the next move follows anew from Lemma 33.

At a fixed value of \(r\), the sequence of chosen colors is reduced in \(W_J\). To prove this, let \(w\) be the reduced word already chosen at this level. If a color \(j\) were a right descent of \(w\), a reduced expression for \(w\) could end in \(j\). Matsumoto moves give the same negative Artin word in this expression. Each intermediate negative operation preserves the upper bound \(r\), regardless of the expression. The final inverse operation in color \(j\) lowers the total test degrees in that color by one. Before it, every test above \(r+2\) vanishes; after it, every test at \(r+2\) in color \(j\) therefore vanishes. Formula (54) excludes a right top there. A color selected for a nonzero right top must consequently be an ascent, proving the assertion by induction on the number of moves.

The length of a reduced word in \(W_J\) is bounded by the length of its longest element. Therefore at a fixed \(r\) the procedure either removes all the indicated tops or lowers the maximum level after finitely many steps. The maximum is an integer bounded below by \(-1\), because the residual simples remain. There are only finitely many possible decreases from its initial value. The procedure terminates with the desired representative. If \(J=\varnothing\), no move is needed. ◻

Proof. The inclusion in degree zero follows from the inclusion of the length-zero vector spaces and compatibility of the length-one differentials. Since there is no incoming term of negative length, their degree-zero cohomologies are kernels, and kernel inclusion gives the asserted injection.

For the degree-two assertion, work first with a single right row and a single left subrepresentation, and retain homogeneous coefficient spaces in their cochain gradings. At every projective vertex of \(\widetilde J\), vanishing of the right top says that every right coefficient is a sum of incoming coefficient images, from projective vertices or from the framing simple. Iterating this statement expresses coefficients in \(\widetilde J\) using ones outside \(\widetilde J\) or in the simple: along an incoming projective path the predecessor cochain placement is smaller by one, and these placements are bounded below.

We make this iteration in the cokernel of the length-one differential, rather than in the coefficient spaces alone. Suppose an incoming right coefficient at \(\beta\) comes from a homogeneous coefficient at \(\gamma\), through a length-one map \(Q_\gamma\to Q_\beta\). Choose a middle path in \(e_\gamma Z^1e_\beta\) dual to this arrow under the perfect backtrack pairing. Tensor this middle path with the coefficient at \(\gamma\) and any homogeneous left coefficient in \(\mathcal S_\beta\). Its differential has two possible token contributions. The contribution at \(\beta\) is the selected incoming right coefficient, tensored with the unchanged left coefficient. The contribution at \(\gamma\) is the original right coefficient tensored with the corresponding left-arrow image, with the ordinary cochain sign. The dual path can be normalized so that the first coefficient is one. Hence, modulo length-two boundaries, the token tensor at \(\beta\) is equal, up to a nonzero scalar and sign, to the latter token tensor at \(\gamma\).

The new left coefficient lies in \(\mathcal S_\gamma\), since \(\mathcal S\) is closed under the left coefficient maps. Varying the dual middle path tests every incoming arrow coefficient by nondegeneracy of the pairing, including when an arrow space has dimension greater than one. Thus this substitution accounts for the full span of incoming images. Each substitution decreases the right cochain placement by one and increases the left placement by one, so the represented total degree is unchanged and the substitution process terminates.

If it terminates at a projective vertex outside \(\widetilde J\), the left coefficient is zero. If an incoming coefficient comes from the simple, its tensor with the same left coefficient is itself an assigned length-one input. The right differential sends that tensor to exactly the token-inclusion image in question. Its left differential is zero because tensoring a radical left map with a right simple gives zero. Such a token tensor is therefore already a boundary. Every length-two generator is a sum of the tensors just treated, so the cokernel vanishes. This proves the first assertion row by row and member by member, hence for the families as well. ◻

Proof. For a family \(\mathcal S\) with folded dimension \(s\), Lemma 35 and the Euler identity give \[\chi(n,s) =\frac{\dim_{\mathrm{wt}}H_0(N,\mathcal S) -\dim_{\mathrm{wt}}H_1(N,\mathcal S)}{D_0} \le\frac{\dim_{\mathrm{wt}}H_0(N,M)}{D_0}.\] The right side is independent of the chosen subfamily. Since \(e\) is a convex combination of these \(s\), and \(\chi(n,\cdot)\) is linear, the same upper bound holds for \(\chi(n,e)\). On the other hand, Lemma 33 gives \[\chi(n,m) =\frac{\dim_{\mathrm{wt}}H_0(N,M) -\dim_{\mathrm{wt}}H_1(N,M)}{D_0}.\] Subtracting yields \[\frac1{D_0}\dim_{\mathrm{wt}}H_1(N,M) \ge\chi(n,e)-\chi(n,m)=-\chi(n,p).\] Finally, \(n\) is supported on \(T\), and \((Cp)_T=-a\). Equation (63) therefore turns the right side into the expression in (64). ◻

Proof. An atom \(s\) right-divides a simple element \(d\) exactly when \(s\in D_R(d)\), and \(sd\) is simple exactly when \(s\notin D_L(d)\). If the inclusion in (65) fails, an atom in its difference can be transferred from the end of \(d_i\) to the beginning of \(d_{i+1}\), giving a larger simple suffix of the corresponding prefix. This proves necessity.

For sufficiency, induct on prefixes. Suppose \(P=d_1\cdots d_{j-1}\) has greatest simple suffix \(d_{j-1}\), and let \(g=t(Pd_j)\). Since \(d_j\) is a simple suffix, it right-divides \(g\); write \(g=cd_j\). If \(c\ne1\), choose a final atom of \(c\), so \(c=c's\). The element \(sd_j\), being a factor of the simple element \(g\), is simple. Also \(Pd_j=bg=bc d_j\) for some positive \(b\), so cancellation gives \(P=bc\). Thus \(s\) right-divides \(P\). Since an atom is simple, it right-divides \(t(P)=d_{j-1}\). We have obtained \(s\in D_R(d_{j-1})\) and \(s\notin D_L(d_j)\), a contradiction. Therefore \(g=d_j\), completing the induction. Removing the final factor successively gives the asserted greedy extraction.

Now take the greedy factorization of a positive original element. Lemma 10 says that reduced-word lifts remain simple after unfolding, and that both descent sets become their full color preimages. Thus the unfolded factors satisfy (65) and form the unfolded greedy factorization. Every extractable final atom of the unfolded element right-divides its greatest simple suffix, hence is a right descent of the last unfolded factor. Descent reflection supplies its color as a right descent of the last original factor. That original atom therefore right-divides the original positive element. ◻

Proof. Fix \(\ell\in\widetilde V\). The only right projective admitting a nonzero map to or from \(U_\ell\) is \(Q_\ell\): a map to the simple uses its top quotient, and a map from the simple uses its token socle. Condition (66) excludes this projective from \(R_\ell(h)\). Its simple term therefore splits off as a complex. Tensoring with \(B_h\) is a homotopy equivalence and preserves indecomposability. Since \(U_\ell\) is indecomposable, the remaining projective summand must be zero in the homotopy category. It follows that \[U_\ell B_h\simeq U_\ell\qquad(\ell\in\widetilde V).\]

Use finite-type fractions to write \(h=h_1h_2^{-1}\) with \(h_1,h_2\in A_T^+\). On setting \(U_V=\bigoplus_{\ell\in\widetilde V}U_\ell\), the preceding equivalences give \[ U_V B_{h_1}\simeq U_V B_{h_2}. \tag{67}\] For a bounded right complex \(R\), use the height \[\max\{t:H^t(Re_\beta)\ne0 \text{ for some }\beta\in\widetilde T\}\] from Lemma 21, with total cohomological degree as there. Its value on \(U_V\) is zero. A positive word avoiding \(V\) fixes \(U_V\). If a positive word contains a color of \(V\), all operations before its first such occurrence fix \(U_V\), and that first occurrence acts on a highest test and raises the height. The height cannot subsequently decrease. Thus the final height is positive exactly when the word contains a color of \(V\).

