A linear isoperimetric inequality is the large-scale face of negative curvature: for finitely presented groups it is equivalent to word hyperbolicity. Local comparison geometry demands more. The realization problem asks whether every hyperbolic group carries a geometric action, meaning a proper cocompact isometric action, on a proper CAT(0) space, or on a proper CAT(-1) space. A finite two-dimensional aspherical complex exists whose fundamental group answers both questions negatively.

There is a finite connected two-dimensional aspherical simplicial complex K and a constant C < infinity such that:

(i) every combinatorial edge loop of length l in the universal cover of K bounds a disk diagram with at most Cl triangular faces;

(ii) the group pi_1 K admits no geometric action on a proper complete CAT(0) space.

Consequently pi_1 K is word hyperbolic, and no finite simplicial complex homotopy equivalent to K admits a locally CAT(0) geodesic length metric inducing its given topology. No such model then admits a compatible locally CAT(-1) geodesic length metric either.

https://doi.org/10.1007/978-3-662-12494-9

The proof is an existence argument and yields no effective size for K. The filling constant may depend on the selected finite presentation.

The question reaches back to the birth of the subject. Gromov introduced linear isoperimetry as a group-theoretic form of negative curvature, then spent years seeking realizations by negatively curved spaces; the interview with Raussen and Skau records the distance between the two.

raussen-skau-2010.pdf

Linear filling became equivalent to word hyperbolicity in Gromov's original account, an equivalence set out in Bridson and Haefliger.

https://doi.org/10.1007/978-1-4613-9586-7_3

https://doi.org/10.1007/978-3-662-12494-9

The CAT(-1) realization question appears as Problem H11 in the 1999 problem list of Baumslag, Myasnikov, and Shpilrain.

baumslag-myasnikov-shpilrain-1999.pdf

Both the CAT(-1) and CAT(0) questions are open in Stark's 2025 survey and in the March 2026 preprint of Duchesne and Simon.

stark-2025.pdf

duchesne-simon-2026.pdf

Several routes do produce such actions. Finitely generated free groups act on trees; closed hyperbolic manifold groups act on hyperbolic space. A finite complex carrying a locally CAT(-1) or locally CAT(0) geodesic length metric inducing its topology has a universal cover with the matching global curvature bound, and the deck action is geometric. Cubulation gives a second route: after Sageev's construction of cube complexes from codimension-one subgroups and Haglund and Wise's theory of special cube complexes, Ollivier and Wise prove that random groups in the density model at density below 1/6 act freely and cocompactly on CAT(0) cube complexes with probability tending to one as relator length grows, generators fixed.

https://doi.org/10.1112/plms/s3-71.3.585

https://doi.org/10.1007/s00039-007-0629-4

https://www.ams.org/tran/2011-363-09/S0002-9947-2011-05197-4/S0002-9947-2011-05197-4.pdf

Dimension already obstructs. Brady and Crisp construct infinitely many torsion-free hyperbolic groups of geometric dimension two admitting no proper isometric action on a proper two-dimensional CAT(0) space, and a hyperbolic group with CAT(0) dimension two and CAT(-1) dimension three. Those conclusions constrain translation lengths in two-dimensional targets. The construction excludes geometric CAT(0) actions in every dimension, and allows the entire simplicial complex to change inside its homotopy type, since any compatible locally CAT(0) metric on such a model recovers a geometric action on the universal cover.

brady-crisp-2007.pdf

Random-presentation proofs of hyperbolicity combine counting for bounded diagrams with a local-to-global step. The treatment of repeated relator occurrences follows the multiplicity and constraint methods of Ollivier and of Antoniuk, Friedgut, and Luczak: within a fixed presentation, small labelled graphs are sampled first, larger labelled graphs whose required small graphs have already appeared are sampled next, and a shared prerequisite is charged once.

ollivier-2004.pdf

antoniuk-friedgut-luczak-2017.pdf

The obstruction to actions rests on the harmonic-energy framework of Gromov and of Izeki and Nayatani. Expansion in a finite graph can force a fixed point once the target obeys a nonlinear spectral inequality. The target hypothesis is essential: Izeki, Kondo, and Nayatani obtain random hyperbolic groups with fixed-point properties for target classes whose tangent-cone invariant is bounded below one, and Toyoda supplies such bounds for a prescribed finite family of geodesically complete CAT(0) spaces admitting proper cocompact actions. A hypothetical target and its action may be chosen after the presentation is fixed, so uniformity must come from elsewhere: bounds on every generator displacement and total-variation couplings of table matches. Sturm's inductive means convert those couplings into the averaging estimate used in the energy argument.

https://doi.org/10.1007/s000390300002

izeki-nayatani-2005.pdf

https://doi.org/10.4171/GGD/171

toyoda-2016.pdf

The construction fixes the combinatorics first; all randomness sits in the labels.

Identify the points of the projective plane over F_2 with the seven nonzero vectors of F_2^3, and each line with the three nonzero vectors of a two-dimensional subspace. Set A=e_1, B=e_2, C=e_3, D=e_1+e_2+e_3; no three of these four are collinear. For distinct points i,j the symbol ij denotes their common line.

The bipartite graph Gamma has a vertex X_i for every point, an extra vertex Y_i for each of A,B,C,D, and a vertex for every line. Both point representatives join every incident line, with a single exception: the edge Y_D--CD is omitted. The edge e_0 = X_D--CD is the loose edge. Every other edge of Gamma lies in the fan F_i at its point i, and at D the fan uses only the two lines other than CD. Each F_i is connected: K_{2,3} at A,B,C, K_{2,2} at D, a three-edge star at the remaining three points. With k_i the number of fan edges,

(k_i) = (6,6,6,4,3,3,3),  sum_i k_i = 31,  |E(Gamma)| = 32.

The graph has eighteen vertices, cycle rank fifteen, and the fan cycle ranks sum to seven. A core hexagon is the six-cycle through three noncollinear points, the three lines joining them, and one available representative over each point for which all six incidences exist; the all-X choice always qualifies. Any three successive edges of a core hexagon meet at most two fields of any one fan, and the loose field lies in no fan.

Fix an orientation of every edge. A field is an edge position in a template or a position in a polygon word; an assignment gives one signed letter to each field. A walk reads that letter with the orientation and its inverse against it. The fields are bookkeeping coordinates: all take values in one alphabet.

For n >= 1 set S_n = {a_1,a_1^{-1},...,a_n,a_n^{-1}}, N = 2n. These are formal signed letters; assignments need not be distinct or freely reduced. Put

f = 2-alpha,  p = 3-beta,  0 < alpha,delta < 1,  7alpha < beta < 1,

with further smallness imposed when the finite family of sampling tests is fixed. Seven mutually independent families of Bernoulli arrays follow.

(i) For each point i, an array indexed by S_n^{E(F_i)} includes every tuple with probability N^{f-k_i}; an included tuple with its type is a fan occurrence b.

(ii) An array indexed by S_n^{E(Gamma)} marks every full core tuple with probability N^{p-7f-1}. A top occurrence a is a marked tuple whose restriction to every F_i is an included fan occurrence; write a -> b when b fills one of the seven fan slots of a.

(iii) For each unoriented nonbacktracking two-edge path tau in Gamma, fix one order and direction and give tau a four-field array whose first two coordinates are the two core fields and whose last two are new; include each row with probability N^{-2-delta}. Its relator reads tau, with orientation signs, followed by the two new letters around a closing path.

(iv) A six-field array includes each word of S_n^6 with probability N^{-3-delta}; its word is a relator.

Every included row of (iii) and (iv) is imposed, not only rows matching some top occurrence. These rows are decorations, of length 2h with h in {2,3}. Opposite path directions cause no ambiguity: reversing a convention inverts or reorders coordinates and keeps the Bernoulli law. There are ninety tables in (iii), since the eleven point-representative vertices give 10*binom(3,2)+binom(2,2)=31 two-edge paths and the line degrees 5,5,4,5,5,5,3 contribute 59.

ollivier-2004.pdf

Every exponent in these laws is negative; in particular p-7f-1 = -12-(beta-7alpha) < 0, so each is a valid probability. Distinct numerical tuples in a fixed table are independent; top occurrences are correlated because they may share fan slots, while top marks stay independent. The normalization matches the expected counts: N^{k_i} candidate tuples per fan type give N^f included fans, and a fixed core tuple needs its mark and seven fan entries, so the expected number of top occurrences is

N^{32} * N^{p-7f-1} * prod_i N^{f-k_i} = N^p.

Each square table has expected size N^{2-delta} and the six-letter table N^{3-delta}. The exponents sit just below 2 for fans and squares and just below 3 for tops and hexagons; the inequality beta > 7alpha leaves a positive margin once shared fan costs are counted once.

The letter presentation has generators a_1,...,a_n. Every included fan and top occurrence imposes the labels of all closed walks in its graph, and the decoration words are imposed. A fixed spanning tree in each connected graph supplies a finite basis of closed walks, so the presentation is finite.

For diagram arguments use a specific finite two-dimensional CW complex P. Begin with the letter bouquet at a vertex o. For each fan occurrence b of type i, attach the cone on F_i along its labelled map; the apex is a new vertex, also written b, and each vertex v of F_i yields a radial edge r_{b,v} from b to o. Every fan edge becomes a triangle with one letter side and two radial sides. Radial edges may share endpoints but keep their distinct labels v. Fan cones already fill the fan cycles.

To impose the remaining relations of a top occurrence a, subdivide the edge D--CD of the Fano incidence graph once and call the result B. It has fifteen vertices and twenty-two edges, cycle rank eight, which matches 15-7. Map a point vertex i to the apex of the fan in slot i of a, and map every line vertex and the subdivision vertex to o. Each ordinary incidence at i maps to the radial edge indexed by its line; the subdivided exceptional incidence maps to r_{b_D,X_D} followed by the loose letter edge, oriented from X_D to CD. Attach the cone on this mapped graph with a new apex, obtaining the top triangles, then attach every decoration polygon. All attachments are finite and P is two-dimensional.

