The Singer conjecture in dimension four

Proof. Lemma 2.1 supplies a finite aspherical integral Poincaré-complex homotopy model of M. A homotopy equivalence lifts to an equivariant homotopy equivalence of universal covers and preserves the singular-chain invariants above. Theorem 1.1 applies to the model.□

Proof. Wall’s dimension theorem [30, Theorem 2.2] gives a four-dimensional CW complex homotopy equivalent to Q. It can be chosen finite because Q is finite, so its reduced finiteness obstruction vanishes. The model is aspherical, and its universal cover is contractible. Its augmented cellular chain complex is consequently a free \mathbb Z\Gamma-resolution of \mathbb Z of length four. In particular, \Gamma is finitely presented and has finite integral cohomological dimension. To see directly why it has no torsion, a finite-order element would yield a subgroup C of prime order. Restricting the resolution to \mathbb Z C preserves freeness and finite length. This contradicts the usual two-periodic resolution of a cyclic group, which gives nonzero groups H^{2j}(C;\mathbb Z) for every j\geq1. If \Gamma were finite, torsion-freeness would make it trivial, so asphericity would make Q contractible. But duality over k gives H_4(Q;k)\cong H^0(Q;k)\cong k. Hence \Gamma is infinite.

For the manifold assertion, let M be a closed connected aspherical topological four-manifold. A compact topological manifold is metrizable. Its coordinate neighborhoods are absolute neighborhood retracts, so Hanner’s locality theorem makes M an absolute neighborhood retract; see [14, Theorems 3.2–3.3]. To check its covering dimension, shrink a finite coordinate cover to a finite closed cover. Each member has small inductive dimension at most four, by the subspace theorem and the dimension of Euclidean space. The closed-sum and coincidence theorems give \dim M\le4; conversely, a coordinate chart contains a copy of an open four-ball, so subspace monotonicity gives \dim M\ge4. These dimension facts are [12, Theorems 1.1.2, 1.5.3, 1.7.7 and 1.8.2]. West’s homotopy dimension estimate [31, Corollary 5.4] therefore supplies a finite four-dimensional simplicial complex homotopy equivalent to M. This complex is a homotopy model, not a triangulation of the manifold. It is aspherical. Transport the orientation character and fundamental class of M to this model. Topological Poincaré duality with the orientation local system gives cap-product isomorphisms for every local coefficient system on M [29, Theorem 1, with R=\mathbb Z]. Since M is compact, compactly supported cohomology there is ordinary cohomology. Transporting these isomorphisms along the homotopy equivalence gives the required absolute integral duality on the model; see also [27, Theorem 10.2 and Remark 10.3], and [13, Theorem A.16] for the oriented chain version.□

Proof. The augmented singular chains of Z are a free resolution of \mathbb Z. The comparison theorem for free resolutions gives chain homotopy equivalences with P_*. Tensoring preserves these homotopies, so it preserves the homology in the definition of b_1^{(2)}; no flatness of \mathcal N(\Lambda) over the group ring is required.

The remaining assertion is the finite-matrix comparison for von Neumann dimension. For a matrix over \mathcal N(\Lambda), the algebraic kernel has dimension equal to the trace of the projection onto its Hilbert kernel. The algebraic image has dimension equal to the trace of the projection onto the closure of its Hilbert image. Additivity of dimension thus identifies the dimension of first algebraic homology with that of

On the other hand, the harmonic representative space for reduced cohomology is

These are the same closed Hilbert submodule. The same argument works in each degree with finite free neighboring terms. This is the comparison underlying [21, Remark 1.31 and Lemmas 6.51–6.53].□

Proof. Use the finite four-dimensional model in Lemma 2.1. A homotopy equivalence with Q lifts to an equivariant homotopy equivalence of universal covers, and cellular and singular chains compute the same invariant; see [21, Definition 6.50 and Lemmas 6.51–6.53]. There are no cellular chains above degree four, proving the assertion for p>4. A harmonic square-summable 0-cochain on the connected universal cover is constant. Since \Gamma is infinite, that constant must be zero. The degree-zero version of Lemma 2.2 gives b_0^{(2)}=0.

Here is the duality argument, including nontrivial orientation characters. Let \Gamma^+ be the kernel of the orientation character \Gamma\to\{1,-1\}, and let Q^+ be the corresponding connected cover. Its degree d=[\Gamma:\Gamma^+] is either one or two. Finite-sheeted covers of Poincaré complexes are again Poincaré complexes of the same formal dimension [28, p. 567, paragraph preceding Corollary 4.4]. The induced orientation character of Q^+ is the restriction to \Gamma^+ and is trivial. Thus Q^+ is a finite connected aspherical oriented integral Poincaré complex of formal dimension four, and Lemma 2.1 supplies a finite four-dimensional homotopy model F.

Transport the fundamental class of Q^+ to F. Poincaré duality with local coefficients applies to Q^+ and hence to F; see [30, Lemma 1.1]. In particular, applying it to the regular group-ring local system shows that the cap-product map from the degree-reversed dual of C_*(\widetilde F;\mathbb Z) to C_*(\widetilde F;\mathbb Z) is a quasi-isomorphism. Here right modules are converted to left modules by the involution g\mapsto g^{-1}. Both complexes are bounded and finitely generated free over \mathbb Z\Gamma^+. The mapping cone is therefore a bounded acyclic complex of free modules, which is contractible: starting in its lowest degree, split each surjection onto the kernel already obtained. Thus cap product is a chain homotopy equivalence.

Extend scalars to \mathcal N(\Gamma^+). The homotopies are preserved, and finite-matrix dimension comparison identifies dual Hilbert complexes with equal von Neumann dimensions. It follows that

This is the finite-chain argument underlying the L^2-duality proof in [21, Theorem 1.35(3) and its proof]; here Poincaré-complex duality supplies the required chain equivalence. Finally, finite-index scaling [21, Theorem 1.35(9)] gives

Dividing by d proves the same duality for Q and completes the reduction.□

Proof. The orientation twist of Q is trivial over k. Since Q is aspherical, its absolute integral Poincaré duality, applied to local systems of k-vector spaces, gives, for every left k\Gamma-module V,

For clarity, the induction used on the right is homological. Indeed, k[\Gamma/J]=k\Gamma\otimes_{kJ}k, and a free right k\Gamma-resolution R_* of k satisfies

Restriction to kJ preserves a free resolution. This proves H_j(\Gamma;k[\Gamma/J])\cong H_j(J;k), the homological Shapiro isomorphism; compare [8, Chapter III, Section 6]. Taking J trivial gives the assertion for k\Gamma, and taking i=1 gives the displayed isomorphism for arbitrary J.