If the common height in (67) is positive, any vertex with highest cohomology gives, by the right version of Lemma 21, an extractable final atom of both unfolded positive words. Lemma 37 supplies a common extractable original color \(s\). Write \(h_i=g_i\sigma_s\), and tensor (67) on the right by \(B_{\sigma_s}^{-1}\). This gives the same equality with \(g_i\) in place of \(h_i\), while \[h_1h_2^{-1}=g_1\sigma_s\sigma_s^{-1}g_2^{-1}=g_1g_2^{-1}.\] The sum of positive lengths decreases by two. Repetition therefore terminates with common height zero. Both remaining positive words then avoid \(V\), so their fraction lies in \(A_{T\setminus V}\). ◻

Proof of Proposition 30. Suppose two candidates are given, and use the notation above. Carry out the top-removal procedure for the first candidate, with the second supplying the positive row weights. If its final highest normalized right level satisfies \(r>-1\), that level contains a nonzero projective coefficient and no framing simple. It must contain a nonzero coefficient in \(\widetilde V\). Indeed, if it were supported entirely on \(\widetilde J\), vanishing of all its right tops there would express every coefficient using incoming ones with strictly smaller cochain placement. Boundedness would force every coefficient to be zero.

All row weights are strictly positive. A nonzero coefficient in \(\widetilde V\) consequently gives \(n_i>0\) for some \(i\in V\). Since \(a_i>0\) there, \(n_T^{\mathsf t}a>0\). Equation (64) then gives nonzero middle cohomology. It survives in actual total degree \(r+q+1>q\), contradicting (58). We conclude that \(r=-1\).

At this level the surviving middle cohomology lies in total degree \(q\), so its weighted dimension is at most that of actual cohomology there. Since \(a'\) is supported on \(V'\), where \(p'_i=m_i^q(z')\), equations (58) and (64) give \[ (a')^{\mathsf t}(p'_T-p_T) \ge n_T^{\mathsf t}a\ge0. \tag{68}\] Interchanging the candidates and repeating the argument gives \[ a^{\mathsf t}(p_T-p'_T)\ge0. \tag{69}\] The two algorithms may choose different representatives. The vectors \(p,p',a,a'\), being data of the candidate cosets, are unchanged by these choices.

The exterior coordinates of \(p\) and \(p'\) outside \(T\) agree. Put \(\delta=p'_T-p_T\). Then \[a'-a=-C_T\delta.\] The sum of the left sides of (68) and (69) is therefore \[(a'-a)^{\mathsf t}\delta =-\delta^{\mathsf t}C_T\delta.\] Both summands are nonnegative, whereas \(C_T\) is positive definite. It follows that \(\delta=0\) and equality holds throughout. Thus \(p=p'\), \(a=a'\), and \[V=\{i\in T:a_i>0\}=V',\qquad J=J'.\] In particular the nonnegative term \(n_T^{\mathsf t}a\) in (68) is zero.

It remains to identify the cosets, rather than just their numerical data. Fix the comparison element \(h_0=z'z^{-1}\) using any initial representatives of the two candidates. Replacing these by \(bz\) and \(cz'\), with \(b\in A_J\) and \(c\in A_{J'}\), changes it to \[c h_0 b^{-1}.\] By Lemma 31, the block with row indices in \(\widetilde V'\) and column indices in \(\widetilde V\) is invariant under all such changes. After the equality just proved, this is the same block \(\widetilde V\times\widetilde V\) in both directions.

In the forward comparison no right projective generator has degree above \(-1\). At degree \(-1\), the equality \(n_T^{\mathsf t}a=0\), formula (62), and strict positivity of all its weights imply \[K^{-1}_{\ell\beta}=0 \qquad(\ell,\beta\in\widetilde V).\] Indeed every summand contributing to a column color in \(V\) has a positive coefficient, and all counts are nonnegative. By block invariance, the fixed block for \(h_0\) is therefore zero in every degree \(d\ge-1\). The reverse comparison proves that the same block for \(h_0^{-1}\) is zero for all \(d\ge-1\), irrespective of the representatives chosen in that direction. Formula (53) transposes these latter vanishings into vanishing of the block for \(h_0\) in all degrees \(d\le-1\). Together the two ranges give \[K^d_{\ell\beta}(h_0)=0 \qquad(d\in\mathbb Z,\ \ell,\beta\in\widetilde V).\] Lemma 38 gives \(h_0\in A_{T\setminus V}=A_J\). Thus \(z'=h_0z\) and \(A_Jz'=A_Jz\), proving equality of the two candidates. ◻

Proof. The symmetric Frobenius trace of \(Z\), normalized by \(\operatorname{tr}(\theta_\alpha)=1\), pairs paths of length \(i\) perfectly with opposite paths of length \(2-i\). On complexes use the supertrace \[\operatorname{Str}_Y(a) =\sum_c(-1)^c\operatorname{tr}(a|_{Y^c}).\] Here \(Y^c\) denotes the cochain-degree-\(c\) term, and the trace includes the ordinary matrix trace on multiplicity spaces. The summandwise pairing \(\langle f,g\rangle=\operatorname{Str}_Y(f\circ g)\) is perfect between \(\mathcal L_i^{n,v}(P,Y)\) and \(\mathcal L_{2-i}^{-n,2-v}(Y,P)\). Cyclicity gives, for the complementary degrees appearing in a differential pairing, \[\langle D_1f,g\rangle+(-1)^n\langle f,D_1g\rangle=0.\] There is no sign from the internal degree. Finite-dimensional duality therefore gives the asserted perfect pairings on cohomology. This argument uses only the linear differentials and the Frobenius pairing, so it also applies to the pure layers associated with graded subrepresentations.

The alternating sum of the dimensions of the three path-length columns is \((\dim P)^{\mathsf t}\widetilde C\dim Y\): there are two equal-vertex paths, of lengths zero and two, and the length-one paths contribute the negative adjacency entries. Euler characteristic and the pairing of columns zero and two give (73).

For the remaining assertion, consider the full bigraded Hom complex \(\mathop{\mathrm{Hom}}_Z(E_L,E_H)\). Minimality writes each projective differential as a sum of its length-one and length-two parts. Thus its Hom differential is \(D_1+D_2\), where \(D_a\) raises path length by \(a\). Filter Hom by paths of length at least \(i\). There are only the three columns \(0,1,2\); the first nonzero differential is \(D_1\). Its cohomology decomposes as the direct sum of the spaces above, indexed by the ordered pair of source and target layers. After this page, the only possible differential goes from column zero to column two. Every column-one class is therefore permanent.