Attaching the cone on a connected graph H mapped into a connected complex changes pi_1 by the normal closure of the image of pi_1 H and adds no generator; this is van Kampen, or elimination of all but one new radial edge along a spanning tree of H, and needs no injectivity of the vertex map. After the fan triangles replace each fan edge by the radial path through the fan apex, close core walks collapse excursions through X_i or Y_i and leave a closed walk in the mapped macro graph; at D the only unreplaced edge is the loose edge, joined to r_{b_D,X_D} in exactly the subdivided incidence, and every macro turn expands back through a connected fan. The two relation sets therefore have the same normal closure modulo the fan relations, so P has the fundamental group of the letter presentation. Gamma and each fan are bipartite, so closed-walk words have even length, and the decorations have even length four or six; sending every generator to the nonidentity element of C_2 respects all defining relations and is surjective, giving a quotient C_2.

The cone description records occurrences, not merely relator words. A fan shared by several tops has one cone, a top has its own, a decoration has its own disk; that distinction drives the multiplicity count.

Define the finite master pattern M as one top occurrence with its seven fan slots together with, for each of the ninety two-edge paths, one matching row of its square table: the two core coordinates of that row are identified with the core fields and its two closing coordinates are new. The six-letter array is excluded. A match is an assignment to the fields of M for which every required entry is present, including the top mark. The set of matches is Omega_n, carrying its uniform law when nonempty; its 32+2*90 = 212 fields count shared core fields once. A set of fields is admissible if it is either at most three distinct core fields with at most two in any fan (the loose field exempt from all quotas), or at most two fields of a single square decoration of M.

A match carries a labelled realization of the core graph and the closing paths of its squares in the group: fix one anchor and label a vertex by the word read from the anchor along any path. All closed-path labels are trivial by the presentation equivalence and the square relators, so the group-valued labelling is path-independent, and any action with a basepoint turns it into a configuration of orbit points.

The combinatorial estimate now yields asphericity and linear filling. Fix small positive alpha,beta,delta with beta > 7alpha, and set f = 2-alpha, p = 3-beta. An occurrence is an included tuple together with its table type; copies of a cell in a diagram need not be distinct occurrences, and every constant may depend on the fixed template and parameters but not on n. The claim is that as n tends to infinity the probability tends to one that P is aspherical and its letter presentation satisfies a linear disk isoperimetric inequality.

The estimate bounds the number of faces in a bounded piece of a normalized diagram by the number of its exposed sides. The occurrence exponents read 2-alpha for fans, 3-beta for tops, and h-delta for decorations of length 2h; the local incidence geometry compares these against 2, 3, and h, and the differences alpha, beta-7alpha, delta leave a strict margin. A probability bound is first obtained for every bounded collection of distinct occurrences; disk and sphere pictures are then normalized so that the bound governs repeated copies of their cells; planar separators finally extend it to all pictures. Every fan slot of a top occurrence is retained in a collection even when a diagram uses only some of those slots.

For a finite collection of top, fan, and decoration occurrences, its field positions are the loose field of each top, the k_i fields of each fan occurrence of type i, and all fields of each decoration. A fan field counts once even when the same fan occurrence fills slots in several tops. The unsigned coincidence rank R is the number of positions minus the number of distinct underlying generators represented there, so a letter and its inverse share an unsigned value.

Lemma. For each fixed integer M, with probability tending to one the following holds at once for every collection of at most M distinct occurrences that contains all fan slots of its tops. If the collection has s tops and u fans, then

R <= (p-7f)s + f u + sum_e (h_e - delta),

the sum over its decorations, 2h_e in {4,6} the length of decoration e.

Proof. Fix the table types, the incidence of fan slots, and a partition of the field positions into unsigned-value classes; finitely many choices for fixed M. With J the number of positions, an assignment to a pattern of rank R has at most C_M N^{J-R} realizations, since one generator per class and a sign for each position, the signs contributing a factor bounded by M. Restrict to choices whose occurrences are distinct in their tables, so all required Bernoulli trials are independent; their product has exponent (p-7f-1)s + sum_b(f-k_b) + sum_e(h_e-delta-2h_e), where a shared fan occurs once in the second sum. Because J = s + sum_b k_b + sum_e 2h_e, the expected number of realizations of the chosen pattern is at most C_M N^{(p-7f)s + f u + sum_e(h_e-delta) - R}. A failure of the displayed inequality makes this exponent negative; there are finitely many exponents for the fixed size bound, and summing expectations proves the claim.

An elementary picture is a finite collection of oriented polygons with orientation-reversing side pairings whose resulting closed oriented surfaces are spheres; vertex links are circles, with a separate source vertex for each link circle. An interior face maps to a fan triangle, a triangle of a top cone, or a decoration disk, with either orientation. The disk problem carries one distinguished outside polygon reading the prescribed letter word, with no extension of its interior required; the sphere problem carries none. Collapsed regions on vertices and free cancellations of edge paths have zero face cost. Every cell has a specified map to P, including the index of each side, so equal edge labels alone do not identify two side indices. The cellular chain of a picture mapped to the universal cover is the sum of its signed lifted two-cells, the outside polygon omitted.

Lemma. Every null letter word has a disk picture, and every nonzero cellular two-cycle c of the universal cover underlies a finite collection of spherical pictures of total cellular chain exactly c.

Proof. The disk assertion is the cellular van Kampen construction on the cone-and-polygon complex. For the sphere assertion, the universal cover is simply connected and has no three-cells, so the Hurewicz isomorphism pi_2 -> H_2 makes the cellular two-chain of a sphere representing the class of c equal to c, not merely congruent to it. Put the map in cellular general position: choose an interior point in each two-cell, make its inverse image a finite set of regular points, take small disks about them, and retract each punctured cell to its boundary, moving the complement into the one-skeleton. Inverse images of interior edge points become strips pairing the corresponding sides; the remaining regions map to vertices. Circular strips meeting no cell side are cut and capped, and a vertex region with several boundary components is replaced by one cap per component. These are compressions along curves in an oriented sphere and so yield spheres rather than positive-genus surfaces, changing no signed cell disk or lift; components with no cell disk are dropped. For a disk map, adjoin the outside polygon first and keep it a single face throughout; its word is preserved and one picture contains it, possibly with sphere components.

At a vertex over a top apex every incident face is a triangle of that cone; grouping them gives a top polygon whose boundary is a closed walk in the macro graph of the top occurrence, with perimeter equal to the number of those triangles. The macro graph is the Fano incidence graph with one subdivided edge, hence fifteen vertices. A simple macro cycle has at most fifteen sides, visits each point at most once, visits at least three points, and at a point turns between two distinct ports, namely the radial labels of the fan there, the exceptional port at D being X_D.

Lemma. A null letter word has a picture of minimum total top perimeter, and subject to this of minimum remaining interior faces, with the following properties; the parallel assertion holds for a fixed nonzero two-cycle c, minimizing among finite sphere collections of total lifted chain c.

(1) Every top polygon is a simple macro cycle.

(2) Adjacent top polygons do not use the same side of the same top occurrence; two copies of one fan triangle are not adjacent along equally indexed sides, and two copies of one decoration are not adjacent along the same position.

(3) At each fan-apex instance the cyclic list of radial labels has no repetition; a circuit with two distinct parallel steps is permitted, each such circuit has at most five steps, and a circuit with no top turn has at least four fan triangles.

Proof. The class of finite pictures is nonempty and carries a lexicographic minimum for the two nonnegative costs; in the sphere problem fix c once in taking the minimum. For (1), a top polygon whose macro walk repeats a template vertex has two boundary points of equal lifted value, both joined to the same lifted apex by the same indexed spoke; split it by a chord collapsed to that vertex, then cancel free backtracks, which removes two oppositely oriented copies of one lifted cone triangle and lowers perimeter by two. Repeating splitting and backtrack removal gives simple macro cycles, and the collapsed chords carry no two-chain. For (2), two top polygons meeting at the same macro side of one occurrence are two copies of one elementary cone triangle with opposite orientations and equal lifts, since the common lifted side fixes the triangle and apex; canceling the pair lowers perimeter by two, an ordinary diagram cancellation with vertex regions capped. A simple macro cycle uses each edge at most once, so same-template contact cannot occur within one polygon; the identical cancellation of two fan triangles or two decorations along equally indexed sides lowers the second cost without raising the first. These cancellations preserve the cellular chain and are forbidden by minimality. For (3), fix a source vertex over a fan apex b; a step of its link is a fan triangle or a corner of a top polygon. Two outgoing spokes of equal radial label are edges to vertices over the bouquet vertex; cut their side pairings and pair crosswise with outgoing orientations agreeing. This preserves each face orientation and face map, splits the link cycle, and at the far ends merges two vertex cycles or splits one. The numbers of faces and edges are fixed. When the far vertices differ, the total vertex count is unchanged and the surface stays connected through the merged far vertex, so Euler characteristic two gives a sphere; when they coincide, the vertex count rises by two and the dual-graph argument shows at most two components, so total Euler characteristic four forces two spheres. All other spheres persist and the outside polygon stays one face. The switch respects the fixed lifted chain: repeated radial label and lifted initial fan vertex fix the same lifted radial edge and far endpoint, so vertex identifications agree in the cover. Each switch increases fan-apex instances while the set of polygon corners is fixed, so the procedure terminates; the minimized costs are unchanged, so a newly created forbidden adjacency would still contradict minimality. The resulting link uses each radial label at most once, at most five occupy any fan, and a two-step triangle-only link is the reverse traversal of one fan edge and would cancel along an equally indexed radial side; every other nonempty triangle-only link is a circuit in the simple bipartite fan graph, of length at least four.