A finite subgroup is trivial by Lemma 2.1. A two-ended group has a finite normal subgroup with quotient infinite cyclic or infinite dihedral; see [30, Lemma 4.1]. If it is torsion-free the finite subgroup is trivial and the dihedral case is impossible; thus a two-ended subgroup of \Gamma is infinite cyclic. Both the trivial group and the infinite cyclic group have zero third homology over k.□

Proof. Choose a finite presentation, and consider its simply connected Cayley two-complex. Because there are finitely many cell orbits in degrees at most two, finite-support cellular cochains in these degrees identify with group-ring-valued equivariant cochains. Explicitly, a finite-support k-cochain a corresponds to

Taking the identity coefficient recovers a, and the identification commutes with coboundaries. The augmented cellular complex is exact in degrees zero and one; adjoining free modules in higher degrees extends it to a resolution without changing its cohomology calculation in degree one.

Suppose that removing finitely many vertices of the Cayley graph leaves two infinite components. Let f be the indicator of one infinite component, with value zero on the deleted vertices. Its gradient df is a finitely supported edge cochain, because the graph is locally finite, and it is a cellular cocycle. By Lemma 2.4, df=du for a finitely supported vertex function u. Connectedness makes f-u constant. This is impossible: f takes each of its values on infinitely many vertices, whereas u has finite support. The infinite connected locally finite Cayley graph has at least one end, so it has exactly one.□

Proof. Fix a vertex o and an edge orbit E whose stabilizer is J. Identify E with \Gamma/J, and define c(g) to be the mod-two sum of the edges of E on the geodesic [o,go]. Cancellation of edges traversed twice gives

If J is finite or two-ended, Lemma 2.4 makes this cocycle principal: c(g)=gv-v for some v\in k[\Gamma/J]. Therefore

For the first assertion, let S be a finite generating set of \Gamma. The paths [o,so], s\in S, contain finitely many edges. Their translates contain every path [o,go], by writing g as a word in S and cancelling backtracking. Thus only finitely many edge orbits occur in these paths. An action without a fixed vertex has unbounded vertex orbits: a bounded tree orbit has an invariant center, which gives a fixed vertex when inversions are absent. Hence one of the finitely many edge-orbit crossing counts is unbounded. If all edge stabilizers were finite or two-ended, this would contradict (2.1).

For the second assertion, fix any edge orbit E and collapse each connected component of the union of edges outside E. The quotient is a nontrivial simplicial tree without inversions. It has no fixed vertex: the preimage of one would be a proper nonempty invariant subtree of the original tree. Its orbits are therefore unbounded, and distance in this quotient counts exactly the edges of E in the original path [o,go]. Again (2.1) excludes a finite or two-ended stabilizer. Apply this to the Bass–Serre tree to obtain the last assertion.□

Proof. Write F(\mathcal C) for the set of two-cells. The cellular coboundary

is bounded: each relator has at most three edges, counted with their cellular multiplicities, and edge–cell incidence is uniformly bounded. An ordinary cellular 1-cocycle on \mathcal C is a gradient: integrate it along edge paths from a fixed vertex, using simple connectivity to show independence of the path. It follows that \mathcal D=\ker d_1, which is closed.

Likewise d_0:\ell^2(\Gamma;\mathbb R)\to\ell^2(E(G);\mathbb R) is bounded by the bounded vertex degree. Since finite-support functions are dense in \ell^2(\Gamma),

This does not assert that \operatorname{im}d_0 itself is closed.

The augmented cellular chains of \mathcal C are exact in degrees zero and one. Extend them to a free resolution by choosing a free module P_3 mapping onto \ker(\partial_2:C_2(\mathcal C)\to C_1(\mathcal C)), and then mapping free modules onto successive kernels. The original cellular terms in degrees zero, one, and two are retained and are finite free. The singular chains of \widetilde Q are free \mathbb Z\Gamma-modules: a deck transformation fixing a singular simplex fixes its image points and is therefore the identity. Thus Lemma 2.2 applies. Over \mathbb C, it identifies b_1^{(2)}(\widetilde Q) with the dimension of \ker d_1\ominus\overline{\operatorname{im}d_0}. All matrices are real, so this complex Hilbert space is the complexification of \mathcal D\ominus\mathcal D_0. If \mathcal D=\mathcal D_0, its dimension would be zero. The asserted strict inclusion follows.□

Proof. Fix a vertex a. For each b\ne a, the functional

is well defined on \mathcal D. A finite path from a to b of length \ell gives |L_b(df)|\le\sqrt{\ell}\,\|df\|_2. The functional is nonzero, as one sees by testing the gradient of the singleton indicator of b. Let r_b\in\mathcal D be its Riesz representer. The unique gradient of minimum norm subject to L_b(df)=1 is

normalize its potential by f_b(a)=0.

The vectors r_b span densely in \mathcal D. Indeed, a gradient orthogonal to all of them has a potential constant at every vertex and is therefore zero. Since \mathcal D_0 is a proper closed subspace, some df_b lies outside \mathcal D_0. Fix this minimizer and write it as df. Replacing f by \min(1,\max(0,f)) preserves the values 0,1 at a,b and cannot increase any edge energy. Uniqueness of the minimizer implies 0\le f\le1.

Suppose f tends to a constant c outside finite vertex sets. For \epsilon>0 define

Each u_\epsilon has finite support. The function (f-c)-u_\epsilon is the truncation of f-c to [-\epsilon,\epsilon], so its edge differences are bounded in absolute value by |df| and tend pointwise to zero. Dominated convergence gives du_\epsilon\to df in edge \ell^2, contrary to df\notin\mathcal D_0.

Fix a regular t\in(0,1). If a component U of \{f>t\} misses b, replace f by t on U. Both prescribed values are preserved. Internal edge energies decrease to zero. At a boundary edge with one endpoint in U, the other endpoint has value below t, so its energy decreases strictly. Such an edge exists because G is connected and a\notin U. This modification is legitimate even when U is infinite: each edge energy weakly decreases, and at least one decreases by a positive amount. It contradicts minimality. Hence every upper component contains b, and the upper side is connected. The lower side is connected by the same argument with a.

Let C be a dual component of S(f,t), and restrict the cochain d1_{\{f>t\}} to C, setting it to zero on other edges. In any cell, all crossing edges lie in one dual component. The signed sum of the restricted cochain around that cell is therefore either the original zero sum or an entirely zero sum. This remains true when an edge has several occurrences on the cell boundary. Simple connectivity of \mathcal C gives a potential for the restricted cochain. That potential is constant on each of the two connected level sides, and an edge of C shows that the constants differ. It consequently has nonzero difference across every edge of S(f,t), so no crossing edge could have been omitted. Thus C=S(f,t).□

Proof. The assertion for translates follows from Lemma 3.1; levels outside [0,1] have empty cuts. The only nonempty cuts of a singleton indicator consist of edges incident to its vertex, a finite set by local finiteness.□

Proof. Write e\sim e' when e=e' or the edges occur on a common cell. This symmetric relation has degree bounded by a constant D, by bounded cell incidence and bounded cell length. For finite-energy coordinates F,F', Tonelli’s theorem and Cauchy–Schwarz give

Thus almost every pair of endpoint parameters has finitely many interactions. The argument also applies when F=F', since the two parameters remain independent. For a fixed finite list of box conditions, intersect the finitely many full-measure conditions for endpoint pairs and exclude the countably many vertex values. The resulting endpoint tuples are dense in their parameter domain. They therefore supply arbitrarily small good boxes about every point of the cube.