By Proposition 13, applying \(B_x\) preserves the bigraded Hom cohomology. Between the initial frame projectives this cohomology is zero when \(L\ne H\). When \(L=H\), it has dimension one in each of bidegrees \((0,0)\) and \((0,2)\) and is zero elsewhere. Both of these dimensions have specific permanent representatives in the filtration. First, the identity of \(E_L\) is a nonzero class in column zero and bidegree \((0,0)\); a boundary cannot have a length-zero component. Second, choose one projective summand in the nonzero minimal complex \(E_L\), and place its normalized token on that diagonal matrix entry, with all other entries zero. This is a cycle in bidegree \((0,2)\), since a token multiplied on either side by a radical entry is zero. If the chosen summand is in cochain placement \(c\), this cycle has supertrace \((-1)^c\ne0\). Every Hom boundary has zero supertrace by graded cyclicity, so the token cycle is nonzero in actual Hom cohomology. It lies in filtration two, and filtration three is zero; hence it supplies a nonzero column-two class in the associated graded cohomology.

The identity and token exhaust the two actual diagonal Hom dimensions. All actual off-diagonal Hom cohomology vanishes. Since every column-one class is permanent, every such class must be zero, proving (74). Finally, a direct summand of a pure layer gives a direct summand of each linear Hom complex. Its column-one cohomology is therefore zero as well. This last argument is on the linear page; it does not assert that the chosen layer summand splits off from the full complex \(E_L\). ◻

Proof. Represent a class on the left of (76) by a homogeneous length-zero linear cocycle \(f\), and extend it by zero on the other source-layer summands. This is a linear cocycle in \(\mathop{\mathrm{Hom}}_Z(E_L,E_H)\) because \(M_L\) is a direct summand for the linear differential. It defines an embedded column-zero class on the page after \(D_1\): the linear differential preserves the ordered source/target layer pair, and column zero has no incoming boundaries.

Write the length-two projective differentials as \(d_{E_L}^{(2)}\) and \(d_{E_H}^{(2)}\). The secondary differential on this class is represented by \[d_{E_H}^{(2)}f-(-1)^n f d_{E_L}^{(2)}.\] The only possible ordered layer pairs in its target are \[ \begin{array}{ccl} (q,d-1)&\text{from}&d_{E_H}^{(2)}f,\\ (q+1,d)&\text{from}&f d_{E_L}^{(2)}. \end{array} \tag{78}\] A length-two entry is diagonal in the unfolded vertex and lowers the layer by one. For the second term in (78) to be nonzero, its source would have to lie at level \(q+1\) on a vertex of \(\widetilde I\). Such generators are excluded by the cap. Vertices outside \(\widetilde I\) cannot contribute through a diagonal length-two entry. Thus the secondary image lies entirely in the first block. The splitting \(Y_L^q=M_L\oplus M_L^{\prime}\) is a splitting for the linear differential, so it splits the target column-two cohomology by source summands as well. The target-side composite is zero on \(M_L^{\prime}\) and therefore lands in the summand \(\mathsf H_2(M_L,Y_H^{d-1})\), giving the map in (76).

If a class were in its kernel, it would be permanent: no differential enters column zero, and there is no possible differential after the secondary one. Its total degree is \(d-q<0\). Actual Hom cohomology between \(E_L\) and \(E_H\) is zero in this degree by the computation in Lemma 40. The class is therefore zero. This proves the injection for each pair of labels before any weighting. Sum the resulting dimension inequalities with positive weights \(w_Lw_H\) and use the dual pairing of Lemma 40; the result is (77). ◻

Proof. Fix one of the subrepresentation families \(\mathcal S\). A nonzero homogeneous coefficient-array map \(S_L\to Y_H\) has a nonzero graded image supported on \(\widetilde J\). Such an image has a socle: arrow coefficients raise cochain placement, so a nonzero coefficient in its highest occupied placement is killed by every outgoing arrow. Since the image is a subrepresentation, this is a socle of \(Y_H\) on \(\widetilde J\). The hypothesis therefore gives \[h_0^{\mathrm{wt}}(\mathcal S,\mathcal Y)=0.\] Also, composition with the inclusion \(S_L\hookrightarrow M_L\) gives injections on length-zero linear cocycles, with no length-zero boundaries to quotient out. Thus \[h_0^{\mathrm{wt}}(\mathcal Y,\mathcal S) \le h_0^{\mathrm{wt}}(\mathcal Y,\mathcal M).\] Apply (75) first to \((\mathcal S,\mathcal Y)\) and then to \((\mathcal M,\mathcal Y)\). The first application uses only \(h_1^{\mathrm{wt}}(\mathcal S,\mathcal Y)\ge0\); the second has \(h_1^{\mathrm{wt}}(\mathcal M,\mathcal Y)=0\) by Lemma 40. Subtracting gives \[ D_0(m-s)^{\mathsf t}Cy \ge h_0^{\mathrm{wt}}(\mathcal M,\mathcal Y)\ge0. \tag{80}\] In particular, no middle-Hom vanishing for \(\mathcal S\) is required.

The left side of (80) is affine-linear in \(s\), so the inequality holds throughout the convex hull. At its interior point \(e=m-c\), it gives \[ D_0c^{\mathsf t}Cy \ge h_0^{\mathrm{wt}}(\mathcal M,\mathcal Y)\ge0. \tag{81}\] On the other hand, symmetry of \(C\), the null equation on \(I\), and nonnegativity of \(y\) give \[ c^{\mathsf t}Cy =-\sum_{a\notin I}\sum_{i\in I}b_{ai}c_i y_a\le0. \tag{82}\] This is strictly negative if \(y_a>0\) for any \(a\in\partial I\). Consequently \(y|_{\partial I}=0\), the two sides of (81) are zero, and \(h_0^{\mathrm{wt}}(\mathcal M,\mathcal Y)=0\).

Because \(e_J\) is an ordinary real interior point, for each \(j\in J\) the hull contains \(e_J\pm\varepsilon\mathbf e_j\) for sufficiently small \(\varepsilon>0\). Substituting these two points into (80), whose value at \(e\) is now zero, forces \((Cy)_j=0\). It follows that \[m^{\mathsf t}Cy =c^{\mathsf t}Cy+(m-c)_J^{\mathsf t}(Cy)_J=0.\] The weighted Euler identity, with middle vanishing for \(\mathcal M\), now forces \(h_0^{\mathrm{wt}}(\mathcal Y,\mathcal M)=0\) as well. When \(J=\varnothing\), one has \(m=c\) and the hull is the singleton in \(\mathbb R^0\); the same argument applies with no coordinate perturbations. ◻

Proof of Proposition 39. Assume (71) holds at every capped representative. Since \(\partial J\subseteq(I\setminus J)\cup\partial I\), Lemma 27 supplies a representative capped on \(J\) and hence on all of \(I\). We shall move only among such representatives. All the exterior data in (70) stay fixed, since operations in \(A_J\) change multiplicities only in colors of \(J\).

At any capped representative, let \(M_L\) be the part of \(Y_L^q\) supported on \(\widetilde I\), and set \(\mathcal M=(M_L)_L\). This is a direct summand of the pure layer: its outside neighbors have zero multiplicity at level \(q\) by (70), so no linear arrow joins it to the rest of that layer. This statement concerns the pure layer, and does not assert a splitting of \(E_L\). By Lemma 24, its weighted generator array is \(\phi m\), where \(m\) equals \(m^q\) on \(I\) and is zero off \(I\).

