Harmonic heights and the Artin K(pi,1) conjecture

Proof. Define

Formula (2.6) 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

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 2.1, write \alpha=\beta h uniquely with h an element of the Artin group on T. Under this correspondence, every face formula becomes

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 (2.8) is its universal-cover face poset. Its realization is contractible by Theorem 2.3. 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

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

which proves the first assertions, including uniqueness.

For the last assertion, every real vector x on G satisfies

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

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 \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,

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

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 (3.4); 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 (3.3) 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

Closing (3.4) 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 (3.5), 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 (3.8), 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

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_\C(P_\alpha)=Q_\alpha\langle-2\rangle and D_\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 (3.8) proves (3.9). The same trace identifies D_\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 (3.2) 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 3.2 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 (3.11) shows that their product on a folded vector is the original reflection (3.1). 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 [4, Chapter 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 (3.12). 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,

in degree zero, and M_\alpha\langle-2\rangle in degree one. Split the middle factor as e_\alpha Ze_\alpha=\C e_\alpha\oplus\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

after rescaling one term if needed. Their evaluation twists agree. Since T_{P_\gamma}=B_\gamma, conjugation gives

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 3.6, and their products are inverse by Lemma 3.5. Moreover

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 3.4, 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 3.6. 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 (3.6), T_H is right tensoring by \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 (3.16) applied to P_{t,Y}, semisimplicity gives the canonical identification

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

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 (4.2), 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 (4.3).□

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 (4.4).

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

Projection onto the token source summand A\langle2\rangle[1] is a chain map with kernel the cone of \id_A. It is therefore a homotopy equivalence, proving (4.6) 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 4.4 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\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 (4.5); outside it the multiplicities are unchanged.

At a vertex of color i, the degree-d test of F_iM is zero by (4.6), 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

Orthogonality implies that F^{-1}X has levels at least -1. If B has levels at least 1, then FB does also, and

It follows that F^{-1}X has levels at most 0. This proves the inverse bound; shifting proves it for every pure level. Formula (4.8) follows by applying (4.6) 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

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

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 (4.10). 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 (4.10) gives

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 (4.11) directly. Set A=e_\beta Z_0, C=e_\alpha Z_0, and choose arrows

The two test complexes are

Let \pi:e_\beta Y\to A\langle2\rangle[1] be the token projection of (4.7). 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 (4.11) with

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

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 4.8 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 4.4 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 3.4, 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 4.6 therefore gives the negative bound.□

Proof. Proposition 3.7 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 4.1.□

Proof. Let \mathbf1 be the unit label. Natural label equivariance gives

For an object X\in\mathscr C and a simple label H, rigidity and semisimplicity give

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 (5.5) and summing over the placements of a fixed level gives (5.2).

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 \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 (5.5) to the cohomology in the two extremal total degrees from Lemma 4.4. 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 (5.2) 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 4.6. 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 \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 (5.3).

Finally, the signed calculation in Lemma 4.5 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 (3.11) folds this to (5.4).□

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

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

Its first term still has levels at most q-1, giving (5.7) by the long exact sequence. This argument does not require that G be a reduced word.□

Proof. Negative preservation follows by applying Lemma 5.4 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 4.5, 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 (4.8) and Lemma 4.4 give

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 4.9.

Put M_L=Y_L^q(x) and N_L=G_wM_L. The latter has levels q-1,q. Its truncation triangle is

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 (5.10). Write its end terms as A_L and C_L, so that the triangle is

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:

in the shifted heart. By Lemma 4.1, 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 (5.11) is

Lemma 5.1 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 (5.9) 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

There is no contribution here from higher layers of the original E_L(x) outside the operated region: Lemma 5.4 identifies its level-q family after G_w with the level-q family of G_wM_L, exactly the high family used in (5.12).

The representative obtained by G_w belongs to A_Jx and retains the joint cap, by Lemma 5.5. Applying the universal hypothesis to (5.12), and using wp=p, gives

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

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, (5.13) implies

The set \mathcal V is finite. If e lay outside its convex hull, a separating linear functional would contradict (5.14). 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_{\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 3.3, adjoint duality on finite graded bimodules satisfies

It is a contravariant equivalence, preserves additive radicals, and sends

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 3.7. As the dual minimal complex is still minimal, the preceding specialization proves (6.2).