Fix a good box and list its endpoint cuts as S_1,\ldots,S_N. Delete all of their edges from G, obtaining a graph G'. On every component of G' all endpoint signs are constant. Let E be the finite set of cut edges which also occur in another cut or share a cell with an edge of another cut. For each i, the dual graph on S_i\setminus E still has finitely many components. Indeed, its original dual graph has finitely many components by Lemma 3.2, has bounded degree, and has had only finitely many vertices deleted.

Consider two dually adjacent edges of S_i\setminus E. On their common triangular cell, their endpoints on a specified side of cut i either coincide or are joined by the remaining cell edge. This joining edge is not in S_i. Moreover, the cell contains no edge of any other endpoint cut, by the definition of E. The joining edge therefore lies in G'. Following a finite dual path shows that all endpoints on either specified side of a dual component of S_i\setminus E belong to one component of G'. Such a dual component is incident to at most two components of G'. Each edge of E is likewise incident to at most two.

If any cut is nonempty, every component of G' is incident to a cut edge, since the original graph is connected. The preceding finite count proves that G' has finitely many components. If all cuts are empty, G'=G is connected. The vertex set of O is a union of those components with the specified signs; its induced subgraph has exactly these components. This proves the required finiteness. The argument also covers repeated vertices or edges on a triangular boundary: the same-side joining path then either has length zero or uses the remaining edge.

There are countably many finite lists of base coordinates. For each list, choose a countable dense set of good endpoint tuples. The resulting boxes form a countable basis. Add all their translates; this remains countable, and goodness is preserved by the action.□

Proof. The operations defining the margin give

Two adjacent vertices with positive margin belong to the same component of O. Thus either both endpoints contribute their margin to h_{O,U}, both contribute zero, or the endpoint outside U has zero margin. In every case

If there are no conditions, the margin and localized coordinate are constant and their gradients vanish.

For the cut inclusion, the endpoint where h_{O,U}>t belongs to U and satisfies every margin inequality strictly above t. At the other endpoint, either it still belongs to U and some margin is below t, or it lies outside U. In the latter case its margin must be zero by adjacency, and again one of the margin inequalities fails. Since t<1, the constant cap in the definition of \psi_O contributes no additional cut. The crossed margin is one of the levels a_i+t or b_i-t, proving (3.2). All exceptional values, for which one of these levels or t itself is a vertex value of its coordinate, form a countable set.□

Proof. A closed subset of X contained in its isolated vertices is finite, since it is compact and discrete. Since \Gamma is infinite, this first implies B\ne\varnothing. If B were a singleton \{x\}, then, for each \epsilon>0, the vertex set where |f(v)-f(x)|\ge\epsilon would be finite. This would make f tend to a constant at infinity, contrary to Lemma 3.1. Thus B has at least two points.

If B were disconnected, write it as a disjoint union of two nonempty compact sets. Choose neighborhoods of those sets in X whose closures have positive distance apart. Their complement is a closed subset of X disjoint from B, hence a finite vertex set. Each neighborhood contains infinitely many vertices. Only finitely many graph edges join the two neighborhoods, by (3.4). Delete their endpoints together with the finite complement. The remaining graph has infinitely many vertices on each side and no path between the sides. A connected locally finite graph minus finitely many vertices has only finitely many components, so each side contains an infinite component. This contradicts one-endedness of G.□

Proof. For a good box O with its finite component list U_1,\ldots,U_r, the identities

hold on vertices and therefore, by continuity, on all of X. Consequently, if \psi_O(x)>0, exactly one component, denoted U_O(x), satisfies

All sufficiently near vertices belong to this component. This is where the finiteness assertion in Lemma 3.3 is used.

Shrink W to a neighborhood defined by finitely many coordinate conditions, each allowing a positive tolerance about its value at x. The base coordinates define a continuous projection from X to [0,1]^{\mathcal F_0}. Choose a good basis box O' about the base profile of x, small enough to have the following properties. First, every tested base coordinate satisfies its prescribed tolerance throughout O'. Second, for every tested h_{O,U} with \psi_O(x)>0, we have O'\subset O and the variation of \psi_O from \psi_O(x) on O' is smaller than the prescribed tolerance. Third, if a tested box has \psi_O(x)=0, its margin is smaller than that tolerance throughout O'. These are finitely many open conditions on the base profile, so Lemma 3.3 supplies such an O'.

We have \psi_{O'}(x)>0. Let U'=U_{O'}(x) be the component selected by (3.6). For any tested box O of positive margin at x, the inclusion O'\subset O and connectedness of U' place U' in one component of O. That component is U_O(x). Indeed, density of the vertices in X supplies a vertex near x at which the two coordinates selecting U' and U_O(x) are both positive.

Every tested localized coordinate therefore satisfies its condition on U'. If its box has positive margin and its component is U_O(x), it equals \psi_O, whose variation was controlled. If its component differs from U_O(x), it is identically zero on U' and at x. This includes the case of zero localized value with positive box margin. Finally, a tested box of zero margin satisfies 0\le h_{O,U}\le\psi_O, so the third condition on O' controls its value. Thus all vertices of U' lie in W.

The open set

contains x, and every vertex in it belongs to U'. Joining two such vertices inside the graph component U' proves the first assertion. Applying that assertion to successively smaller metric balls about x, and choosing their radii slowly enough for both endpoint sequences, proves the sequential formulation. Conversely, failure for one fixed W would give two sequences converging to x with no joining paths in W, contradicting that formulation.□

Proof. For each edge, the set of straddled levels has Lebesgue measure |dF(e)|. Tonelli’s theorem gives (4.2). Bounded incidence implies that

belongs to \ell^2(E(G)). The right side of (4.2) is \sum_{e'}a(e')\lambda(ge'), so it is finite by Cauchy–Schwarz. Translates of a fixed finite edge set eventually avoid any other fixed finite edge set. Approximate both a and \lambda in \ell^2 by finitely supported functions and apply Cauchy–Schwarz to the errors. Their translated pairing therefore tends to zero.

To prove (4.3), join any two edges of D by a simple path in the dual adjacency graph. For each consecutive pair, traverse the boundary of a cell containing both edges. This joins their endpoints using edges near D. The simple dual path uses each of its edges at most once; bounded cell size and incidence give a uniform bound on the multiplicity with which any near edge is used. The triangle inequality proves the estimate. Sum it over the components to obtain the last assertion.□

Proof. For a finite list of base functions, Lemma 4.1 permits a subsequence along which the sum over n of all the integrals in (4.2) is finite. For almost every level, the corresponding nonnegative length sums then tend to zero. Each regular base cut has finitely many dual components. Choose one endpoint in each nonempty component. The translates of these finitely many fixed vertices tend to a by (4.1). The diameter estimate in Lemma 4.1 therefore makes every endpoint of the translated cut tend uniformly to a.