Harmonizing the full level-\(q\) vector on \(J\) gives the coordinates \(c_J\). Indeed, only neighbors of \(J\) enter these equations, their prescribed values are those of \(c\), and \(C_Ic_I=0\); uniqueness follows from the sphericity of \(J\). Remote coordinates outside \(I\cup\partial I\) do not enter these equations. Capping a representative on \(J\), with the fixed exterior cap, is equivalent to capping it on \(I\). Proposition 28 therefore applies with its strict, universally quantified hypothesis. It shows that \[ e=m-c,\qquad \mathop{\mathrm{supp}}(e)\subseteq J, \qquad e_J\in\operatorname{int}_{\mathbb R^J} \operatorname{conv}\{s_J\}, \tag{83}\] where each \(s\) is the folded weighted dimension of subrepresentations \(\mathcal S\subseteq\mathcal M\) supported on \(\widetilde J\). Such subrepresentations, initially obtained in the full layer, lie in \(\mathcal M\) because of their support.

The interior in (83) is the ordinary real interior supplied by Proposition 28, even when folded multiplicities are not integers. For \(J=\varnothing\), it means \(e=0\) in \(\mathbb R^0\).

Thus Lemma 42 applies at every capped representative. This uniformity is essential: both \(\mathcal M\) and its convex hull may change during the following operations, but the hypothesis (71) supplies (83) anew after each cap-preserving move.

Let \(d_b\) be the lowest layer that meets \(\partial I\). This level exists because the frame \(o\) belongs to \(\partial I\) and \(m_o^0=1\). It is fixed throughout the coset, and \[ d_b\le0<q. \tag{84}\] Choose an integer \(d_{\min}\) below or equal to every initially occupied layer of every member of the family. Positive operations cannot create layers below \(d_{\min}\) by their amplitude. We shall proceed through \(d=d_{\min},\ldots,d_b\). At the beginning of a stage, the induction conditions are, for every \(k<d\), \[ \begin{aligned} \operatorname{soc}_{\widetilde J}\mathcal Y^k&=0, & (Cm^k)_J&=0,\\ h_0^{\mathrm{wt}}(\mathcal Y^k,\mathcal M)&=0, & h_0^{\mathrm{wt}}(\mathcal M,\mathcal Y^k)&=0. \end{aligned} \tag{85}\] Here \(m^k\) is the folded vector of the full layer \(\mathcal Y^k\), and \(\mathcal M\) always denotes the current level-\(q\) summand. The conditions hold initially since all lower layers are zero.

Since \(d\le d_b<q\), Lemma 41 and the condition at \(d-1\) give \[ h_0^{\mathrm{wt}}(\mathcal M,\mathcal Y^d)=0. \tag{86}\] Suppose that \(\mathcal Y^d\) has a nonzero socle in a color \(j\in J\). Then \(\mathcal M\) has zero top in that color. To see this, a top simple quotient of some \(M_L\) and a socle simple of some \(Y_H^d\) at the same unfolded vertex compose to a nonzero homogeneous length-zero map. Lemma 24 ensures that if a top occurs anywhere in color \(j\) in the family, then it occurs at every label in that color in some family member. It can therefore be matched with the label of the chosen socle. The resulting map would contradict (86), since every weight is positive.

Apply \(F_j\) to this representative. The top of \(\mathcal M\) at color \(j\) is the top of the full level-\(q\) layer there, because \(\mathcal M\) is a direct summand containing that color. Its vanishing makes this move cap-preserving by Lemma 27. Concretely, the only new level-\(q+1\) contribution from the level-\(q\) layer is this top; higher layers have no generators on \(I\cup\partial I\), and the twist acts identically on that remote part. Other colors keep their multiplicities.

We next prove that this move preserves (85) and changes the level-\(d\) layer only by its unshifted positive operation. Denote the data immediately before and after the move by “old” and “new”. For this single move, ascend again through the integers \(k<d\), starting below all occupied layers. Suppose that the old top at color \(j\) in level \(k-1\) has already been shown to vanish. This is vacuous at the starting level. Lemma 15 then identifies \[ \mathcal Y^k_{\mathrm{new}} \cong\text{the level-$k$ part of }F_j\mathcal Y^k_{\mathrm{old}}. \tag{87}\] Indeed, the level-\(k\) term from the old level \(k-1\) is exactly its top in color \(j\), and no other input layer contributes at \(k\). By Lemma 18, the vanishing of the old socle on \(J\) implies the vanishing of the new socle on \(J\) in (87).

The new representative is capped, so Lemma 42 applies using its new family \(\mathcal M_{\mathrm{new}}\). As \(k<d\le d_b\), the layer has no boundary support. The lemma yields all the new conditions (85) at this \(k\), including \((Cm^k_{\mathrm{new}})_j=0\). The old row is also zero, and all multiplicities off color \(j\) are unchanged. Since \(C_{jj}=2\), comparison of the two rows gives \[2\bigl(m^k_{\mathrm{new},j}-m^k_{\mathrm{old},j}\bigr)=0.\] Let \(t_j^k\ge0\) be the folded weighted top multiplicity of \(\mathcal Y^k_{\mathrm{old}}\) in color \(j\). Formula (30), together with (87), says \[m^k_{\mathrm{new},j} =m^k_{\mathrm{old},j}-(Cm^k_{\mathrm{old}})_j+t_j^k.\] Thus \(t_j^k=0\). Positivity of the weights makes every old top in this color zero, providing the premise at \(k+1\). This completes the upward verification. In particular, the old top at level \(d-1\) is zero, so (87) also holds at \(k=d\).

The order of this verification is important: the absence of a shifted-in contribution at \(k\) uses the top vanishing already established at \(k-1\); the top vanishing at \(k\) is then deduced from the new row equation and the pure-layer dimension formula. Hence no assertion about the level being checked is assumed in advance. The verification also proves that (86) can be used again after the move, with the new \(\mathcal M\), by Lemma 41.

Continue applying \(F_j\) whenever a socle in a color \(j\in J\) remains in level \(d\). This process terminates. To prove it, consider only the unshifted operations on that fixed pure layer. By Lemma 16, a composite is the bottom cohomology of the corresponding positive word, so these operations satisfy the positive braid relations. A last-action color has zero socle after its operation by Lemma 18. Inductively suppose the colors used so far form a reduced expression for \(w\in W_J\), with the most recent operation written on the left. If the next chosen color \(j\) were a left descent of \(w\), a reduced expression for \(w\) would begin with \(j\). Braid moves would therefore express the current pure layer as the result of an unshifted \(F_j\) operation acting last. Its socle in color \(j\) would be zero, contradicting the choice of \(j\). Each choice is consequently a Coxeter ascent. Since \(W_J\) is finite, the number of choices at level \(d\) is bounded by the length of its longest element.

At termination, the socle on \(J\) vanishes in the current layer \(d\). If \(d<d_b\), Lemma 42 supplies (85) also at \(k=d\), and we proceed to the next level. The range from \(d_{\min}\) to \(d_b\) is finite, and the earlier conditions were preserved during every move. At \(d=d_b\), the same lemma asserts that the layer has no support on \(\partial I\), contradicting the definition of \(d_b\).