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

which proves row invariance. Apply (6.2) to (ha)^{-1}=a^{-1}h^{-1}, for a\in A_J, to obtain column invariance.□

Proof. The inverse bimodule complex is

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

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

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

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

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 (6.5). 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 4.5 is (6.4). Equivalently, it follows by canceling the token–unit component of D_\alpha e_\alpha. Its total shift is -2+1=-1, so in particular

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

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 (6.7) forces it to be zero. The middle degree is r+q+1, proving (6.8).

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 5.2, 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 6.5 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 5.5 applied with spherical set T. Replace

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 6.4, 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 6.5.

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 (6.3) 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 6.7 and the Euler identity give

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 6.5 gives

Subtracting yields

Finally, n is supported on T, and (Cp)_T=-a. Equation (6.12) therefore turns the right side into the expression in (6.13).□

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 (6.14) 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 3.4 says that reduced-word lifts remain simple after unfolding, and that both descent sets become their full color preimages. Thus the unfolded factors satisfy (6.14) 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 (6.15) 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

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

For a bounded right complex R, use the height

from Lemma 4.8, 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 (6.16) is positive, any vertex with highest cohomology gives, by the right version of Lemma 4.8, an extractable final atom of both unfolded positive words. Lemma 6.9 supplies a common extractable original color s. Write h_i=g_i\sigma_s, and tensor (6.16) on the right by B_{\sigma_s}^{-1}. This gives the same equality with g_i in place of h_i, while

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 6.2. 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 (6.13) then gives nonzero middle cohomology. It survives in actual total degree r+q+1>q, contradicting (6.7). 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 (6.7) and (6.13) give

Interchanging the candidates and repeating the argument gives

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

The sum of the left sides of (6.17) and (6.18) is therefore

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

In particular the nonnegative term n_T^{\mathsf t}a in (6.17) 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

By Lemma 6.3, 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 (6.11), and strict positivity of all its weights imply

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 (6.2) transposes these latter vanishings into vanishing of the block for h_0 in all degrees d\le-1. Together the two ranges give

Lemma 6.10 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

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,

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 (7.4).

For the remaining assertion, consider the full bigraded Hom complex \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 3.7, 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 (7.5). 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 (7.7) 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 \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

The only possible ordered layer pairs in its target are

A length-two entry is diagonal in the unfolded vertex and lowers the layer by one. For the second term in (7.9) 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 (7.7).

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 7.2. 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 7.2; the result is (7.8).□

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

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

Apply (7.6) 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 7.2. Subtracting gives

In particular, no middle-Hom vanishing for \mathcal S is required.

The left side of (7.11) is affine-linear in s, so the inequality holds throughout the convex hull. At its interior point e=m-c, it gives

On the other hand, symmetry of C, the null equation on I, and nonnegativity of y give

This is strictly negative if y_a>0 for any a\in\partial I. Consequently y|_{\partial I}=0, the two sides of (7.12) 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 (7.11), whose value at e is now zero, forces (Cy)_j=0. It follows that

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 \R^0; the same argument applies with no coordinate perturbations.□

Proof of Proposition 7.1. Assume (7.2) holds at every capped representative. Since \partial J\subseteq(I\setminus J)\cup\partial I, Lemma 5.5 supplies a representative capped on J and hence on all of I. We shall move only among such representatives. All the exterior data in (7.1) 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 (7.1), 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 5.2, 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 5.6 therefore applies with its strict, universally quantified hypothesis. It shows that

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 (7.14) is the ordinary real interior supplied by Proposition 5.6, even when folded multiplicities are not integers. For J=\varnothing, it means e=0 in \R^0.

Thus Lemma 7.4 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 (7.2) supplies (7.14) 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

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,

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 7.3 and the condition at d-1 give

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 5.2 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 (7.17), 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 5.5. 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 (7.16) 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 4.2 then identifies

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 4.5, the vanishing of the old socle on J implies the vanishing of the new socle on J in (7.18).

The new representative is capped, so Lemma 7.4 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 (7.16) 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

Let t_j^k\ge0 be the folded weighted top multiplicity of \mathcal Y^k_{\mathrm{old}} in color j. Formula (4.5), together with (7.18), says

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 (7.18) 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 (7.17) can be used again after the move, with the new \mathcal M, by Lemma 7.3.

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 4.3, 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 4.5. 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 7.4 supplies (7.16) 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, (7.14) says m=c, and Lemma 7.4 already contradicts the boundary layer. Thus the dimension-zero case is included. The contradiction proves the proposition.□

Proof. A chain containing v consists of a chain below v, the vertex v, and a chain above v. Every element of the first chain precedes every element of the second in the poset order. Restricting all other vertices to those earlier than v proves the asserted join description. The join of two nonempty contractible complexes is contractible. Together with the stated conventions for empty factors, this proves the last assertion from the first one.