The outside word uses only letter edges and never meets a fan-apex instance, so each full link in the normal-form lemma is a cyclic link of interior faces.

Fix a normal picture. A fragment is any selection of its interior faces, need not connected or simply connected; a side pairing is kept precisely when both incident faces are selected, and all other selected sides are fragment boundary sides. Let T, Q, E be the numbers of selected top polygons, fan triangles, and decoration disks, and A = T+Q+E. For a top occurrence a let t_a be its number of selected top polygons, and for a decoration e let t_e be its selected multiplicity. At a fan-apex instance z represented in the fragment, d_z counts selected top turns and q_z selected fan triangles, and z is partial if not every step of its full cyclic link is selected. For a fan occurrence b write r_b for the number of represented apex instances over b, m_b for the largest t_a over slots a->b, S_b for the sum of those t_a, and w_b = max{r_b, m_b}, empty maxima zero, with all fan slots of every selected top included whether or not a polygon visits the fan; then sum_b S_b = 7T. Write J_b for the number of partial instances over b, and P_b^* for the number of selected loose top sides in turns at this fan, nonzero only when b has type D. Let I count internal contacts between two letter sides, radial contacts excluded, let B_raw and B_rad be the letter and radial boundary-side counts, and B = B_raw + B_rad. The coincidence bound controls internal letter contacts, and the local estimate at fan apices charges partial-link errors to radial boundary sides.

Lemma. If the rank inequality holds for collections of at most 8A occurrences, then I <= pT + f sum_b(w_b - S_b) + sum_e(h_e - delta) t_e.

Proof. Give each fan field of b weight w_b, each loose field of a weight t_a, and each field of a decoration e weight t_e; these bound the side multiplicities, since a fan field is used at most once at a normalized apex link and so at most r_b, a simple top polygon uses its loose field at most once, and a decoration copy uses each indexed field once. For each integer j >= 1 apply the rank inequality to the positions of weight at least j; this is eligible because t_a >= j forces w_b >= m_b >= t_a >= j, and its size is at most 8A since there are at most T tops, 7T+Q fans, and E decorations. Summing over j, and noting that within an unsigned-value class the ranks sum to the total weight minus the largest position weight, gives sum over classes of (total weight minus largest weight) bounded by (p-7f)T + f sum_b w_b + sum_e(h_e-delta) t_e. Every internal letter contact joins two positions of different field type or, for fan triangles, decorations, and loose top sides, is distinct by the normalization cancellations. In each class choose one maximum-weight position and charge each contact to an endpoint other than it, possible always; different contacts use different side copies, so the charge to any position is at most its multiplicity and hence its weight. Thus I is bounded by the left side, and substituting sum_b S_b = 7T gives the claim.

Lemma. For every fan occurrence b in a fragment, sum_{z->b} d_z + 2(w_b - S_b) <= (1/2)(sum_{z->b} q_z + P_b^*) + 2 J_b, and 2 sum_b J_b <= B_rad.

Proof. Suppose w_b = r_b. A simple top polygon has at most one turn at this fan, so S_b >= sum_{z->b} d_z and the left side is at most sum(2 - d_z). A full link with d_z = 0 has at least four fan triangles; with d_z = 1 the triangles form a path between the two distinct ports of the turn, ordinary ports being distinct line vertices, so at least two edges, the only exception being a path incident to X_D, which may have one edge paid by its loose side, giving triangle count plus payment at least two; d_z >= 2 makes 2 - d_z nonpositive. A partial link with d_z = 0 has at least one selected fan triangle, deficit at most 3/2; d_z = 1 gives deficit at most one; d_z >= 2 nonpositive; so 2 J_b covers it. Suppose w_b = m_b > r_b, and choose a dominant top a_0 with t_{a_0} = m_b. Let x, y be the totals of selected dominant and nondominant turns over b; nondominant selected faces have at most one turn each at this fan, so y <= S_b - m_b and sum_{z->b} d_z + 2(m_b - S_b) = x + y - 2(S_b - m_b) <= x - y. In a full link the consecutive dominant-to-dominant interfaces number at least dominant minus other turns, since each dominant turn is followed by one turn and at most as many dominant turns as nondominant ones are followed by a nondominant turn, and a single turn is followed by itself. Between a counted pair lies a path of fan triangles of nonzero length, because length zero would put two top polygons on the same radial side of a_0, forbidden; a length-one path must use an exceptional port and its loose-side payment. Thus path length plus payment is at least two for each interface, the paths are disjoint portions of the link, and no loose-side payment is used twice. At a partial link the selected steps form a nonempty proper subset of a cyclic list of at most five steps, splitting into at most two maximal runs. In a run with k dominant and l nondominant turns the linear turn list has at most l+1 dominant blocks, so the count H of consecutive dominant-to-dominant interfaces satisfies H >= k - l - 1 and k - l <= H + 1; each such interface path is entirely selected, and the same indexed-side and exceptional-port argument gives length plus payment at least two, with disjoint paths and distinct payments. A turn-free run contributes zero. Summing over the at most two runs bounds the dominant-minus-nondominant count by half the selected triangle count and loose-side payments plus two. Choosing a_0 once for the whole fan occurrence b and applying the estimates at all its apex instances proves the first inequality in the second case. For the boundary count, a represented partial cyclic link has at least two transitions between selected and unselected steps, each a radial edge with exactly one selected incident face, hence a radial fragment boundary side, and a radial side belongs to one apex instance; this gives 2 sum_b J_b <= B_rad.

Lemma. Put kappa = min{beta-7alpha, alpha/5, delta} > 0. If the rank inequality holds for collections of at most 8A occurrences, then every fragment of area A satisfies A <= C_0 B with C_0 = 1/kappa.

Proof. Write W = sum_b w_b and P^* = sum_b P_b^*. Every simple top polygon has at least three turns, so summing the apex inequality and using sum_b S_b = 7T gives 3T + 2(W - 7T) <= (Q + P^*)/2 + 2 sum_b J_b. The right side of the contact bound equals 3T + 2(W - 7T) + sum_e h_e t_e - Delta, with Delta = (beta-7alpha)T + alpha W + delta E. There are at most five selected triangles at a represented apex instance, so Q <= 5 sum_b r_b <= 5W and Delta >= kappa(T+Q+E) = kappa A. The raw side count is R_raw = Q + P^* + sum_e 2h_e t_e, and the contact and nominal-apex bounds give I <= R_raw/2 - Delta + 2 sum_b J_b. Since B_raw = R_raw - 2I, the partial and defect bounds give 2 kappa A <= 2 Delta <= B_raw + 4 sum_b J_b <= B_raw + 2 B_rad <= 2B, as required.

Lemma. For every D < infinity and eta > 0 there is an integer K = K(D,eta) such that the vertices of any finite planar multigraph of maximum degree at most D admit a partition into sets of size at most K with at most eta times the total number of vertices of edges crossing between sets.

Proof. Ignore loops and apply the planar separator theorem to the underlying simple graph: at most c sqrt q vertices of a q-vertex graph can be removed to leave two edge-disjoint sets of size at most 2q/3, with c = 2 sqrt 2 by the Lipton-Tarjan bound. Recurse inside every set larger than K, making each removed separator vertex a singleton part. Charge c/sqrt q to every vertex of every recursive set of size q; these charges cover the removed separators, and a fixed vertex travels through set sizes shrinking by a factor at most 2/3 while above K, so its total charge is at most c/((1 - sqrt(2/3)) sqrt K). With A the original vertex count, at most C A / sqrt K separator vertices are removed, C = c/(1 - sqrt(2/3)); every crossing edge has a removed separator vertex as an endpoint, and the degree bound limits crossings, with multiplicity, to DC A / sqrt K. Take K so large that DC/sqrt K <= eta, treating disconnected graphs component by component.

https://epubs.siam.org/doi/pdf/10.1137/0136016

Proof of the diagram theorem. With the parameters fixed, kappa and C_0 are positive constants. Choose eta > 0 with 2 C_0 eta <= 1/2 and apply the separator lemma with D = 15, fixing a finite threshold K. By the rank lemma, with probability tending to one the rank inequality holds for all collections of at most 8K occurrences; work with any realization having that property, the rest being deterministic and simultaneous over all pictures. For a normal picture the dual graph of interior faces and contacts is planar of degree at most fifteen, since fan triangles have three sides, decorations at most six, and simple top polygons at most fifteen; parallel and self-contacts are harmless. Partition its A interior faces into fragments of size at most K, cutting at most eta A contacts, and let B_0 count contacts with the outside polygon, setting B_0 = 0 in the sphere problem. The fragment boundary lengths sum to at most B_0 + 2 eta A, so summing the fragment bound gives A <= C_0(B_0 + 2 eta A), hence A <= 2 C_0 B_0. A disk picture whose outside word has length l has B_0 <= l, since outside-to-outside pairings are free cancellations adding no interior contact, so it uses at most 2 C_0 l interior faces.

A nonzero two-cycle c of the universal cover would have a normal spherical picture with B_0 = 0, forcing A = 0 and a zero cellular chain, a contradiction; thus H_2 of the universal cover vanishes. The cover is a simply connected two-dimensional CW complex with all reduced homology zero; any first nonzero homotopy group would equal a nonzero homology group by Hurewicz, so the cover is weakly contractible and, by Whitehead, contractible, proving asphericity.