If \(J=\varnothing\), the procedure contains no firing steps: the socle condition on \(J\) is vacuous, (83) says \(m=c\), and Lemma 42 already contradicts the boundary layer. Thus the dimension-zero case is included. The contradiction proves the proposition. ◻

Proof. We first dispose of nonemptiness and the rank-zero convention. If \(T=\varnothing\), there is no proper subset \(J\) in (88), so the strict sublevel is empty. The closed exterior hypotheses would prescribe a cap and \(m^q=c\) on \(I\cup\partial I\). This is precisely the rank-zero situation excluded by Proposition 39.

Now let \(T\ne\varnothing\) and assume the closed exterior hypotheses. Lemma 27 gives a representative capped on \(T\), and hence on \(I\). If the closed sublevel were empty, then at every such representative \(z\) one would have \(m_j^q(z)>c_j\) for every \(j\in T\). Indeed, any \(j\) with \(m_j^q(z)\le c_j\) would give the allowed pair \((T\setminus\{j\},A_{T\setminus\{j\}}z)\). The resulting strict inequalities at every capped representative contradict Proposition 39, applied with its spherical subset equal to \(T\). Thus the closed sublevel is nonempty.

We prove the contractibility assertions together by induction on \(|T|\), with the conventions just established as the base. Let \(\mathcal R\) denote either of the two sublevels, and suppose that the proposition holds for all smaller spherical subsets. For \(X=(J,A_Jz)\in\mathcal R\), define \(p_X\) by \[(p_X)_i=m_i^q(z)\quad(i\notin J),\qquad (Cp_X)_J=0, \qquad u_X=(p_X)_T-b_T.\] This vector is independent of the representative of \(A_Jz\). On \(V=T\setminus J\) it satisfies \((u_X)_i=m_i^q(z)-b_i\le0\). Harmonic extension gives \[ (u_X)_J=-C_J^{-1}C_{J,V}(u_X)_V\le0, \qquad (Cp_X)_T=C_Tu_X. \tag{91}\] The inequality uses nonnegativity of \(C_J^{-1}\) and nonpositivity of the off-diagonal entries of \(C\); the formula includes the empty \(J\) convention. In particular \(u_X\le0\) coordinatewise. In the strict sublevel \(u_X\ne0\), since \(V\) is nonempty and all of its inequalities are strict.

Filter the vertices by increasing height \[h(X)=\sum_{i\in T}(u_X)_i\le0.\] Only finitely many height values occur. For each possible type \(J\), the open counts belong to bounded intervals \(0\le m_i^q(z)\le b_i\), and each such interval contains only finitely many possible layer counts, by Section 5. The exterior vector is fixed, and harmonic extension determines the remaining coordinates. There are only finitely many types. At equal negative height put larger \(|J|\) first; at height zero put smaller \(|J|\) first. Break all remaining ties by any well-order. This gives a well-order of \(\mathcal R\). Remaining ties occur only between incomparable distinct vertices, so they do not affect any earlier lower or upper part.

The upper part at nonzero height. Fix \(X=(J,A_Jz)\) with \(u=u_X\ne0\), and put \(V=T\setminus J\). Every upper extension has the unique form \((K,A_Kz)\) with \(J\subsetneq K\subsetneq T\). It is allowed because its open colors are a subset of \(V\) and retain the same data. Writing \(p_K\) for its harmonic vector, the change is supported on \(K\) and is \[ (p_K-p_X)_K=-C_K^{-1}(C_Tu)_K. \tag{92}\] The rows \((C_Tu)_J\) vanish. Set \[D=\{i\in V:(C_Tu)_i\ge0\}.\] If \(K\setminus J\) avoids \(D\), all its rows in (92) are strictly negative. The nonnegative inverse, with positive diagonal, makes the total change in height strictly positive. Such an extension is later. If instead \(\varnothing\ne K\setminus J\subseteq D\), the change is coordinatewise nonpositive, so the extension is earlier: a strict height decrease suffices, and equality is resolved in its favor by the larger type at the negative height \(h(X)\).

It follows that every earlier upper extension meets \(D\) in a new color, whereas every extension using only new colors in \(D\) is earlier. Also \(D\ne V\). Otherwise \(C_Tu\ge0\), and \(u=C_T^{-1}C_Tu\ge0\), contradicting \(u\le0\) and \(u\ne0\). If \(D=\varnothing\), the earlier upper part is empty. If \(D\ne\varnothing\), define on that part \[(K,A_Kz)\longmapsto \bigl(K\cap(J\cup D),A_{K\cap(J\cup D)}z\bigr).\] The new type properly contains \(J\), uses only new colors in \(D\), and is a proper subset of \(T\). Hence the image is an earlier upper extension. This order map is below the identity and retracts onto the subposet of extensions whose types are contained in \(J\cup D\). That subposet has maximum \((J\cup D,A_{J\cup D}z)\), which is itself earlier. Thus the earlier upper part is contractible.

The lower part at nonzero height. Since equal negative heights place larger types first, every earlier lower pair has strictly smaller height. Write such a pair as \(Y=(R,A_Rw)\), where \(R\subsetneq J\) and \(w\in A_Jz\). Its newly open colors are \(U=J\setminus R\). The vector \(\delta=p_Y-p_X\) vanishes outside \(J\), equals \(m_i^q(w)-(p_X)_i\) on \(U\), and is harmonic on \(R\). If every one of these new open differences were nonnegative, nonnegative harmonic interpolation on \(R\) would give \(\delta\ge0\) and preclude a strict height decrease. Therefore the set \[N(Y)=\{i\in J\setminus R:m_i^q(w)<(p_X)_i\}\] is nonempty. All its colors are capped, since \(Y\) is allowed. Close the other newly open colors by defining \[ \rho(Y)=\bigl(J\setminus N(Y),A_{J\setminus N(Y)}w\bigr). \tag{93}\] This pair is well-defined on the coset \(A_Rw\): both the tests on \(J\setminus R\) and the larger coset are unchanged if \(w\) is replaced by another element of \(A_Rw\).

The boundary data on \(\partial J\) are capped. Indeed a neighbor outside \(J\) lies either in \(V\) or in \(\partial T\), where the cap is already known. The harmonic bound for the residue \((J,A_Jz)\) is exactly \(p_X\). Consequently (93) belongs to the strict sublevel \(\mathcal R_{<}(J,A_Jz;q)\). Viewed as a pair inside \(A_Tx\), it is allowed: its new open counts are strictly below \(p_X\le b\), and its old open colors \(V\) retain their original inequalities. Its harmonic difference from \(p_X\) is strictly negative on the nonempty set \(N(Y)\) and is a nonnegative interpolation of these differences on the remaining colors of \(J\). It is therefore coordinatewise nonpositive with strictly negative sum. Thus \(\rho(Y)\) remains an earlier lower pair.

For completeness, this closure respects actual coset inclusions when representatives differ. Suppose \[(R_1,A_{R_1}w_1)\le(R_2,A_{R_2}w_2).\] Then \(R_1\subseteq R_2\) and \(w_1\in A_{R_2}w_2\). On their shared open colors \(J\setminus R_2\) the counts agree, whence \[N(Y_2)=N(Y_1)\cap(J\setminus R_2)\subseteq N(Y_1).\] Writing \(K_a=J\setminus N(Y_a)\), we obtain \(K_1\subseteq K_2\). Moreover \(R_2\subseteq K_2\), so \(w_1\in A_{K_2}w_2\) and \[A_{K_1}w_1\subseteq A_{K_2}w_2.\] This proves that \(\rho\) is an order map. It lies above the identity because \(R\subseteq J\setminus N(Y)\).