For an added coordinate associated with a box, its straddling set is contained in a finite union of base straddling sets at levels of the form a_i+t and b_i-t. Include those base coordinates in the list. Translations and reflections of the level parameter preserve null sets, so almost every t has the required estimates for all the covering cuts. Levels outside [0,1] give empty cuts. This proves the assertion for added coordinates. To obtain both directions, choose the summable subsequence simultaneously for g_n and g_n^{-1}.□

Proof. Suppose (4.4) fails. After taking a subsequence, there are x_n\to r\ne b with g_nx_n\to s\ne a. For each n, density of the vertices and continuity of the particular map g_n give a vertex v_n such that

Thus v_n\to r and g_nv_n\to s.

Choose a coordinate separating r from b, and another separating s from a. Apply Lemma 4.2 to these two coordinates. Choose regular levels strictly between the respective values and outside its null exceptional sets. Call the resulting cuts the first and second cuts. The translated first cut has all endpoints uniformly near a, while the inverse-translated second cut has all endpoints uniformly near b.

Let C_n be the graph component of the r-side of the first cut containing v_n. Write F for its coordinate and t for its level; for definiteness suppose F(b)<t<F(r). A neighborhood of b then misses the entire r-side, whose vertices satisfy F>t. Both sides have vertices. A path from v_n to the other side has a first straddling edge; its endpoint w_n on the r-side lies in C_n. Since the inverse-translated second cut is uniformly near b, none of its edges can have an endpoint on the r-side for large n. Consequently no edge of g_nC_n straddles the second cut. Yet g_nv_n lies on its s-side, whereas g_nw_n is uniformly near a and lies on the opposite side. The image of a path in C_n between v_n and w_n must straddle that cut, a contradiction.

For proper discontinuity, suppose infinitely many distinct group elements take a point of one compact set of distinct triples to a point of another such compact set. Pass to convergent source and target triples and to subsequences with g_n\to a and g_n^{-1}\to b. The latter limits belong to B, since group vertices are isolated. At least two coordinates of the limiting source triple differ from b. Equation (4.4) sends both to a, contradicting distinctness of the limiting target triple.□

Proof. The powers h^n escape every finite vertex set as n\to+\infty or n\to-\infty. If a subsequence tends to z\in B, (4.1) gives \rho(h^n,h^{n+1})\to0, whereas continuity of left multiplication by h gives h^{n+1}\to hz. Thus hz=z, and uniqueness of the parabolic fixed point gives z=p. Compactness proves convergence of the entire sequences. Proposition 4.3 now supplies their compact-uniform convergence off p; they fix p itself.

If (4.5) fails, take a subsequence with g_n\to a and g_np\to c\ne a. The cosets g_n\langle h\rangle escape every finite vertex set. Otherwise some fixed w belongs to infinitely many of them, so g_n=wh^{j_n} on a subsequence, with |j_n|\to\infty. Then both g_n and g_np tend to wp, a contradiction.

For fixed n, the sequence u_{n,j}=g_nh^j, j\ge0, starts at g_n and tends to g_np. By whole-coset escape and (4.1),

Choose three distinct numbers strictly between 0 and \rho(a,c). For each number choose the first index j at which \rho(a,u_{n,j}) reaches it. The vanishing step size and compactness provide, after a common subsequence, three selected vertex sequences with distinct limits z_1,z_2,z_3\in B.

The conjugate k_n=g_nhg_n^{-1} sends each selected vertex to its successor, so asymptotically fixes these three sequences. If k_n escapes finite sets, pass to limits of k_n and k_n^{-1} in B. At least two of the z_i avoid the repelling limit; Proposition 4.3 would send both to the same attracting limit, a contradiction. Otherwise take a constant subsequence of the k_n. By continuity, that fixed conjugate of h fixes all three z_i, again a contradiction. This proves (4.5).□

Proof. Every invariant subset with at least two points is dense. Indeed, given a\in B, choose vertices g_n\to a and pass to g_n^{-1}\to b\in B. An invariant subset with at least two points contains z\ne b, and Proposition 4.3 gives g_nz\to a.

It remains to exclude a global fixed point c. Choose a\in B different from c and vertices g_n\to a. After passing to a subsequence, Proposition 4.3 and g_nc=c force g_n^{-1}\to c. Choose disjoint nonempty closed neighborhoods A,C of a,c in B. For large n,

Fix one such nonidentity g_n. Since \Gamma is torsion-free, it has infinite order. It cannot be parabolic: Lemma 4.4 would make the forward orbit of a point of A and the backward orbit of a point of C tend to the same point, which would belong to both disjoint closed sets. It is therefore loxodromic. Since all of \Gamma fixes c, the stabilizer statement cited above would make \Gamma virtually cyclic. An infinite virtually cyclic group has two ends, contrary to the established one-endedness of \Gamma.□

Proof. First suppose that F is a base coordinate. The level-length estimate of Lemma 4.1, with the identity translate, gives

for almost every t. Here the fixed neighborhood of an edge includes the boundaries of its incident cells. Traversing these boundaries along a simple dual path shows that the diameter of the endpoint set of a dual component D is at most

where C depends only on the finite presentation. Bounded local incidence bounds the multiplicity with which a nearby edge is used.

Fix a level with the finite sum above. The wall has finitely many dual components, and the dual adjacency graph is locally finite. Delete finite subsets D_n\subset S(F,t) increasing to the entire wall. At each stage only finitely many dual components remain: deleting finitely many vertices from a connected locally finite graph creates only finitely many components. Their endpoint-set closures cover E(F,t), since deleting finitely many isolated graph vertices does not change boundary accumulation. The sum of their diameters is at most

Their maximum diameter also tends to zero. These finite covers prove that \mathcal H^1_\rho(E(F,t))=0.

For a localized coordinate, its straddling edges are contained in a finite union of base-coordinate walls at the shifted levels appearing in its definition. The corresponding boundary accumulation set is therefore contained in the finite union of their accumulation sets. Each shifted level is admissible for almost every original level; translations and reflections of the level parameter preserve null sets. The result follows.□

Proof. We construct arbitrarily fine finite open covers of multiplicity at most two. Choose a finite grid of levels in finitely many coordinates. Each level avoids vertex values and satisfies Lemma 5.1. Let E be the union of the corresponding sets E(F,t). It is compact and has \mathcal H^1_\rho-measure zero.

The set E has arbitrarily small finite clopen partitions. Indeed, for x\in E, the distance map y\mapsto\rho(x,y) is Lipschitz, so its image on E has Lebesgue measure zero. There are arbitrarily small positive radii r for which the sphere of radius r about x misses E. The sets E\cap B_\rho(x,r) are then clopen in E. Compactness gives a finite small clopen cover, which becomes a disjoint partition by successively subtracting its previous members.

Each selected cut has a locally constant sign on the complement of its accumulation set. To define it at x\notin E(F,t), choose a neighborhood W of x in X that contains no endpoint of a straddling edge. Lemma 3.6 joins all vertices sufficiently near x by paths with vertices in W. Along such a path the sign of F-t cannot change. Hence all sufficiently near vertices have the same sign. This sign also applies at all sufficiently nearby boundary points, so it is locally constant. This definition is valid even when F(x)=t: it uses the signs on graph vertices approaching x.