The linear inequality over letters is now explicit. Choose a vertex of each fan template and, for each of its vertices, a fixed path of length at most two to it. Send every fan-apex vertex to the bouquet vertex and replace its indexed radial edge by the label word of the corresponding fan path, oriented appropriately. A fan triangle boundary becomes a closed fan walk of length at most five. A top polygon boundary becomes a closed walk in its copy of Gamma: at an ordinary point the two radial paths concatenate to a fan path between the two ports, and at the exceptional point the path to X_D is followed by the loose edge, for expanded length at most thirty. A decoration boundary is unchanged. Take as a finite presentation all simple cycle relators in each included fan and top graph together with the decoration rows; by the cone construction this presents the same group, and any closed graph walk of length at most thirty splits at repeated vertices into at most thirty simple cycles after free backtracks are erased. Each expanded face thus has a filling with at most thirty relator disks, and replacing the faces of the normal picture by these fillings leaves the outside letter word unchanged and gives a disk diagram of area at most 60 C_0 l, sphere components discarded. Any other fixed finite set of defining cycle relators gives the same linear inequality after bounded relator substitutions.

hatcher-2002.pdf

The separator threshold is chosen after the positive parameters are fixed and before n grows, with no uniformity asserted as alpha, beta, delta approach zero. For each fixed parameter choice the single finite-collection event controls all disk words and all cellular two-cycles at once.

The relations must now be tested against an action chosen after the tables are sampled, so the statements concern total variation of finite distributions on matches and never a target metric space; they apply at once to all uniformly bounded functions of the relevant assignments, including functions built from a later action.

Recall the master pattern with field set F of 212 fields. Its requirements are the seven fan entries, the top mark, and one entry from each of the ninety four-letter tables, the first two coordinates of each such entry being core fields and its other two new; the six-letter table is excluded. Write R for the set of 98 array entries and E_rho within F for the support of rho in R, every rho sitting in a different Bernoulli array, its coordinate order fixed throughout. A match is a map sigma from F to S_n for which every required entry is present, with no injectivity demanded of one match; Omega_n is the random set of matches and mu_N its uniform law when nonempty. A conditional test is a triple (I,e,t) with I within F, e in F minus I, the union I with {e} admissible, and t a positive integer.

For theta = (alpha,beta,delta) let c_rho(theta) be defined by inclusion probability N^{-c_rho(theta)}, with critical value c_rho = c_rho(0):

fan i: c_rho(theta) = k_i - 2 + alpha, c_rho = k_i - 2;
top mark: c_rho(theta) = 12 + beta - 7alpha, c_rho = 12;
four-letter entry: c_rho(theta) = 2 + delta, c_rho = 2.

For W within F put h_theta(W) = |W| - sum_{rho: E_rho within W} c_rho(theta) and h = h_0, so that N^{h_theta(W)} is the expected number of assignments to W for which every entry supported in W is present. The function h is submodular, since each indicator of E_rho within W is supermodular and the costs are nonnegative. Moreover h(F) = 212 - (17 + 12 + 180) = 3, hence E|Omega_n| = N^{h_theta(F)} = N^{3 - beta - 90delta}, the expectation being exact because each assignment uses one entry from each array and all its inclusion requirements are independent.

Lemma. For every W within F, h(W) >= min{2, |W|}; and if S is admissible with S within W, then h(W) >= |S|.

Proof. Take the core part of W, with a_i fields in F_i and a loose indicator ell. Before the top mark the exponent is sum_i(a_i - (k_i - 2) of the indicator of a_i = k_i) + ell, which is at least sum_i min{a_i,2} + ell. The top-mark cost falls only on the complete core, where the exponent is 32 - 17 - 12 = 3; so every core subset has exponent at least the size of any admissible core subset it contains and at least the smaller of two and its own size. Adding the fresh fields of a four-letter entry adds one per field, and adds two as a pair only when that entry's two core fields are also present, so each fresh pair contributes nonnegatively; this proves the admissible bound for core S and the budget whenever W has at least two core fields. With fewer than two core fields no four-letter, fan, or top entry is charged and h(W) = |W|, proving the budget throughout. An admissible subset of one four-letter entry has size at most two, covered by the budget.

Fix a conditional test (I,e,t). To couple t matches agreeing on I, require agreement on a possibly larger set U that absorbs the flat directions, so every further extension raises the critical exponent. When I is two fields of one fan F_i, h(I) = 2 = h(E(F_i)); once the fan fields are present, adjoining both fresh fields of any square whose two core fields lie in that fan also leaves the exponent unchanged. Since |I| <= 2 and every entry support has at least three fields, h(I) = |I|. Choose an inclusion-maximal U containing I with h(U) = |I|. Two properties follow:

e not in U, and h(U with V) - h(U) >= 1 for every nonempty V within F minus U.

The first is the admissible budget applied to I with {e}. For the second, every superset of U has exponent at least |I| by the same lemma applied to I, maximality rules out equality for a strict superset, and all critical exponents are integers.

Form a star from t ordered copies of the master pattern by identifying all copies of each field in U; entries supported in U are identified and every other entry keeps one copy per branch. Assignments to the star are required to be injective in each field: for s in F minus U the t branch values are pairwise distinct, while values in different fields may coincide and all fields take values in S_n. Injectivity makes copies of any entry not supported in U distinct tuples, since some coordinate lies outside U, and prevents an entry of one branch not supported in U from equalling the corresponding entry of a fixed branch. Hence all variables counted are products of distinct Bernoulli trials once conditioned trials are removed.

An overlap type between two star assignments records exactly which entries coincide across the two stars; there are finitely many for fixed t. Each equality equates corresponding coordinates, and for s not in U injectivity turns these into a partial matching between the t branch variables of the two stars; different arrays may use different partial matchings whose union must still be a partial matching when the type is realizable, else two different branch values of one field would be forced equal. For s in U there is one variable in each star, and an equality never compares two different coordinate fields even when numerical letter values coincide. The counting rule: reveal the first assignment, and suppose the equalities prescribe q distinct, previously unpinned variables of the second. For each field the unpinned variables number at most t and are chosen uniformly without replacement, possibly avoiding one pinned value, so prescribing q_s of them has probability at most C_t N^{-q_s}, the remaining choices being at most N^{r - q_s} against a falling factorial (N-a)_r with r <= t and a in {0,1}. Multiplying over the finitely many fields gives, for the fraction of pairs with the specified overlap,

<= C_t N^{-q}.

If the coincident unconditioned entries have total actual cost C, their joint inclusion probability divided by the product of the two individual ones is N^C, and all other trials are independent, so the overlap type contributes at most C_t N^{-q + C} to the relative variance. Pairs with no coincident unconditioned entry have zero covariance. Since all candidate assignments of a given counting problem share one inclusion probability, the relative-variance normalisation is exactly the sum of these contributions over nonempty overlap types.

For a single master pattern, with W the fields of the shared entries, q = |W| and the recovered cost is at most the sum of the costs of all entries supported in W, so the critical excess is at least h(W) > 0; small actual parameters keep every one of these finitely many excesses positive, giving

Var|Omega_n| / (E|Omega_n|)^2 = o(1),

so the master count is asymptotic to its expectation in probability, and the parameters are always chosen so the exponent in the master mean is positive.

For the star counts put C_U(theta) = sum_{rho: E_rho within U} c_rho(theta), C_out(theta) = sum_{rho: E_rho not within U} c_rho(theta), and d = |F minus U|. With Z_N the number of present injective stars and (N)_t = N(N-1)...(N-t+1),

E Z_N = N^{|U|} (N)_t^d N^{-C_U(theta) - t C_out(theta)}.

Lemma. For each fixed conditional test, sufficiently small actual parameters give the following as N tends to infinity.

(1) Fix an assignment sigma, place it in a specified branch, and condition on it being a match. The count A_sigma of its extensions to a present injective star has conditional mean independent of sigma, equal to

a_N = ((N-1)_{t-1})^d N^{-(t-1) C_out(theta)},

and conditional relative variance o(1) uniformly in sigma.

(2) Fix an ordered list b of t distinct letters as the branch values of field e. The count B_b of present injective stars with those values has mean independent of b, equal to

b_N = N^{|U|} (N)_t^{d-1} N^{-C_U(theta) - t C_out(theta)},

and relative variance o(1) uniformly in b.

Proof. For (1), condition only on the trials of sigma; entries in U and the other entries of the specified branch are then fixed present, and every other branch uses new entries distinct from those trials, its t-1 values per outside-U field being distinct and avoiding the value in sigma, giving (N-1)_{t-1} choices and the stated mean. For an overlap of two extensions let V_j be the outside-U fields of the shared entries in extra branch j; the assignment penalty is sum_j |V_j|, and charging each shared entry to its branch in the first extension makes the recovered critical cost at most sum over entries E_rho within U with V_j but not within U of c_rho. The critical excess is at least sum_j [h(U with V_j) - h(U)] >= 1 over nonempty V_j by the closure gap, so the bounds stay strict for small actual parameters and the variance rule gives the conditional relative variance, uniform in sigma; for t = 1 the extension count is one. For (2), no Bernoulli trial is conditioned, only letter values are pinned, and since e lies outside U its t pinned values already satisfy field injectivity, the other fields admitting N^{|U|}(N)_t^{d-1} choices. For an overlap let W be the variables of the first star in shared entries, A = W with U, V_j its outside-U fields in branch j, and m(W) the number of e-variables; a realizable overlap matches an e-variable only to the same pinned value, imposing no further penalty, so q = |W| - m(W). With h_star(W) computed from the star's critical costs,

h_star(W) >= h(A) + sum_{j: V_j nonempty} [h(U with V_j) - h(U)]

when A is nonempty, the term h(A) >= 1 by the budget and each summand >= 1 by the closure gap, and each branch holds at most one pinned e-variable, so the excess h_star(W) - m(W) is at least one. When A is empty, no shared entry meets U, entries in different branches use disjoint variables, and h_star(W) = sum_j h(V_j); an active branch contains the full support of a shared entry, at least three fields, so |V_j| >= 3, the budget gives h(V_j) >= 2, and subtracting at most one pinned leaf per branch leaves a positive excess. All nonempty overlap types thus carry critical excess at least one, preserved under small actual parameters, and the variance rule summed yields the relative variance.