Conversely, any pair in \(\mathcal R_{<}(J,A_Jz;q)\) is an allowed earlier lower pair in \(\mathcal R\): the same interpolation argument gives a strict decrease of height, and \(p_X\le b\) gives the outer inequalities. All its newly open colors pass the strict test, so \(\rho\) fixes it. Thus \(\rho\) is an order retraction onto exactly that strict sublevel. Since \(J\subsetneq T\), induction says that the earlier lower part is empty or contractible. This includes \(J=\varnothing\), when both lower posets are empty.

The zero-height case. If \(u_X=0\), the pair lies in the special closed sublevel. All open colors have \(m_i^q(z)=c_i\) and are capped. An upper extension retains \(p_X=b\), so its height remains zero and its larger type makes it later. There is no earlier upper part. Every allowed lower pair is earlier: its height is either negative, or zero with a smaller type.

The lower part is exactly \(\mathcal R_{\le}(J,A_Jz;q,I,c)\). Its exterior hypotheses follow from the original ones together with the equalities and caps on \(V=T\setminus J\): these sets supply all of \((I\setminus J)\cup\partial I\). If \(J=\varnothing\), those hypotheses contradict the rank-zero case. Hence \(J\ne\varnothing\), and induction makes this earlier lower part nonempty and contractible.

Sources. We have shown that every earlier lower and upper part is empty or contractible. Lemma 43 therefore makes each component contractible with exactly one source. The zero-height case has a nonempty earlier lower part, so a source must have \(u_X\ne0\). The upper calculation then forces \(D=\varnothing\), and the strict lower sublevel for \((J,A_Jz)\) must be empty.

These facts give all the hypotheses of a candidate in Proposition 30, for the fixed ambient coset \(A_Tx\) and cap \(q\). The colors \(V\) are capped by definition. Equation (91) gives \[-(Cp_X)_i=-(C_Tu_X)_i>0\quad(i\in V), \qquad (Cp_X)_J=0.\] Finally consider any representative \(w\in A_Jz\) capped on \(T\). If \(m_j^q(w)<(p_X)_j\) for some \(j\in J\), the pair \((J\setminus\{j\},A_{J\setminus\{j\}}w)\) would belong to the strict lower sublevel. Its emptiness therefore proves \[m_j^q(w)\ge(p_X)_j\qquad(j\in J)\] at every such capped representative, as required. For \(J=\varnothing\) this condition is vacuous.

Proposition 30 gives at most one source in \(\mathcal R\). A nonempty \(\mathcal R\) has a first vertex in the filtration and hence a source. It therefore has one component and is contractible. The special closed sublevel was already shown to be nonempty. This completes both inductions. ◻

Proof. The weighted layer counts of Section 5 have only finitely many possible values in a bounded interval, uniformly over the level. Indeed, for any fixed unfolded label the count is a nonnegative integer combination of finitely many fixed positive weights, up to a fixed positive proportionality factor. A bound on the count bounds each integer coefficient. This proves the asserted finiteness without requiring the real weights to lie in a lattice.

If \(\sum_i p_i^k(P)\le B\), every exterior coordinate \(m_i^k(x)=p_i^k(P)\) for \(i\notin T\) lies in \([0,B]\). There are therefore only finitely many exterior vectors. For a fixed type \(T\), (94) determines the interior vector uniquely. There are only finitely many spherical types. Consequently there are only finitely many possible primary vectors with sum at most \(B\), uniformly over \(P\) and \(k\). The secondary integer in (100) lies between \(-|S|\) and \(|S|\).

The set of possible level blocks thus has only finitely many predecessors below any one of its members. Adjoin the zero block if necessary. It is the least block: a zero primary vector has all signs zero, whereas every nonzero primary vector has positive sum. Assign to each block its rank \(n(\mathsf b)\in\mathbb Z_{\ge0}\) in this order, with \(n(0)=0\). For a finite-support sequence, the ordinal \[\omega^N n(\mathsf b_N)+\omega^{N-1}n(\mathsf b_{N-1}) +\cdots+\omega n(\mathsf b_1)+n(\mathsf b_0) <\omega^\omega\] has exactly the prescribed highest-level-first comparison. The block sequences are therefore well ordered. A well-ordered refinement of each fiber is again a well-order. Such a refinement may also be obtained from an enumeration of words in the finitely generated group and the finitely many types.

It remains to examine comparable ties. For every \(i\in T\), the frame and (94) give \[ p_T^0(P)\ge C_T^{-1}\mathbf1>0. \tag{101}\] All signs on positive coordinates at level zero are \(-1\), so the secondary value at that level is exactly \(-|T|\). Suppose \(A_Tx\subseteq A_Ux\) and all nonnegative primary vectors agree. If \(T\ne U\), their level-zero secondary values differ. If \(T=U\), the two nested cosets are equal. Hence distinct comparable vertices are never fully tied. Negative levels may distinguish further incomparable cosets, but their omission cannot change this conclusion or the well-ordering argument. ◻

Proof. Let \(P'=A_Ux\) with \(T\subsetneq U\) spherical, and put \(H=\max_{i\in U\setminus T}H_i\). At levels above \(H\), the coordinates at every added color and all of its neighbors vanish. The \(T\)-rows of \(Cp^k(P)\) vanish by harmonicity, and the added rows vanish by this incidence condition. Formula (97) shows that the vectors agree above \(H\). All added signs are zero there, so the secondary values agree as well.

At level \(H\), an added color with \(H_i<H\) has zero coordinate, zero row, and zero sign. Suppose first that no added color is active. An added color with \(H_i=H\) has row at most zero. Thus all rows on \(U\) are nonpositive, and (97) gives a nonnegative change of vector. If any such row is negative, the positive diagonal of \(C_U^{-1}\) makes this change nonzero, and the sum increases strictly. If all these rows vanish, the vectors agree at \(H\). Each added color with \(H_i=H\) then has positive coordinate and sign \(+1\). There is at least one such color. Since the signs for the two cosets agree whenever their vectors at and above \(H\) agree, the secondary value increases strictly. In either case \(P'\) is later.

If all added colors are active, the same argument reverses the inequalities. The rows at level \(H\) are nonnegative, so the vector change is nonpositive. A nonzero row decreases the sum strictly. If all rows are zero, every added color with \(H_i=H\) has sign \(-1\), and their nonempty contribution decreases the secondary value strictly. Hence \(P'\prec P\). ◻

Proof. Let \(G\) be a connected component of this subset, and choose the highest level \(k\ge0\) having a positive coordinate on \(G\) or on one of its neighbors. Such a level exists by the frame. For a vertex of \(T\) the full \(C\)-row is zero. At an active vertex whose incidence maximum is less than \(k\), its coordinate and neighboring coordinates are all zero at \(k\). At an active vertex with incidence maximum \(k\), the full row is nonnegative by (103). Therefore \[ (Cp^k(P))_G\ge0, \qquad C_Gp_G^k(P)\ge0. \tag{104}\]

The vector \(p_G^k(P)\) is nonzero. Otherwise a positive incident exterior coordinate, whose existence would be forced by the choice of \(k\), would give a negative full row at its adjacent vertex of \(G\). Moreover, a zero coordinate of \(p_G^k(P)\) cannot neighbor a positive coordinate in \(G\), since its \(C_G\)-row would be negative. Connectedness implies \(p_G^k(P)>0\).