The finitely many sign tuples partition B\setminus E into finitely many disjoint sets open in B. Within any one such set, each tested coordinate belongs to the closure of one grid interval. First choose enough coordinates that the remaining tail of the metric \rho=\sum_j c_j|F_j(x)-F_j(y)| is small, and then choose the grids sufficiently fine. Every sign-tuple set consequently has arbitrarily small diameter.

Partition E into finitely many small clopen sets. They are disjoint compact subsets of B, so they admit pairwise disjoint open neighborhoods in B, still of arbitrarily small diameter. These neighborhoods, together with the sign-tuple sets, cover B. At each point at most one neighborhood and at most one sign-tuple set occur. The cover therefore has multiplicity at most two. A Lebesgue number for any prescribed finite open cover shows that a sufficiently fine cover of this form refines it. This is the assertion \dim B\le1.□

Proof. Let d=(d_{ij}) be a simplicial 1-cocycle on the nerve of a finite open cover \{U_i\}_{i=1}^N of B, with d_{ii}=0. We show that it becomes a coboundary after refinement.

Labeling graph vertices. Choose \delta>0 so that every ball of radius 3\delta in B is contained in some U_i. For each vertex u\in\Gamma, choose a nearest point b(u)\in B and a label i(u) with

If u\to x\in B, then \rho(b(u),x)\le2\rho(u,x)\to0. Thus all labels occurring sufficiently near x contain a fixed neighborhood of x in B.

Define a cellular 1-cochain on \mathcal C by

when that pair occurs in the cover nerve, and by zero otherwise. Its coboundary has finite support. Otherwise choose infinitely many distinct cells where the coboundary is nonzero. Local finiteness forces their vertices to escape every finite set. Each cell boundary has at most three edges, whose lengths tend to zero, so a subsequence of the vertex sets converges to one point x\in B. All labels on every sufficiently late cell contain x. The nerve cocycle identity then gives zero around its boundary, a contradiction. Repeated labels or boundary edges cause no difficulty: the identity is a cellular sum, and d_{ii}=0.

The group-ring cocycle. We explain why the finite-support cochain \omega=d\alpha represents a group 2-cocycle. Start a free k\Gamma-resolution with the cellular chains of \mathcal C through degree two. This is possible because \mathcal C is simply connected: choose the next free module to map onto \ker\partial_2, and continue. For any finite-support scalar cellular cochain \omega, the formula

defines a k\Gamma-module cochain. The sum is finite for every finite cellular chain c, and its identity coefficient is \omega(c). Conversely, since the quotient has finitely many cells in each relevant degree, every module cochain through degree two is obtained in this way.

For every finite cellular 2-cycle z and every g\in\Gamma,

Thus \Phi_\omega vanishes on \ker\partial_2, so it is a cocycle on the extended free resolution. Lemma 2.4 gives H^2(\Gamma,k\Gamma)=0. There is therefore a module 1-cochain whose coboundary is \Phi_\omega. Taking identity coefficients yields a finite-support scalar 1-cochain \eta with

Ordinary simple connectivity of \mathcal C now supplies a scalar function l on its vertices such that

Returning to the boundary. Fix x\in U_i. Choose a neighborhood W of x in X avoiding all endpoints of edges in the support of \eta, and small enough that every vertex label occurring in W contains x. If uv is an edge with vertices in W, the sets U_i,U_{i(u)},U_{i(v)} meet at x. The cocycle identity and the equation dl=\alpha on this edge imply

By Lemma 3.6, this expression has one constant value on all vertices sufficiently near x. Denote that value by l_i(x). It is independent of the chosen neighborhood, since graph vertices approach every boundary point.

The function l_i is locally constant on U_i. Indeed, a smaller neighborhood on whose graph vertices the displayed expression is constant also computes the same germ at every sufficiently nearby boundary point. On an overlap U_i\cap U_j, the cocycle identity gives

Refine the cover by the open sets U_i\cap l_i^{-1}(a), for a\in k, omitting empty sets. On this finite refinement the original cocycle is the coboundary of the assignment a to the corresponding cover member. Its Čech class vanishes.□

References for the construction. The complete median pretree \Phi is the flow completion of P in [6, Theorem 3.19]. The embedding preserves the separation betweenness of the cut set. Adjacency and the simplicial-tree description are Lemmas 3.28 and 3.34. The compact supports are constructed in Section 5; their formula is Lemma 6.8, and the absence of a fixed point is Lemma 6.11. Full equivalence relations and their successive finite-interval collapses are treated in Lemmas 4.2–4.4. Their terminal quotient is identified with the dendritic quotient in Section 5, and Theorem 5.23 makes it a dendrite. Lemma 6.5 gives the induced convergence action; minimality follows from equivariance and surjectivity. Finally, the fixed-point assertions are Lemmas 6.9 and 6.10. All these constructions apply to a continuum before any local connectivity is known.□

Proof. Bowditch’s dendrite obstruction [5, Lemma 1.4] applies to a finitely presented one-ended group with no infinite torsion subgroup acting minimally by convergence on a dendrite, provided every two-ended subgroup over which the group splits is parabolic on that dendrite. Here there are no such splittings, so the last condition is vacuous; torsion-freeness implies the torsion condition. Proposition 6.1 therefore gives D(B) a single point.

By [6, Theorem 6.1], a minimal convergence action of a finitely generated one-ended group with no infinite torsion subgroup on a continuum has a nontrivial dendritic quotient whenever there is a cut point which is not parabolic. Its contrapositive shows that every point of P is parabolic. Here “parabolic point” means that its stabilizer is infinite and contains no loxodromic element. Any nonidentity element of this stabilizer has infinite order because \Gamma is torsion-free; the convergence-action classification then makes it parabolic.□

Proof. Every nonempty compact \gamma-invariant subset of B contains a, because the iterates of each point different from a converge to a. If q\in\Phi is fixed, R_B(q) is such a subset. For q\in P, uniqueness of the fixed point in B gives q=a. If q\notin P, then a\in R_B(q) and \phi(a)=a\ne q. Formula (6.1) therefore gives a\in\Lambda(q), as required.

Any two points equal or adjacent to a have a finite interval between them, contained in the union of their intervals to a. Thus a parabolic element fixing a point of P cannot have the two fixed points in the last assertion. A parabolic element whose fixed point is outside P cannot have them either, by Proposition 6.1(iv). The infinite-order classification and the loxodromic assertion in that proposition complete the proof.□

Proof. Lemma 6.2 says that the terminal quotient in (6.2) is a point. Let \alpha be the least ordinal for which \Phi/\sim_\alpha has a point fixed by \Gamma. Proposition 6.1(ii) gives \alpha\ne0. We shall rule out limits and successors greater than one.