For probability measures lambda, lambda' on a finite set put d_TV(lambda,lambda') = (1/2) sum_z |lambda(z) - lambda'(z)|. All table-convergence statements hold in probability as N tends to infinity, equivalently with deterministic bounds tending to zero.

Lemma. For each fixed conditional test and sufficiently small actual parameters, with probability tending to one there is a probability law on t matches with: (1) identical restrictions to I; (2) every branch marginal at total-variation distance o(1) from mu_N; (3) the joint law of the t values of field e at total-variation distance o(1) from the uniform product on S_n^t.

Proof. Use the uniform law on present injective stars. With L_N = |Omega_n|, lambda_N = E L_N, and D_N = sum_{sigma in Omega_n} |A_sigma - a_N|, the conditional variance bound and Cauchy-Schwarz give E[|A_sigma - a_N| given sigma in Omega_n] = o(a_N) uniformly; multiplying by the probability sigma is present and summing yields E D_N = o(lambda_N a_N), so D_N = o(lambda_N a_N) in probability, and the concentration of L_N gives L_N/lambda_N -> 1 in probability. Because Z_N = sum_sigma A_sigma and |Z_N - L_N a_N| <= D_N, Z_N is positive and Z_N/(L_N a_N) -> 1 in probability, and sum_sigma |A_sigma/Z_N - 1/L_N| <= D_N/Z_N + |L_N a_N/Z_N - 1| -> 0, giving the marginal for each of the finitely many branches and exact agreement on I through U containing I. For the leaves, with B_N the ordered distinct lists of length t and |B_N| = (N)_t, the second variance bound gives E sum_b |B_b - b_N| = o((N)_t b_N); Markov and sum_b B_b = Z_N make the leaf law B_b/Z_N uniformly close in total variation to the uniform law on B_N. An iid uniform list repeats a value with probability at most binom(t,2)/N, and conditioned on distinctness is uniform on B_N.

The two marginal conclusions serve separate ends. The branch comparison moves averages of bounded tests between the coupled branches and a uniform match: a pointwise norm estimate bounds the average over the t branches by a function of the e-labels alone, the shared I-labels entering only through a uniform bound, and comparison with independent uniform letters controls that function of the leaves, so no independence between the common I-tuple and the test labels is needed.

Every critical overlap margin is at least one, and each shared trial is charged once to its occurrence in the first star, which holds at most t copies of each array entry. Since beta > 7alpha all cost increments are positive, so replacing critical costs by actual costs raises any recovered-cost sum by at most t[7alpha + (beta - 7alpha) + 90delta] = t(beta + 90delta); conditioning only removes trials, and a single master match is covered by t = 1. For a finite family of tests with t_max the largest requested t, choosing t_max(beta + 90delta) < 1/2 keeps every overlap margin at least 1/2, compatible with beta > 7alpha > 0 by taking beta = 8alpha and then alpha, delta small. The parameters are fixed while N grows, and no variance argument asserts uniformity for unbounded t at fixed parameters.

Let D_N within S_n^6 be the six-letter table and Q_{rq}(u,v) the number of its rows with r-th coordinate u and q-th coordinate v. Each Q_{rq}(u,v) is a sum of N^4 independent Bernoulli variables of parameter N^{-3-delta}, of mean N^{1-delta}. With eps_N = N^{-(1-delta)/4} the Chernoff bound gives P(|Q_{rq}(u,v) - N^{1-delta}| > eps_N N^{1-delta}) <= 2 exp(-eps_N^2 N^{1-delta}/3), a union bound over the 15 N^2 choices of positions and letters yields max_{r 7alpha has, with probability tending to one, nonempty match set with |Omega_n| = (1+o(1)) N^{3 - beta - 90delta}, every single-field marginal of mu_N at total-variation distance o(1) from the uniform law on S_n, the coupling of the previous lemma for every test in T, and the six-letter pair estimate. All these properties concern only the tables and are independent of any action or metric space.

Proof. Adjoin to T the finitely many tests (empty set, e, 1). For such a test the budget forces U empty and the one-branch star is exactly a master match, so the leaf assertion gives single-field uniformity. Choose parameters meeting all finite overlap margins, beta + 90delta < 3, and delta < 1, and combine the single-copy concentration, the coupling lemma, and the six-letter estimate over this finite set of events.

The proposition is compatible with the diagram theorem: for each finite family of tests choose positive parameters small for that family and the theorem, then the fragment threshold, and finally N, so both the diagram and sampling events hold with probability tending to one, and a deterministic diagonal sequence follows with parameters tending to zero and every fixed test valid in the limit.

The sampling statements concern finite tables and are now applied to functions drawn from an isometric action chosen after those tables are known. The order matters: a bound on every generator displacement is extracted first from the pointwise relative counts of the six-letter table, and only then are total-variation estimates used for action-dependent functions.

For a complete CAT(0) space X the completed tangent cone at x is written T_x X, with vertex 0. For y in X the vector log_x y has length d(x,y) and the direction of the geodesic from x to y. In a Euclidean metric cone set  = (||u||^2 + ||v||^2 - d(u,v)^2)/2, so  = ||u|| ||v|| cos of the angle, truncated at pi. Multiplication by a nonnegative real number is radial dilation, and no additive vector-space structure is assumed. The tools are strong convexity of squared distance, first variation, the flat-triangle equality theorem, and the tangent-cone theorem, none of which needs properness; the tangent cone in this generality follows Kleiner and Leeb, who credit Nikolaev. In particular angle comparison gives

d(y,z)^2 >= d(x,y)^2 + d(x,z)^2 - 2 .

kleiner-leeb-1997.pdf

https://doi.org/10.1007/978-3-662-12494-9

Lemma. Let an infinite finitely generated group G act properly and cocompactly by isometries on a proper complete CAT(0) space X. For a finite symmetric generating alphabet S counted with multiplicities, E(x) = (1/|S|) sum_{s in S} d(x,sx)^2 attains a positive minimum; at any minimizer the vectors v_s = log_x(sx) satisfy (1/|S|) sum_{s in S}  <= 0 for every h in T_x X; equivalently 0 is their barycenter, and (1/|S|) sum_{s in S} d(h,v_s)^2 >= ||h||^2 + (1/|S|) sum_{s in S} ||v_s||^2.

Proof. Take a minimizing sequence x_j and a compact C with GC = X; choose g_j with y_j = g_j x_j in C. Bounded energy bounds each displacement d(x_j,sx_j), hence d(y_j,(g_j s g_j^{-1}) y_j). The closed neighborhood of C of any fixed finite radius is compact by properness, and properness of the action leaves only finitely many group elements carrying a point of C into that neighborhood, so each g_j s g_j^{-1} ranges over finitely many values. After a subsequence the conjugated tuple is constant and y_j tends to a point y of C, where the energy of that tuple is the original infimum; fixing one element of the subsequence, g_{j_0}^{-1} y attains the infimum for the original tuple. Zero energy would fix a point, and a proper action has finite stabilizers, so the minimum is positive. For the first variation, let h be represented by a geodesic x_t of initial speed ||h||; convexity at the midpoints of crossed geodesics gives d(x_t,sx_t)^2 <= (1/2)d(x_{2t},sx)^2 + (1/2)d(x,sx_{2t})^2, and by symmetry the two terms on the right have equal average over the alphabet after replacing s by s^{-1} in the second distance. Minimality then forces the right derivative at 0+ of (1/|S|) sum_s d(x_{2t},sx)^2, equal to -(4/|S|) sum_s , to be nonnegative. Represented vectors are dense in the completed cone, so continuity gives the barycenter condition, and the cone cosine formula gives the variance inequality.

For an aspherical instance P the group G is infinite: the parity quotient makes G nontrivial, and a finite G would make the universal cover a finite contractible complex with 1 = chi = |G| chi(P), forcing |G| = 1. Any geometric action considered therefore admits the lemma. Rescale its metric so that

(1/N) sum_{s in S_n} ||v_s||^2 = 1,

a rescaling preserving completeness, properness, and the CAT(0) condition.

Lemma. Suppose the six-letter table has pointwise relative error at most eps < 1 in the counts for positions 1 and k, for every 2 <= k <= 6. Then every action normalized at a minimizing basepoint satisfies max_{s in S_n} ||v_s||^2 <= 25 (1+eps)/(1-eps), with a bound independent of the action.

Proof. Let Q_{1k}(a,b) count the rows with first coordinate a and k-th coordinate b, so (1-eps)N^{1-delta} <= Q_{1k}(a,b) <= (1+eps)N^{1-delta}. Summing over b makes the rows beginning with a fixed a nonempty, with conditional k-th-coordinate distribution bounded pointwise by ((1+eps)/(1-eps))(1/N). The relation of a row (a,s_2,...,s_6) reads a^{-1} = s_2...s_6, and since d(x,a^{-1}x) = d(x,ax), the triangle inequality and Cauchy-Schwarz give ||v_a||^2 <= 5 sum_{k=2}^6 ||v_{s_k}||^2. Averaging over the rows beginning with a and using the pointwise bound and the normalization, each of the five conditional averages is at most (1+eps)/(1-eps). Only nonnegative weights enter, so the estimate survives weights chosen from an action after the table is revealed.