Index the vertices as v_\alpha, \alpha<\kappa, and let K_\alpha be the induced order complex on \{v_\beta:\beta<\alpha\}. At a successor stage we attach the cone on the earlier link L_\alpha:

An empty link adds an isolated vertex and starts one component. A nonempty contractible link lies in a single component of K_\alpha. The inclusion of that link in its cone is a homotopy equivalence and a subcomplex cofibration. The homotopy extension property therefore makes the induced inclusion of the old component into the new one a homotopy equivalence. No other component changes, and two old components cannot merge. Thus the asserted description of components and their sources persists at successor stages.

At a limit ordinal \lambda, the complex is the union of the earlier subcomplexes, with the CW topology. We recall why compact images cause no difficulty in this union. A compact subset of a closure-finite CW complex lies in a finite subcomplex. Otherwise it meets infinitely many open cells; choose one point in each of a countable set of these cells. Every subset of the chosen points meets each closed cell in a finite set, by closure-finiteness, and hence is closed in the whole complex by the weak topology. The chosen points would therefore form an infinite closed discrete subspace of the compact subset, a contradiction. The closures of the finitely many cells met by a compact subset form the required finite subcomplex.

Consequently a path or a sphere map in K_\lambda has image in some K_\alpha with \alpha<\lambda: only finitely many vertex indices occur in its finite carrier. Paths cannot merge previously distinct components at the limit. Every sphere map within a component is null-homotopic in an earlier component, which is contractible by induction. Each component of K_\lambda is thus weakly contractible and has the same unique source as any of its earlier vertices. Components are CW subcomplexes, so the CW Whitehead theorem makes them contractible. Transfinite induction, including the final union, proves the result.□

Proof. We first dispose of nonemptiness and the rank-zero convention. If T=\varnothing, there is no proper subset J in (8.1), 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 7.1.

Now let T\ne\varnothing and assume the closed exterior hypotheses. Lemma 5.5 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 7.1, 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

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

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

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

The rows (C_Tu)_J vanish. Set

If K\setminus J avoids D, all its rows in (8.5) 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

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

is nonempty. All its colors are capped, since Y is allowed. Close the other newly open colors by defining

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 (8.6) 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

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

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

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 8.1 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 6.2, for the fixed ambient coset A_Tx and cap q. The colors V are capped by definition. Equation (8.4) gives

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

at every such capped representative, as required. For J=\varnothing this condition is vacuous.

Proposition 6.2 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, (9.1) 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 (9.7) 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\Zbb_{\ge0} in this order, with n(0)=0. For a finite-support sequence, the ordinal

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 (9.1) give

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 (9.4) 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 (9.4) 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 (9.10). Therefore

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 3.1. 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:

Together with (9.11), 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 (9.6), 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 9.3. Otherwise put K=T\cup\operatorname{Act}(P), which is spherical by Lemma 9.4. On the earlier upper part the map

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

This is a finite maximum over a nonempty set. Above it all open coordinates of both vectors vanish. Formula (9.5) 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 (9.5).

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

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

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

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 (9.14) 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 (9.6) 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

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 8.2. 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

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 (9.15). 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 8.2.□

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 ath 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

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 9.7 retracts it onto the all-eligible part. Lemma 9.8 identifies that part with the product in Lemma 9.9, 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 9.2. Propositions 9.5 and 9.10 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 8.1 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 1.1. Lemma 2.4 gives a homotopy equivalence from the universal cover of the full Artin Salvetti complex to |\mathcal D|. The latter is contractible by Theorem 9.11, 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 1.1 and (2.5). 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 \Zbb-coefficients alternate between zero and multiplication by p, gives

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

as in (2.9). If S_j is spherical, Deligne’s center theorem for a finite Coxeter system with nonempty connected diagram gives Z(A_{S_j})\cong\Zbb [12, Theorem 4.21]. If S_j is not spherical, its connectedness means that A_{S_j} has no spherical factor. Theorem 1.1 applied to the submatrix on S_j verifies the K(\pi,1) hypothesis of Jankiewicz–Schreve’s theorem [21, Theorem 3], which gives Z(A_{S_j})=\{1\}. Taking centers of the direct product proves (ii).

For (iii), Dobrinskaya’s equivalence [13, Theorem 2 and the following paragraph], reproved by Ozornova [24, Section 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 1.1 supplies that condition.□

Proof. This is precisely Boyd’s sequence of diagrams [5, 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 1.1 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.□