A limit cannot be first. Suppose \alpha is a limit ordinal, and choose x in an invariant \sim_\alpha-class. For each member s of a finite generating set, x\sim_\alpha sx holds at some stage \beta_s<\alpha. Let \beta be the maximum of these finitely many stages. Invariance and transitivity of \sim_\beta give x\sim_\beta gx for every word g in the generators and their inverses. Thus the \sim_\beta-class of x is invariant, contrary to the choice of \alpha.

A higher successor creates a simplicial tree. Suppose \alpha=\beta+1 with \beta\ge1, and let \Xi\subset\Phi be an invariant \sim_\alpha-class. Its \sim_\beta-classes form a simplicial tree \Sigma=\Xi/\sim_\beta. The action on this tree has no fixed vertex, by the minimality of \alpha.

For adjacent classes U,V\in\Sigma, their union is full in \Phi. The facing-point observation supplies a point on at least one side. There cannot be facing points on both sides: they would be adjacent in \Phi, hence \sim_1-equivalent and therefore \sim_\beta-equivalent. This contradicts U\ne V. Exactly one side has a facing point, so an automorphism cannot interchange U and V. The action on \Sigma has no edge inversions.

We next show that every edge of \Sigma with infinite stabilizer is a leaf edge. This is the only step in which the dynamics on B enters the ordinal argument.

Let UV have infinite stabilizer, and write p\in V for its facing point. Choose a nonidentity stabilizing element \gamma; it has infinite order. Since the edge is not inverted, \gamma preserves U,V and fixes their uniquely specified point p. We claim that \gamma fixes another point outside V. For y\in U put

Fullness gives x\in[y,\gamma y]\subset U, and both x,\gamma x lie in the linearly ordered interval [p,\gamma y]. If x=\gamma x, the claim is proved. Otherwise, after replacing (\gamma,x) by (\gamma^{-1},\gamma x) if needed, we have x\in(p,\gamma x). The intervals [p,\gamma^n x], n\ge0, are then strictly increasing. Their union is a full directed arc with first point p and no last point. Completeness supplies an endpoint z with

The endpoint is unique in a median pretree [6, Lemma 3.13 and the following observation]. The union in (6.4) is \gamma-invariant, so z is fixed. Since x\in(p,z) but x\notin V, fullness of V implies z\notin V. This proves the claim.

Let q\notin V be the second fixed point. The points p,q belong to different \sim_\beta-classes; as \beta\ge1, their original interval is infinite. Lemma 6.3 makes p terminal in \Phi. For every v\in V and u\in U, formula (6.3) puts p\in[u,v]. Since p\ne u, terminality forces v=p. Thus V=\{p\}. This vertex is a leaf of \Sigma: if it had two neighbors, p would lie strictly between representatives of them in \Phi, again contradicting terminality.

Delete every edge with infinite stabilizer, together with its leaf endpoint. The remaining vertices form a nonempty invariant subtree. Indeed, removing selected leaves preserves all paths between remaining vertices. If nothing remained, the original tree would have at most two vertices. Such an action without inversions has a fixed vertex, which has already been excluded. All remaining edges have finite stabilizer, and a fixed vertex of the remaining tree would also be fixed in \Sigma.

A finitely generated one-ended group acting without inversions on a simplicial tree with finite edge stabilizers fixes a vertex [6, Lemma 6.6]. Applying this fact gives the desired contradiction.

The least ordinal is therefore \alpha=1, which is exactly the assertion of the lemma.□

Proof. Let C\subset\Phi be the invariant finite-interval class supplied by Lemma 6.4. It contains more than one point, because \Phi has no fixed point. Therefore its simplicial tree has an edge, and Proposition 6.1(i) places one endpoint of that edge in P.

Choose p\in C\cap P. The orbit \Gamma p is dense in B and lies in C. If q\in P, take an open separation of B\setminus\{q\}. Both sides are nonempty open subsets of B, so each meets \Gamma p. For orbit points a,b on opposite sides, separation gives q\in[a,b]_{\Phi}. Since C is full, q\in C. We have proved P\subset C.

Let \mathcal T be the subtree of C spanned by P, that is, the union of the intervals between its points. Its intervals agree with those of \Phi, proving the asserted separation compatibility. The adjacency property gives the indicated bipartition and rules out edge inversions. A vertex outside P cannot be a leaf of a tree spanned by P.

To prove minimality, let \mathcal S\subset\mathcal T be a nonempty invariant subtree. A singleton would be a fixed point of \Phi. Otherwise \mathcal S contains an edge, hence a point of P. Its orbit is again dense in B; the separation argument of the preceding paragraph gives P\subset\mathcal S. Thus \mathcal S=\mathcal T.

Choose a vertex o and a finite generating set S for \Gamma. The union of the finitely many finite paths [o,so], s\in S, is a finite subtree after adjoining o. Its translates form a connected invariant subtree and therefore cover \mathcal T by minimality. Consequently the quotient has finitely many vertex and edge orbits.

Finally suppose an edge had finite stabilizer. Collapsing the edges outside its orbit gives a nontrivial tree with one edge orbit and a minimal action without inversions. Bass–Serre theory then yields a nontrivial splitting over that edge stabilizer, contrary to the hypotheses. Thus every edge stabilizer is infinite. The parabolic assertion is Lemma 6.2.□

Proof. The finite quotient of \mathcal T gives finitely many \Gamma-orbits in P. Choose representatives p_0,\ldots,p_k and a finite symmetric generating set S of \Gamma. Start with the graph formed by all translates of the edges from p_0 to sp_0, for s\in S, and from p_0 to p_i, for 1\leq i\leq k. Omit loops. The generator edges connect the orbit of p_0, and the other edges connect every orbit to it, so this graph is connected. It has finitely many edge orbits.

Replace each edge by the consecutive pairs of P-vertices along the tree interval between its endpoints. These intervals are finite, and the replacement is equivariant. Thus the resulting graph K^* is still connected, still has finitely many edge orbits, and has the required distance-two property.

For an edge ab of K^*, let a-s-b be its tree interval. If s=r, then a,b\in R and \pi fixes both endpoints. If s\ne r, then a and b lie in the same component of \mathcal T\setminus\{r\}, so \pi(a)=\pi(b). This proves the projection assertion. Projecting a path between two members of R and deleting stationary steps gives a path in K, proving its connectedness.

If a group element carries one edge of K to another, it carries their unique tree midpoints to one another. Both midpoints are r, so that element lies in H. Consequently, each \Gamma-orbit of edges in K^* meets K in at most one H-orbit. There are therefore finitely many H-edge orbits in K. Since K is connected and has at least two vertices, every vertex is incident to an edge; hence there are also finitely many H-vertex orbits. Finally, for p\in R, the stabilizer H_p is the stabilizer of the tree edge rp, which is infinite.□

Proof. Using the finitely many orbit representatives chosen in Lemma 7.1, write a_n=g_np_{i_n} and b_n=h_np_{j_n}. We may choose g_n and h_n to escape every finite subset of \Gamma: each representative has infinite stabilizer, so there are infinitely many choices for each representing element. Equation (7.1), applied to this finite list of representatives, gives g_n\to x and h_n\to x in X.