For a geodesic from u to v in a CAT(0) Euclidean cone write [u,v]_t for the point at parameter t; its squared radial norm satisfies

||[u,v]_t||^2 = (1-t)||u||^2 + t||v||^2 - t(1-t)d(u,v)^2,

immediate when an endpoint is the vertex and otherwise following from the Euclidean sector or passage through the vertex. Scalar-product concavity along geodesics is due to Gigli and Nobili, and the inductive-mean bound is the unit-variance case of Sturm.

gigli-nobili-2021.pdf

https://doi.org/10.1090/conm/338/06080

Lemma. In a complete CAT(0) metric cone, u ->  is concave along every geodesic for fixed h. Let a finite distribution of vectors V have barycenter 0 and E||V||^2 = 1, and let V_1,...,V_t be independent samples with m_1 = V_1 and m_j = [m_{j-1},V_j]_{1/j}. Then E||m_t||^2 <= 1/t, and for any deterministic list V_1,...,V_t one has (1/t) sum_{j=1}^t  <= .

Proof. Subtract strong convexity of d(h,.)^2 from the radial identity in the formula 2 = ||h||^2 + ||u||^2 - d(h,u)^2; the quadratic terms cancel and concavity follows, and induction gives the mean comparison. For the independent samples condition on m_{j-1} = u; the barycenter variance inequality gives E[d(u,V_j)^2 given u] >= ||u||^2 + 1, so strong convexity of squared distance from the vertex gives a_j <= (1-1/j)^2 a_{j-1} + 1/j^2 with a_j = E||m_j||^2, and a_1 = 1 yields a_j <= 1/j by induction.

Now take a sequence of table instances nu for which the sampling conclusions hold, with errors tending to zero for every fixed test and every fixed positive integer t, and suppose their groups have geometric actions with normalized minimizing basepoints x_nu. Let Omega_nu be the nonempty finite match set and mu_nu its uniform law. The displacement lemma gives an absolute constant M with ||v_s|| <= M for all letters in all sufficiently large instances, and discarding finitely many instances changes no limit. The actions, their spaces, and the associated test functions may depend on the complete tables.

Lemma. Let I and {e} be disjoint sets of fields with admissible union. For each nu let h_I be a function of the labels on I with values in T_{x_nu} X_nu and ||h_I|| <= B, for B independent of nu and the match. For either fixed choice sigma(s) = s or sigma(s) = s^{-1}, limsup_nu E_{mu_nu}  <= 0.

Proof. Fix t. The coupling gives t matches agreeing on I, each branch marginal at total-variation distance at most eta_{nu,t} from mu_nu, and their e-labels at distance at most eta_{nu,t} from independent uniform letters, with eta_{nu,t} tending to zero; a single bound covers all comparisons after enlarging the error, and a fixed inversion preserves the assertion. The vector h_I has the same value in every branch. Form the inductive mean m_t of the branch vectors v_{sigma(e_j)}; the mean comparison gives, for every coupled tuple, (1/t) sum_j  <=  <= B||m_t||, the last step using only the norm bound and not independence of h_I from the leaves. Under independent uniform leaves the inductive-mean bound applies through the variance inequality and the normalization; under the coupling ||m_t|| <= M. The total-variation comparison changes the expected squared norm by at most 2M^2 eta_{nu,t} and each scalar-product expectation by at most 2BM eta_{nu,t}, so E_{mu_nu} <= B sqrt(1/t + 2M^2 eta_{nu,t}) + 2BM eta_{nu,t}. Taking the upper limit in nu at fixed t and then letting t grow gives the claim, uniformly over the action-dependent functions.

Enlarge Gamma by adding, for each prescribed two-edge path, the complementary two-edge path of its decoration square, obtaining a fixed finite graph Gamma-hat whose extra midpoint vertices are distinct abstract vertices. A match gives a consistent orbit realization of Gamma-hat in X_nu: choose one anchor at x_nu and send a vertex reached by the successive letters s_1,...,s_k to s_1...s_k x_nu. Core closed walks are relations and the decoration squares give consistency on the added paths, and fixing one path per vertex keeps every orbit point within LM of x_nu, where L depends only on Gamma-hat. Take the finite product over the matches with metric

d_nu((y_m),(z_m))^2 = (1/|Omega_nu|) sum_{m in Omega_nu} d(y_m,z_m)^2,

based at the tuple of minimizing basepoints; it is complete CAT(0), the vertex tuples lie at bounded distance from the basepoint, and a nonprincipal ultrafilter on the index set produces a pointed ultralimit Z, complete and CAT(0). The bounded vertex tuples give a configuration in Z indexed by the vertices of Gamma-hat; midpoints and fixed-fraction points of the products form coordinatewise, and their ultralimits are the corresponding points of Z by uniqueness of geodesics, properness and a dimension bound being unnecessary.

Lemma. Every edge of Gamma-hat has length one in Z. Every two-step path through two distinct fields of one decoration square has squared endpoint distance at least two. A path of k <= 3 distinct core fields whose field set is admissible has squared endpoint distance at least k.

Proof. For an edge with letter s, translation at its initial orbit vertex identifies its squared length with ||v_s||^2 or ||v_{s^{-1}}||^2, and single-field total-variation convergence with the uniform bound and normalization make the squared product length tend to one. For a path with oriented readings a_1,...,a_k, translate the penultimate orbit vertex to x_nu; the earlier endpoint becomes a_{k-1}^{-1}...a_1^{-1} x_nu and the next becomes a_k x_nu. The vector h_I = log_{x_nu}(a_{k-1}^{-1}...a_1^{-1}x_nu) is a function of the preceding labels with ||h_I|| <= (k-1)M, and log comparison gives d(u_0,u_k)^2 >= d(u_0,u_{k-1})^2 + ||v_{a_k}||^2 - 2 in every factor. Averaging, applying the conditional test, and taking the ultralimit add at least one at each step by induction; every prefix field set in the statement is admissible, so the conditional test is available, and distinct fields need not carry distinct numerical letters.

Lemma. For four points z_1,...,z_4 of a CAT(0) space with m = mid(z_1,z_3) and n = mid(z_2,z_4), d(z_1,z_3)^2 + d(z_2,z_4)^2 + 4 d(m,n)^2 <= sum_{i=1}^4 d(z_i,z_{i+1})^2, with z_5 = z_1. Equality of the diagonal-square sum with the side-square sum forces the two diagonal midpoints to coincide.

Proof. Apply midpoint convexity to d(m,z_2)^2 and d(m,z_4)^2 on the segment z_1 z_3, and then to d(m,n)^2 on z_2 z_4; substituting the first two into the third and multiplying by four gives the inequality, and equality of the two diagonal squares with the side squares forces d(m,n) = 0.

Each decoration square has unit sides, and both diagonals have squared length at least two by the path bounds, so the quadrilateral lemma makes both equal two and, in particular, d(u,w)^2 = 2 for every nonbacktracking two-edge core path, whose table exists even when the two fields lie in one fan.

A core hexagon comes from three noncollinear Fano points i,j,k, the joining lines, and a choice of X or an available Y representative at each of the three points with all six incidences present. Its vertices z_0,...,z_5 have squared distances, at cyclic separation one and two, equal to one and two; each three-edge half-path uses at most two fields of any fan, so opposite endpoints have squared distance at least three. Apply the quadrilateral lemma to (z_0,z_1,z_3,z_4): the side squares sum to 1+2+1+2 = 6 and each diagonal square is at least three, so both are three and their midpoints coincide; applying the same argument to (z_1,z_2,z_4,z_5) identifies the third opposite-pair midpoint. This common center c satisfies d(c,z_i) = r = sqrt(3)/2 for every vertex.

Lemma. For a core hexagon every triangle of its center with two of its vertices is Euclidean flat, including the degenerate case of opposite vertices, and  = r^2 - (1/2) d(z_i,z_j)^2.

Proof. Opposite pairs are geodesic segments through c, so their directions in T_c Z are antipodal and their radial length is r. For antipodal v^+,v^- the angular triangle inequality gives  +  <= 0 for every u, the two angles summing to at least pi. With u = log_c z_i, v^+ = log_c z_j, v^- = log_c z_{j+3}, log comparison gives  >= r^2 - (1/2)d(z_i,z_j)^2 and  >= r^2 - (1/2)d(z_i,z_{j+3})^2; the two prescribed squared distances sum to 3 = 4r^2, so the lower bounds sum to zero and both are equalities. The angle at c then equals its Euclidean comparison angle, and the flat-triangle equality theorem makes the triangle flat; an opposite or repeated pair is a segment or a point.

Lemma. All core hexagons share one center c, and every core vertex, lying in some core hexagon, is at distance r from c.

Proof. With X representatives only, a noncollinear ordered triple of Fano points is a basis of F_2^3; reordering a basis leaves its hexagon unchanged, and replacing one basis vector i by i+j preserves the line ij and the third point k, so the old and new hexagons share the opposite pair (X_k,ij) and have the center of that pair as common midpoint. Permutations and elementary additions generate all basis changes over F_2, so all X hexagons share a center. Replacing one Y representative by its X representative keeps a core hexagon, and either unchanged point representative together with its opposite line is an opposite pair shared by the two, so their centers agree; repetition connects every core hexagon to one using only X representatives. Every X_i lies in a hexagon obtained by extending i to a basis, every line in one by taking two points spanning it and a third outside, and every Y_i, having two distinct incident lines, in one through a choice of another point on each line together with i; the case of Y_D uses its two full lines. The radius statement follows.

The square and hexagon comparisons now fix one further angle. Put H_0 = CD, p = X_C, q = Y_C, z = X_D, J = mid(p,q), and choose another line H through C different from H_0. The core four-cycle (p,H_0,q,H) has unit sides and both diagonals of squared length two by the two-step equality, so its diagonal midpoints coincide and J = mid(H_0,H), d(H_0,J)^2 = 1/2, d(H_0,z) = 1. In each orbit factor let J_m be the midpoint of the realizations of p and q, and translate the realization of H_0 to x_nu; the realizations of p and q are then given by the two letters on the adjacent F_C edges with their orientation signs, so the translated vector toward J_m is a function h_I of just those two fields with ||h_I|| <= M, and the translated vector toward z is v_{sigma(e_0)} with e_0 the loose field. These three fields are admissible, and the conditional test bounds the upper limit of their averaged scalar product by zero.