Apply Lemma 7. Either \(C_G\) is positive definite, or \(C_Gp_G^k(P)=0\). In the latter case the full rows equal minus the exterior incidence: \[(Cp^k(P))_i=-\sum_{j\notin G}b_{ij}p_j^k(P) \quad(i\in G).\] Together with (104), this forces every displayed sum to vanish. There is no positive exterior neighbor, and the full rows are zero. Since \(p_G^k(P)>0\), the set \(G\) is exactly one positive-support component of \(p^k(P)\).

The level \(k\) cannot be zero: the frame lies outside \(G\subseteq S\), has positive level-zero coordinate, and neighbors every vertex of \(G\). By the maximality of \(k\), all higher coordinates vanish on \(G\cup\partial G\). Thus \(G\) satisfies every condition in (99), and all its signs at level \(k\) are \(+1\). Each active vertex in \(G\) would now have \(H_i=k\), positive coordinate, and zero full row, contradicting its active condition. Hence there are no active vertices in \(G\), so \(G\subseteq T\). This contradicts the assumption that \(C_G\) is not positive definite, because \(T\) is spherical. We conclude that every component \(G\) has positive-definite Cartan matrix, proving the assertion. ◻

Proof. If \(\operatorname{Act}(P)=\varnothing\), there is no earlier upper extension by Lemma 47. Otherwise put \(K=T\cup\operatorname{Act}(P)\), which is spherical by Lemma 48. On the earlier upper part the map \[A_Ux\longmapsto A_{U\cap K}x\] is an order map pointwise below the identity. Its image is still in the earlier strict upper part: \(U\cap K\) contains an added active color and adds only active colors. The map is a retraction onto the upper extensions adding only active colors, and that image has largest element \(A_Kx\). The order homotopy observation in Section 8 proves contractibility. ◻

Proof. Let \(Q=A_Rz\subsetneq P\) and \(V=T\setminus R\). Test the comparison at the level \[H=\max\left( \{q_i:i\in V\}\ \cup\ \{k\ge0:m_i^k(z)>0\text{ for some }i\in V\} \right).\] This is a finite maximum over a nonempty set. Above it all open coordinates of both vectors vanish. Formula (98) gives equality of the full vectors there, and the removed colors have zero signs, so the secondary values also agree.

Suppose first that \(E(Q)=\varnothing\). At level \(H\), a color with \(q_i<H\) has prescribed value zero and actual value nonnegative. A color with \(q_i=H\) has no actual data above \(q_i\) by the choice of \(H\); ineligibility therefore implies \(m_i^H(z)\ge p_i^H(P)\). Every open difference at \(H\) is nonnegative. If one is positive, the full vector has a strictly larger sum by (98).

If all open differences vanish, no color with \(q_i<H\) has actual data at \(H\). In particular \(H\) is the prescribed maximum of at least one open color; it could not have arisen solely from actual data above a prescribed maximum. The open colors with \(q_i<H\) contribute zero signs at \(H\). Each with \(q_i=H\) has equal positive actual and prescribed coordinates, and ineligibility forces sign \(-1\). The full vectors now agree at and above \(H\), so the signs are common to the two cosets. The change in the secondary value is \[\sum_{i\in R}\eta_i^H(Q)-\sum_{i\in T}\eta_i^H(P) =-\sum_{i\in V}\eta_i^H(P)>0.\] Thus \(Q\) is later in this case as well. This proves the first assertion by contraposition, including the possibility of actual data above the prescribed maximum on a different open component.

If every open color is eligible, the actual caps imply \(H=\max_{i\in V}q_i\). Open colors with smaller \(q_i\) have zero actual and prescribed values at \(H\); all other open differences are nonpositive. A strict difference decreases the sum. If every difference vanishes, the nonempty set of open colors with \(q_i=H\) has sign \(+1\), and its removal strictly decreases the secondary value. Hence \(Q\prec P\).

For an earlier lower coset define \[ \rho(Q)=A_{T\setminus E(Q)}z. \tag{107}\] Since \(E(Q)\) is nonempty and contained in \(T\setminus R\), this is a proper lower coset above \(Q\). Its open colors are exactly \(E(Q)\) and remain eligible, so it is earlier. Changing the representative of \(Q\) by \(A_R\) does not change the displayed coset, since \(R\subseteq T\setminus E(Q)\).

We verify order preservation with representatives included. If \(Q_1=A_{R_1}z_1\subseteq Q_2=A_{R_2}z_2\), then all colors outside \(R_2\) have the same actual data at \(z_1\) and \(z_2\). Hence \[E(Q_2)=E(Q_1)\cap(T\setminus R_2).\] Putting \(K_a=T\setminus E(Q_a)\) gives \(K_1\subseteq K_2\) and \(R_2\subseteq K_2\). Moreover \(z_1\in Q_2\subseteq A_{K_2}z_2\). Thus \(A_{K_1}z_1\subseteq A_{K_2}z_2\), as required. Finally, if every open color of \(Q\) is eligible, then \(T\setminus E(Q)=R\) and \(\rho(Q)=Q\). So (107) is the stated order retraction. The order homotopy observation in Section 8 identifies the homotopy type of the earlier lower part with that of its all-eligible image. ◻

Proof. Consider first a connected component \(B\) with sign \(-1\) and highest level \(q_B\). Its boundary is outside \(T\), because \(B\) is an entire component of \(T\). It has a cap at \(q_B\): if some exterior neighbor had a positive coordinate at a level \(d>q_B\), the harmonic equation on \(B\) and positivity of \(C_B^{-1}\) would give a positive coordinate throughout \(B\) at that level, contrary to its definition. On this boundary the actual data \(m^d(x)\) equal \(p^d(P)\). Their harmonic extension on \(B\) therefore equals \(p_B^{q_B}(P)\) at the test level. Eligibility is exactly a cap followed by the strict inequality against this harmonic vector. These are all the hypotheses of the strict residue sublevel.

Now take a positive-sign component, with support component \(I\) at level \(q\). Condition (99) gives \(q>0\), a positive vector \(c_I=p_I^q(P)\), and vanishing of all higher coordinates on \(I\cup\partial I\). The coordinates on \(\partial I\) at level \(q\) are zero by the definition of a support component. Hence \((Cp^q(P))_I=0\) also gives \(C_Ic_I=0\).

Every component of \(T\) meeting \(I\) lies wholly in \(I\), by harmonic positivity on that component at level \(q\). Its highest level is \(q\) because the higher coordinates vanish on \(I\), and it has the same positive sign. We must group all these components, giving \(B=T\cap I\). If a vertex \(j\in T\setminus I\) neighbored \(I\), then \(p_j^q(P)=0\), since otherwise it would lie in the same support component. An edge to a positive coordinate in \(I\) would give \((Cp^q(P))_j<0\), contradicting harmonicity on \(T\). Therefore \[ \partial I\cap T=\varnothing, \qquad (I\setminus B)\cup\partial I \subseteq\widehat S\setminus T. \tag{108}\] All exterior values required by the special closed sublevel are thus actual, fixed values on the entire coset \(A_Tx\). They are capped at \(q\) and equal the zero-extended vector \(c\) at that level. The block \(B\) is spherical and its harmonic prescription is \(c_B\). It cannot equal \(I\), because a positive-definite principal matrix cannot have the positive null vector \(c_I\). Eligibility on \(B\) is exactly the closed inequality in Proposition 44. This verifies every special closed hypothesis, including when several disconnected components of \(T\) meet the same \(I\).