By Lemma 3.6, there are Cayley graph paths from g_n to h_n whose vertices tend uniformly to x. Fix finite K^*-paths from p_0 to sp_0 for each Cayley generator s, and from p_0 to each representative p_i. Replace a Cayley step g\longrightarrow gs by the g-translate of the first appropriate path. At the endpoints, use the translates of the paths to p_{i_n} and p_{j_n}. The result is a K^*-path from a_n to b_n.

Only finitely many points t\in P occur in the fixed path templates. For each such point, \rho(gt,g)\to0 as g escapes, by (7.1). The Cayley path vertices escape finite sets uniformly, because they tend uniformly to the boundary point x. Hence all vertices of the transferred paths tend uniformly to x as well.

For the last assertion, Lemma 7.1 says that every nonconstant projected edge was already an edge of the original path. When the projected endpoints differ, every vertex of the projected path after stationary steps are removed is incident to one of these edges. It therefore occurred in the original path.□

Proof. By the minimality established in Proposition 4.5, the nonempty invariant set P is dense in B. We first determine the limit of \pi(p_n) whenever p_n\in P tends to a point of B.

Let x\in B\setminus L, and choose a neighborhood O of x in B disjoint from L. The map \pi is constant on P\cap V for some neighborhood V\subset O of x. Otherwise, we could choose a_n,b_n\to x with \pi(a_n)\ne\pi(b_n). Lemma 7.2 would join them by paths whose vertices eventually all lie in O. Their nonconstant projected paths contain vertices in R that also belong to the original paths. This contradicts O\cap R=\varnothing.

Now let x\in L and a_n\in P tend to x. Choose b_n\in R tending to x, and join a_n to b_n by the paths supplied by Lemma 7.2. Whenever \pi(a_n)\ne b_n, the same Lemma places \pi(a_n) on the original path. Therefore \pi(a_n)\to x. The indices for which \pi(a_n)=b_n satisfy the same conclusion directly.

Take the closure of the graph of \pi in B\times L. Density of P and compactness of L show that this closed set projects onto B. The preceding two paragraphs show that it has exactly one point over each point of B. Its projection onto B is consequently a continuous bijection from a compact space to a Hausdorff space, hence a homeomorphism. The second coordinate defines the desired continuous map \mathfrak r. It is the identity on L, and the first paragraph makes it locally constant on B\setminus L.

Since B is connected, its image L is connected. The set R has at least two points, so L is nondegenerate. The dimension bound follows because L is a closed subset of the compact metric space B. If i:L\hookrightarrow B denotes inclusion, then \mathfrak r\circ i=\operatorname{id}_L. Functoriality of Čech cohomology gives i^*\mathfrak r^*=\operatorname{id}_{\check H^1(L;\mathbb F_2)}. Thus \mathfrak r^* injects this group into \check H^1(B;\mathbb F_2)=0.

It remains to rule out a cut point of L. Suppose that L\setminus\{q\}=U\sqcup V is a separation into nonempty open sets. Density supplies a\in R\cap U and b\in R\cap V. The extra fiber

is closed in B\setminus\{q\} by continuity. It is also open there: it lies in B\setminus L, where \mathfrak r is locally constant. Consequently,

is a separation, with a and b on opposite sides. This would make q a cut point of B separating two members of R, contrary to the separation property of the tree noted above. Hence L has no cut points.□

Proof. If P=\varnothing, take H=\Gamma, Y=X, L=B, and let K be the original Cayley graph, with I=H. The continuum properties are those already proved for B, and absence of cut points is the definition of this case. The closure, convergence, finite orbit, and near-path assertions follow from the construction of X, Proposition 4.3, and Lemma 3.6. Every boundary point is approached by distinct group vertices, since the group vertices are isolated and dense in X.

For completeness, the comparison assertion in this case also follows from convergence. If it failed for a fixed w\in H, choose escaping h_n with \rho(h_nw,h_n) bounded away from zero. After passing to a subsequence, compactness gives h_n\to a\in B and h_n^{-1}\to b\in B. Since w\in H is different from b, Proposition 4.3 gives h_nw\to a, a contradiction. The last assertion has no content in this case.

If P\ne\varnothing, use the objects constructed above:

and let K be the graph of Lemma 7.1. The continuum and finite orbit assertions follow from that Lemma and Proposition 7.3. We verify the remaining interfaces explicitly.

First, Y=\overline H^{\,X}. Fix p\in R. If an escaping sequence h_n\in H converges to a boundary point, then (7.1) and h_np\in R show that its limit belongs to L. Conversely, for any p\in R, the infinite stabilizer H_p supplies distinct elements h_n with h_np=p. Equation (7.1) gives h_n\to p. Thus R\subset\overline H^{\,X}, and taking closures gives L\subset\overline H^{\,X}. No point of \Gamma\setminus H belongs to this closure, since all group vertices are isolated. This proves the equality and compactness of Y. The subgroup H preserves both L and Y, and the convergence assertion is the restriction of Proposition 4.3.

The set R is dense in the nondegenerate continuum L, which has no isolated points. It follows that every point of L is approached by distinct members of R: otherwise a sufficiently small neighborhood would meet R in at most one point and make that point isolated in L. Equation (7.2) is exactly (7.1) restricted to H.

For the near-path assertion, take two sequences in R tending to the same point of L. Lemma 7.2 joins them by K^*-paths whose vertices tend uniformly to that point. Project to R. If the endpoints differ, the projected paths use only vertices of the original paths; if they agree, use the constant path. Thus the projected K-paths still tend uniformly to that point. In metric spaces this sequential statement gives the stated neighborhood formulation: a failure for some W would yield endpoints in successively smaller balls that cannot be joined inside W.

Finally, H_p is infinite by Lemma 7.1. Since \Gamma is torsion-free, it contains an infinite-order element. The cut-point stabilizer conclusions in Lemma 6.2 make this element parabolic with fixed point p. Its powers have the asserted dynamics by Lemma 4.4.□

Proof. Put U=L\setminus\{x\}. We prove directly, using finite-cover cohomology, that there are locally constant maps f_U:U\to k and f_V:V\to k such that

Choose open sets U_0,V_0 covering L whose closures are contained in U,V, respectively. The set C=\overline{U_0}\cap\overline{V_0} is a compact subset of U\cap V. Its two subsets C\cap u^{-1}(0) and C\cap u^{-1}(1) are disjoint compact sets. Choose \delta>0 smaller than their distance when both are nonempty; if either is empty, any positive \delta suffices.

Take a finite open cover (W_i) of L refining (U_0,V_0), with each W_i of diameter less than \delta, and give it a parent P_i\in\{U,V\} according to this refinement. On every nonempty W_i\cap W_j, define c_{ij}=0 if P_i=P_j, and define c_{ij} to be the value of u there otherwise. This latter value is constant: the intersection lies in C and cannot meet both of its two compact parts. At a triple intersection either all parents agree, or exactly two of the three transition values equal the same value of u. Thus (c_{ij}) is a 1-cocycle on the nerve of (W_i).