Fix a,b in (0,1] and, in each factor, the points at fractions a,b along the segments from H_0 to J_m and to z; their tangent vectors are scalar multiples of the two tangent vectors just described. Log comparison, averaging, and the ultralimit give, for the limiting fractional points J_a and z_b,

d(J_a,z_b)^2 >= a^2 d(H_0,J)^2 + b^2 d(H_0,z)^2 = a^2/2 + b^2,

so every comparison angle at H_0 is at least pi/2, and letting both fractions decrease to zero in Z yields angle_{H_0}(J,z) >= pi/2. Only fixed-fraction distances pass through the ultralimit; the angle is computed afterwards in the limit.

Proposition. Let P_nu be a sequence of aspherical instances whose master-match statistics and six-letter relative counts satisfy the sampling conclusions with errors tending to zero for every fixed test and every fixed t. If every pi_1(P_nu) has a proper cocompact isometric action on a proper complete CAT(0) space, then there are a complete CAT(0) space Z, a configuration indexed by the core vertices of Gamma, and a point c in Z with three properties. (1) Every core edge has length one, every nonbacktracking two-edge core path has squared endpoint distance two, and every three-edge core path with distinct admissible fields has squared endpoint distance at least three. (2) Every core hexagon has its three opposite-pair midpoints equal to c, all core vertices lie at distance r = sqrt(3)/2 from c, and for any two vertices occurring in one core hexagon the central triangle is flat with tangent scalar product r^2 - (1/2)d(z_i,z_j)^2. (3) With H_0 = CD and J = mid(X_C,Y_C), d(H_0,J)^2 = 1/2, d(H_0,X_D) = 1, angle_{H_0}(J,X_D) >= pi/2, and all the fractional-point inequalities hold.

Proof. Use the normalized minimizing actions and the uniform bound. The product construction, the path bounds, and the decoration-square argument give (1); the hexagon-flatness and common-center lemmas give (2); the fixed-fraction argument gives (3).

The configuration just obtained cannot exist in a complete CAT(0) space. The task is carried out first in that space and then in its completed tangent cone at the common hexagon center, which is not a vector space, so sums are defined metrically and every additive inequality comes from concavity.

For a CAT(0) Euclidean cone T with vertex 0 let delta_a be radial dilation by a >= 0, write a u = delta_a(u), and put ||u|| = d_T(0,u) and  = (||u||^2 + ||v||^2 - d_T(u,v)^2)/2, which for nonzero vectors is ||u|| ||v|| cos of the truncated angle. Define

u oplus v = 2 mid(u,v),

commutative but with no associativity or cancellation law assumed. When the directions of u,v are joined by a geodesic shorter than pi their cone is a Euclidean sector and u oplus v is the ordinary sum there, and all two-vector computations either lie in such a sector or follow from the definition. Scalar-product concavity gives

 >=  + ,

an inequality in a general cone, never replaced by equality without rigidity. A triangle is flat when one vertex angle equals its Euclidean comparison angle or an interior comparison inequality is an equality, and flat then means the convex hull is isometric to the Euclidean triangle.

https://doi.org/10.1007/978-3-0348-9240-7

Lemma (square rigidity). Let p_0,p_1,p_2,p_3 be points of a CAT(0) space, indices modulo four, with d(p_i,p_{i+1}) = 1 and d(p_0,p_2) = d(p_1,p_3) = sqrt 2. Their convex hull is a unit Euclidean square in the given cyclic order, and the diagonals share a midpoint.

Proof. Let o be the midpoint of p_0,p_2. Squared-distance convexity gives d(o,p_1)^2 <= 1/2 and d(o,p_3)^2 <= 1/2. Since sqrt 2 = d(p_1,p_3) <= d(p_1,o) + d(o,p_3) <= sqrt 2, both inequalities are equalities and o is also the midpoint of p_1,p_3. The equality for d(o,p_1) is an interior comparison equality in the triangle p_1,p_0,p_2, so that triangle is flat; with each vertex as the vertex opposite a diagonal midpoint, every triangle on three consecutive vertices is flat, and the angle at each corner is pi/2. The flat quadrilateral theorem sums the four angles to 2pi, and the Euclidean quadrilateral with those sides and diagonals is a unit square.

Suppose Z is a complete CAT(0) space holding the configuration, use the graph vertex names for its points, let c be the common hexagon center, and put r = sqrt 3 / 2, T = T_c Z. The completed tangent cone is CAT(0) and its completed space of directions is CAT(1). Write x_i = log_c X_i, y_i = log_c Y_i, h_H = log_c H, and h_ij = h_H when H = ij; all these vectors have norm r.

Any two distinct line vertices H,H' occur together in a core hexagon: take their intersection point i, choose a point j on H and a point k on H' different from i, and use the X representatives of the noncollinear triple i,j,k. The central triangle is flat and the distance is sqrt 2, so

 = -1/4 for distinct lines H,H'.

A line through a doubled point i in {A,B,C,D} is full at i if it is adjacent to both X_i and Y_i; all lines through A,B,C are full and the full lines at D are AD,BD.

Lemma (fan identities). For i in {A,B,C,D} there is a unit vector M_i in T with

M_i = x_i oplus y_i = h_H oplus h_H' whenever H,H' are distinct full lines at i,

each sum lying in a nondegenerate Euclidean sector, and

 = -1/4,  =  = 1/2.

Proof. The four-cycle X_i,H,Y_i,H' has unit sides and diagonals of length sqrt 2, so square rigidity makes it a flat square with center J_i the midpoint both of X_i,Y_i and of H,H'. The triangle c,H,H' is flat, so its median gives d(c,J_i)^2 = r^2 - (1/4)d(H,H')^2 = 1/4. Squared-distance convexity in c,X_i,Y_i has right-hand side (1/2)d(c,X_i)^2 + (1/2)d(c,Y_i)^2 - (1/4)d(X_i,Y_i)^2 = 1/4, so this is also an equality and c,X_i,Y_i is flat. Taking its midpoint in the tangent cone gives x_i oplus y_i = 2 log_c J_i = h_H oplus h_H'; calling this common vector M_i makes it independent of the full-line pair and of norm one. The cosine law gives  = -1/4, and h_H,h_H' have angular separation arccos(-1/3) < pi, so in their Euclidean sector  = ||h_H||^2 +  = 3/4 - 1/4 = 1/2, and symmetrically for h_H'.

A contradiction comes from evaluating  two ways. At A,B the fan identities give M_A = h_AB oplus h_AC and M_B = h_AB oplus h_BC; applying sum-concavity in both entries, with AB,AC,BC distinct, yields

 >= ||h_AB||^2 +  +  +  = 0.

The rest of the argument forces M_A,M_B to be antipodal, of scalar product -1. First the exceptional incidence makes M_C,M_D antipodal; their axis then splits a factor R x Q off the cone, the second coordinates h_AC,h_BC,h_AD,h_BD form a flat square in Q whose opposite side midpoints make M_A,M_B antipodal through the fan identities at A,B.

Set H_0 = CD, h_0 = h_CD, p = X_C, q = Y_C, z = X_D, and J the midpoint of p,q; the omitted edge Y_D--H_0 drives the asymmetry.

Lemma (exceptional-incidence identity). The vectors M_C,x_D lie in a Euclidean sector of T, and

h_0 = M_C oplus x_D,  = -1/2,  = 1/2,  = 1/4.

Proof. Apply the square construction at C with H_0 and another line through C; the center J gives d(H_0,J)^2 = 1/2 and d(c,J)^2 = 1/4. Also d(H_0,z) = 1, d(c,z) = d(c,H_0) = r, and the paths p,H_0,z and q,H_0,z give d(p,z)^2 = d(q,z)^2 = 2. The ternary-midpoint conclusion makes the angle at H_0 between J,z at least pi/2, so angle comparison and these distances give d(J,z)^2 >= d(H_0,J)^2 + d(H_0,z)^2 = 3/2. The reverse follows from J being the midpoint of p,q: d(J,z)^2 <= (1/2)d(p,z)^2 + (1/2)d(q,z)^2 - (1/4)d(p,q)^2 = 3/2. Hence the angle and distance comparisons are equalities, and H_0,J,z is flat with a right angle at H_0. The point B_0 of [J,z] at parameter 1/3 from J satisfies d(H_0,B_0)^2 = (4/9)(1/2) + (1/9) = 1/3 in that right triangle, and squared-distance convexity toward c gives d(c,B_0)^2 <= (2/3)(1/4) + (1/3)(3/4) - (2/9)(3/2) = 1/12. The triangle inequality then reads sqrt 3 / 2 = d(H_0,c) <= d(H_0,B_0) + d(B_0,c) <= 1/sqrt 3 + 1/(2 sqrt 3) = sqrt 3 / 2, so every inequality is an equality, B_0 lies on [H_0,c] in that order, and c,J,z is flat by its interior comparison equality. With j = log_c J, the vector to B_0 in the Euclidean sector containing j and x_D is (2/3)j + (1/3)x_D, so collinearity and d(c,H_0) = 3 d(c,B_0) give h_0 = 3 log_c B_0 = 2j + x_D = M_C oplus x_D, using M_C = 2j and ordinary addition in that sector. The norm identity 3/4 = ||h_0||^2 = 1 + 3/4 + 2 gives  = -1/2, and pairing the sector identity with either summand gives the other two products.