These blocks partition \(T\) into unions of its connected components. The presentation and the standard embeddings give \(A_T=\prod_B A_B\). For \(z\in A_Tx\), write uniquely \(zx^{-1}=\prod_B g_B\) with \(g_B\in A_B\). A subcoset \(A_Rz\) corresponds to the tuple \[\bigl(A_{R\cap B}g_Bx\bigr)_B.\] Coset equality and inclusion are coordinatewise under this correspondence. Operations in other blocks leave every multiplicity in \(B\) unchanged, so the eligible-color conditions in \(B\) are precisely those for \(g_Bx\) in the displayed block coset. The boundary data just verified lie outside \(T\): for a strict block this follows because it is a full component of \(T\), and for a closed block it is (108). They are therefore fixed under every operation in \(A_T\), including operations in the other blocks. Opening nothing in a block gives its single largest element \(A_Bx\); opening somewhere gives the stated proper residue sublevel. The requirement \(R\subsetneq T\) deletes just the all-largest tuple. This proves the product description. Its contractibility assertions follow from Proposition 44. ◻

Proof. We first record the product fact in the topology of order complexes. If \(\mathcal P\) and \(\mathcal Q\) are nonempty posets with contractible realizations, projection \(\mathcal P\times\mathcal Q\to\mathcal P\) has lower fiber \(\mathcal P_{\le p}\times\mathcal Q\) over \(p\). The order map \((p',q)\mapsto(p,q)\) is pointwise above the identity and retracts that fiber onto \(\{p\}\times\mathcal Q\). The fiber is contractible. Quillen’s Theorem A, in the form recalled in Section 2, shows that the product poset has contractible realization. Iteration proves this for any finite number of nonempty contractible factors. This argument does not require identifying an infinite order-complex realization with an ordinary topological product.

Discard indices for which \(\mathcal E_a\) is empty, since their augmented factor has only its largest element. If no indices remain, the deleted product is empty. Otherwise cover its realization by the subcomplexes \(X_a\) consisting of chains whose \(a\)th coordinate lies in \(\mathcal E_a\). They cover every chain: its largest vertex has a nonlargest coordinate, and every earlier vertex has a nonlargest coordinate at that same index. For a nonempty set \(J\) of indices, the intersection \(\bigcap_{a\in J}X_a\) is the realization of \[\left(\prod_{a\in J}\mathcal E_a\right) \times \left(\prod_{a\notin J}\mathcal E_a^+\right).\] It is nonempty and contractible by the product fact: the required factors are contractible, and each augmented factor has a largest element.

There are finitely many subcomplexes in this cover. Their union is contractible by induction on their number. Indeed the intersection of the last subcomplex with the preceding union has a cover by the corresponding intersections; all of its nonempty finite intersections are again nonempty contractible. Induction makes both the preceding union and this intersection contractible. Gluing the last contractible subcomplex along the contractible intersection preserves the homotopy type of a point, since subcomplex inclusions are cofibrations. This completes the proof. ◻

Proof. For type \(T=\varnothing\) the lower part is empty. Otherwise Lemma 51 retracts it onto the all-eligible part. Lemma 52 identifies that part with the product in Lemma 53, whose factors satisfy that lemma’s hypotheses. The conclusion follows. ◻

Proof. If \(S=\varnothing\), the poset has a single vertex. Otherwise use the well-order of Lemma 46. Propositions 49 and 54 show that the earlier strict upper and strict lower parts at each vertex are empty or contractible. Their join is the earlier attaching link. The empty-part convention and Lemma 43 therefore show that each component of the final order complex is contractible.

Finally \(\mathcal D\) is connected. The singleton cosets \(x\) and \(\sigma_sx\) both lie in \(A_{\{s\}}x\), since rank-one types are spherical. Left generator steps therefore connect all singleton cosets. Every other coset contains a singleton coset and hence joins this same component. The preceding componentwise conclusion proves contractibility of \(|\mathcal D|\). ◻

Proof of Theorem 1. Lemma 5 gives a homotopy equivalence from the universal cover of the full Artin Salvetti complex to \(|\mathcal D|\). The latter is contractible by Theorem 55, so the universal cover is contractible. Since covering maps induce isomorphisms on homotopy groups in degrees at least two, the specified Salvetti complex is aspherical. ◻

Proof. When \(S=\varnothing\), both \(A\) and \(A^+\) are trivial and \(k=0\), so all three assertions hold. Assume that \(S\ne\varnothing\).

The complex \(X(W,S)\) has dimension at most \(|S|\), and it is a \(K(A,1)\) by Theorem 1 and (5). If \(A\) had a nonidentity element of finite order, it would contain a subgroup \(P\cong C_p\) for some prime \(p\). The corresponding covering space \(\widetilde X(W,S)/P\) would be a \(K(P,1)\) of dimension at most \(|S|\). This is impossible: the standard two-periodic resolution of \(C_p\), whose cochain maps with trivial \(\mathbb Z\)-coefficients alternate between zero and multiplication by \(p\), gives \[H^{2q}(C_p;\mathbb Z)\cong\mathbb Z/p\mathbb Z\qquad(q\ge1),\] whereas the cohomology of a CW complex of dimension at most \(|S|\) vanishes in higher degrees. This proves (i).

Let \(S_1,\ldots,S_r\) be the connected components of the Coxeter diagram. All cross labels are \(2\), so the presentation gives \[A\cong\prod_{j=1}^r A_{S_j},\] as in (9). If \(S_j\) is spherical, Deligne’s center theorem for a finite Coxeter system with nonempty connected diagram gives \(Z(A_{S_j})\cong\mathbb Z\) (Deligne 1972, Theorem 4.21). If \(S_j\) is not spherical, its connectedness means that \(A_{S_j}\) has no spherical factor. Theorem 1 applied to the submatrix on \(S_j\) verifies the \(K(\pi,1)\) hypothesis of Jankiewicz–Schreve’s theorem (Jankiewicz and Schreve 2023, Theorem 3), which gives \(Z(A_{S_j})=\{1\}\). Taking centers of the direct product proves (ii).

For (iii), Dobrinskaya’s equivalence (Dobrinskaya 2002, Theorem 2 and the following paragraph), reproved by Ozornova (Ozornova 2017, sec. 5, especially Conjecture 5.9 and Theorem 5.10), states that the natural map \(BA^+\to BA\) is a homotopy equivalence exactly when the \(K(\pi,1)\) conjecture holds for \(A\). Theorem 1 supplies that condition. ◻

Proof. This is precisely Boyd’s sequence of diagrams (Boyd 2020, Introduction, Definition 6.1 and Remark 6.2). Deleting \(\sigma_1\) from the initial diagram gives the diagram for her \(A_0\), while \(\sigma_1,\ldots,\sigma_n\) form her type-\(A\) subdiagram. In particular, the new vertices have no other edges to the fixed core, and our parameter \(n\) is Boyd’s parameter. The inclusions \(s_n\) are the standard parabolic inclusions.

Every \(\Sigma_n\) is finite, so Theorem 1 proves the \(K(\pi,1)\) conjecture for every \(A_n\) in this sequence. This is exactly the hypothesis of Boyd’s Corollary B, whose conclusion is the displayed stabilization map and the stated constant-coefficient range. ◻