Lemma. The unit vectors M_C,M_D are antipodal: d_T(M_C,M_D) = 2 and  = -1.

Proof. Take a full line H at D. The path Y_D,H,X_D,H_0 uses two fields of F_D and the loose field, so it is admissible and d(Y_D,H_0) >= sqrt 3 = d(Y_D,c) + d(c,H_0); equality holds, the segment from Y_D to H_0 passes through c, and the directions of y_D,h_0 are antipodal. Let Sigma be the completed space of directions of Z at c and hat a direction of a nonzero vector; set a = arccos(1/sqrt 3) and b = arccos(1/3), both in (0,pi/2) with 2a+b = pi. The exceptional products give d_Sigma(M_C hat, h_0 hat) = a and d_Sigma(h_0 hat, x_D hat) = b, and these arcs occur consecutively in the Euclidean sector, so their concatenation is its geodesic direction arc. The fan products give d_Sigma(x_D hat, y_D hat) = arccos(-1/3) = pi - b, and the direction of M_D = x_D oplus y_D bisects that arc because the summands have equal length, so d_Sigma(x_D hat, M_D hat) = a. The concatenation h_0 hat, x_D hat, y_D hat has length pi and antipodal endpoints, hence is minimizing and its two arcs continue geodesically through x_D hat, making M_C hat, h_0 hat, x_D hat, M_D hat a local geodesic of length a+b+a = pi. In the CAT(1) space Sigma a local geodesic is globally minimizing, including length exactly pi, so its endpoints are antipodal; with unit norms this gives the stated distance and scalar product.

Proposition. No complete CAT(0) space contains the configuration of the energy section.

Proof. Suppose otherwise and keep the notation. The bound  >= 0 is to be contradicted by splitting the axis of the antipodal pair. The fan identities give M_C = h_AC oplus h_BC and M_D = h_AD oplus h_BD with four pairwise distinct lines, so sum-concavity gives  >= -1/2 and  >= -1/2. Antipodality forces equality in -1 =  >=  +  >= -1, and interchanging the two sums gives the symmetric equalities; with the fan products this yields

 =  = 1/2,  =  = -1/2,
 =  = -1/2,  =  = 1/2.

The opposite rays M_C,M_D form a complete geodesic line l through the cone vertex. With a = arccos(1/sqrt 3), each of the four vectors has angles a and pi - a to the two ends of l in some order, its two direction arcs concatenate to a minimizing semicircle whose Euclidean cone is a flat half-plane with boundary l containing the vector, so each vector lies on a complete line parallel to l. The parallel-set product decomposition gives P(l) = R x Q with l = R x {q_0}, 0 = (0,q_0), M_C = (1,q_0), M_D = (-1,q_0), and Q CAT(0). The axial coordinate is the scalar product with M_C, so

h_AC = (1/2,u), h_BC = (1/2,u'), h_AD = (-1/2,v), h_BD = (-1/2,v'),

with u,u',v,v' in Q, each at distance 1/sqrt 2 from q_0 since each original vector has squared norm 3/4. Distinct line vectors have squared cone distance two, so the product metric gives d_Q(u,u') = d_Q(v,v') = sqrt 2 and d_Q(u,v) = d_Q(u,v') = d_Q(u',v) = d_Q(u',v') = 1. The identity h_AC oplus h_BC = M_C makes the midpoint of the two line vectors equal (1/2,q_0), the half-dilation of M_C on the axis, so product midpoints give q_0 = mid(u,u'); the identity at D gives q_0 = mid(v,v').

In cyclic order u,v,u',v' these points satisfy square rigidity and span a unit flat square in Q centered at q_0. Its opposite side midpoints m = mid(u,v) and m' = mid(u',v') satisfy d_Q(q_0,m) = d_Q(q_0,m') = 1/2 and d_Q(m,m') = 1, and the square lies in the transverse factor only, no planar realization of the incidence configuration being assumed.

The full-line choices AC,AD at A and BC,BD at B give M_A = h_AC oplus h_AD = delta_2(0,m) and M_B = h_BC oplus h_BD = delta_2(0,m'), the dilation delta_2 doubling distances in T; by the opposite-midpoint distances, ||M_A|| = ||M_B|| = 1 and d_T(M_A,M_B) = 2, so  = -1. This contradicts  >= 0 and proves that the configuration cannot occur.

The pieces assemble with the order of all choices kept explicit.

Lemma (deterministic diagonal sequence). There is a sequence of table realizations with n_j tending to infinity and positive parameters alpha_j,beta_j,delta_j tending to zero such that: (i) the complexes P_j are aspherical and their letter presentations carry a linear disk-filling inequality; (ii) the match sets are nonempty and their single-field marginals tend to uniformity in total variation; (iii) for every admissible pair I,{e} and every fixed integer t, the common-I coupling has both total-variation errors tending to zero along the sequence; (iv) the six-letter pair counts have uniformly vanishing relative error. The groups G_j = pi_1 P_j are infinite.

Proof. At stage j request all admissible tests of the fixed master pattern with 1 <= t <= j, a finite list. Choose eps_j > 0 small enough for the sampling proposition on this list and for the diagram theorem, with eps_j < 1/(400j), and set alpha_j = delta_j = eps_j and beta_j = 8 eps_j. Then beta_j > 7 alpha_j, the exponent 3 - beta_j - 90 delta_j of the master count is positive, and j(beta_j + 90 delta_j) = 98 j eps_j < 98/400 < 1/2, so the overlap margins of the finite list are preserved. Fixing these parameters, the diagram theorem selects a finite fragment threshold depending possibly on j, its coincidence inequalities hold with probability tending to one, and the sampling proposition delivers the finite collection of sampling properties with probability tending to one. Choose n_j > n_{j-1} so large that there is positive probability all these events hold and all sampling errors are at most 1/j, and select one realization in the intersection; no comparison between the fragment threshold and n is needed before fixing them. This gives (i)-(iv). Each G_j has a quotient C_2, and a finite G_j would make its universal cover a finite contractible CW complex with 1 = chi = |G_j| chi(P_j), forcing |G_j| = 1 against the quotient, so G_j is infinite.

Proposition. For all sufficiently large j the group G_j admits no geometric action on any proper complete CAT(0) space.

Proof. Suppose infinitely many G_j admitted such actions. Restrict to that subsequence and choose one action for each group; all fixed-test limits remain valid and the spaces, their dimensions, and their local geometries may vary. Normalizing the minimizing generator energy makes it positive by infinitude, and the displacement estimate together with the uniform conditional test holds for these choices of actions. The configuration proposition therefore supplies the limiting configuration, which the no-configuration proposition forbids. Only finitely many of the selected groups can then admit such actions, and an instance with no action exists.

Lemma (finite-model filling transfer). Let P be a finite connected CW complex with contractible universal cover, and suppose a finite presentation of pi_1 P has a linear disk-filling inequality. Then P has a finite connected simplicial homotopy model K whose universal cover is contractible and satisfies a linear combinatorial disk-filling inequality.

Proof. A finite CW complex has a finite simplicial homotopy model of the same dimension, by approximating attaching maps with simplicial maps after finite subdivisions and using simplicial mapping cylinders. The homotopy equivalence lifts to the universal covers, so the cover of K is contractible. The linear estimate is independent of the finite presentation: choose substitutions phi from new generators to old words and psi from old to new, inducing inverse group identifications, let L bound the lengths of the phi(b), let M bound the new-presentation filling areas of the finitely many substituted old relators psi(r), and let D bound the filling areas of the null words b psi(phi(b))^{-1}; the last two lists are finite null lists, so the bounds exist. A new null word of length l substitutes to an old word of length at most L l, and if the old filling constant is C_old, filling it and replacing its relator cells costs at most M C_old L l new cells while the boundary substitutions cost at most D l, so the new filling constant is at most M C_old L + D. Apply this to the maximal-tree presentation of K, whose generators are its non-tree edges and whose relators are its triangular faces: a null-homotopic combinatorial edge loop of length l gives a null word of length at most l after inserting paths in the fixed maximal tree, which have uniformly bounded lengths, and filling that word linearly, replacing each relator cell by its triangular face and canceling tree paths, gives a diagram with at most C l triangular faces for a fixed finite C. An edge loop in the universal cover projects to a null loop in K, and its disk diagram lifts.

hatcher-2002.pdf

Proof of the main theorem. Take an instance P_j supplied by the no-action proposition. Its finite letter presentation has a linear disk bound, and the transfer lemma yields K, of the same dimension, so K is two-dimensional and aspherical with the asserted linear filling estimate; its fundamental group is G_j, carrying the no-action property. A linear Dehn function is equivalent to word hyperbolicity. Suppose a finite simplicial complex L homotopy equivalent to K admitted a compatible locally CAT(0) geodesic length metric. The space L is compact, hence complete and locally compact. Give its universal cover the induced length metric, for which the covering map is a local isometry. This cover is complete, since a Cauchy sequence upstairs projects to a convergent one downstairs and sufficiently short connecting paths force a tail into one sheet over an evenly covered neighborhood of the limit. The Cartan-Hadamard theorem makes the cover globally CAT(0); it is locally compact, so the length-space Hopf-Rinow theorem makes it proper, and its deck transformations act freely, properly, and cocompactly by isometries. The isomorphism pi_1 L = pi_1 K = G_j would then give a geometric action of G_j on a proper complete CAT(0) space, a contradiction. Compatible locally CAT(-1) geodesic length metrics are excluded as well, since a CAT(-1) space is CAT(0).

No curvature assumption is made on the random complexes; curvature comparison enters only in a hypothetical action space, after the combinatorial filling and asphericity statements are settled. No common filling constant is needed across the diagonal sequence, since only the constant of the single chosen finite complex appears in the conclusion.