Since \check H^1(L;k)=0, some finite open refinement (Z_\alpha), with refinement map r, admits constants a_\alpha\in k satisfying

For z\in U\cap Z_\alpha, set f_U(z)=a_\alpha if P_{r(\alpha)}=U, and set f_U(z)=a_\alpha+u(z) otherwise. The displayed identity makes these definitions agree on overlaps. They are locally constant. Define f_V by the same rule, with U,V interchanged. Then f_U+f_V=u on U\cap V.

Since x is not a cut point, U is connected, so f_U has one constant value, say c. The map f_V+c is the required extension. Uniqueness follows because a nondegenerate continuum has no isolated points, so V\setminus\{x\} is dense in V.□

Proof. There is a neighborhood W of x such that (I\setminus T)\cap W\subset\{x\}. Since x is a limit of distinct vertices of I, it is consequently a limit of distinct vertices of T as well.

Suppose first that x is not a deleted vertex. After shrinking W, it contains no deleted vertex. The near-path property then puts all vertices sufficiently near x in one component of K[T]. There are allowed vertices arbitrarily near x, so this component is unique. The same neighborhood, together with approximation by allowed vertices, shows that the component assignment is locally constant near x. When x itself is an allowed vertex, it belongs to this component.

The remaining case has I\subset L and x\in I\setminus T. Choose an open neighborhood V of x in L containing no other deleted vertex. By the preceding case, every y\in V\setminus\{x\} has a component label, and these labels form a locally constant map

Every map from this label set to k, composed with q, extends across x by Lemma 8.1. We claim this forces q itself to be constant sufficiently near x.

To see this, call a label recurrent if it occurs arbitrarily near x. If two labels are recurrent, a coloring separating them contradicts the extension property. If exactly one label is recurrent but q is not eventually constant, color that label 0 and all others 1 to obtain the same contradiction. If no label is recurrent, a sequence of points tending to x can be chosen with pairwise distinct labels: at each step, first exclude the finitely many labels already chosen. Color these successive labels alternately 0 and 1. This again contradicts extension as a locally constant map. Thus one component label occurs throughout a small punctured neighborhood of x. Every allowed vertex there belongs to its own label by the first case, so all these vertices belong to that component.

This proves existence at a deleted vertex too. Approximation by allowed vertices proves uniqueness and, on shrinking the neighborhood once more, local constancy of Q_T there.□

Proof. We first record a uniform consequence of convergence. For every r,\varepsilon>0, all but finitely many g\in H admit a point b_g\in L such that

Otherwise choose distinct exceptions g_n and pass to a subsequence with g_n\to a\in L and g_n^{-1}\to b\in L. Uniform convergence on the compact set Y\setminus B_\rho(b,r) gives a contradiction.

Choose r_i\downarrow0 with r_i<\operatorname{diam}(L)/3, and put \varepsilon_i=2^{-i}. Apply the preceding observation with r=r_i/4 and \varepsilon=\varepsilon_i, obtaining a finite exceptional set E_i\subset H. Uniform continuity of the finitely many maps in E_i permits a common choice 0<s_i<r_i/4 such that

Choose finitely many centers y_{ij}\in L whose s_i/2-balls cover L. At each center take the seed annulus

Its sides are disjoint and have distance at least r_i-s_i. The positive side contains its center. The negative side meets L because every point of L is at distance at least \operatorname{diam}(L)/2 from some point of L. The sides cannot cover L, since their nonempty closed intersections with L would disconnect it.

Let \mathcal A consist of all distinct translates of these annuli and their reversals. A ball of radius r_i/4 cannot meet both sides of a seed annulus, whose mutual distance exceeds 3r_i/4. For g\notin E_i, one side therefore lies outside the exceptional ball for g and has image of diameter less than \varepsilon_i. For g\in E_i, the positive side has that property by the choice of s_i. Consequently every translated level-i annulus has at least one side of diameter less than \varepsilon_i.

For a fixed seed and fixed \delta>0, only finitely many distinct translates can have both sides of diameter at least \delta. If not, choose corresponding distinct group elements and a convergence subsequence with repeller b. At least one of the two closed seed sides excludes b, so its images have diameter tending to zero. There are only finitely many seeds at each level, and levels with \varepsilon_i<\delta contribute no annuli with two such large sides. This proves the first assertion.

For the second assertion, let F be closed and x\in L\setminus F. At a sufficiently fine level, choose a center y_{ij} with \rho(x,y_{ij})<s_i/2 and r_i+s_i/2<\operatorname{dist}(x,F) when F is nonempty. The negative side contains F, and the positive side contains a neighborhood of x. Suppose an annulus already chosen has positive side containing B_\rho(x,\eta). Choose a finer level with r_j+s_j/2<\eta, and a center y within s_j/2 of x. Then B_\rho(y,r_j)\subset B_\rho(x,\eta), so the new annulus is larger in the order <, and its positive side again contains a neighborhood of x. Repeating gives any prescribed finite length.□

Proof. Translation carries annuli and chains bijectively to annuli and chains, which proves equivariance. Suppose h\in F\cap H and v\in H. As L is closed and excludes h,v, the number

is positive. Throughout a chain from F to v, every negative side contains h and meets L, and every positive side contains v and meets L. Both sides therefore have diameter at least \delta. Lemma 8.3 supplies only finitely many such annuli, and a strict chain cannot repeat one. This proves finiteness. A finite nonzero supremum of possible integer lengths is attained. The final assertion is precisely the second part of Lemma 8.3, using its terminal neighborhood.□

Proof. By equivariance, D(F_n,1)=D(A,g_n). In any chain from F_n to 1, every positive side contains 1. If the chain has more than N(\epsilon,d_0) members, one of its negative sides has diameter less than \epsilon. This side contains F_n. Hence \operatorname{diam}F_n\to0.

Lemma 8.4 gives D(A,g)<\infty for each fixed g\in H. Thus g_n escapes every finite subset of H, as does g_n^{-1}. Since 1\in A, we have g_n^{-1}\in F_n. Compactness of Y and isolatedness of its group vertices give a subsequence g_n^{-1}\to q\in L. The diameter conclusion now gives the assertion.□

Proof. Choose n large enough that F_n\subset B(q,r/2) and \ell=D(F_n,1)>N(r/2,d_0)+1. Take a chain \sigma_1<\cdots<\sigma_\ell of maximum length from F_n to 1, and put U=B(q,r). If \sigma_j^-\not\subset U, its negative side contains a point of F_n and a point outside U, so has diameter at least r/2. Its positive side has diameter at least d_0. There are therefore at most N_-=N(r/2,d_0) such annuli. Because negative sides increase along a chain, these form a terminal segment. Remove that segment and one further annulus. The retained initial chain has at least \ell-N_--1 members. The negative side of the next annulus lies in U, so the last retained positive side contains Y\setminus U. In particular, this retained chain already ends on the side containing any b under consideration.

Let t=D(\{1\},b), and take a maximum chain \tau_1<\cdots<\tau_t from 1 to b when t>0. Put V=Y\setminus\overline{B(q,2r)}. A positive side not contained in V contains both b and a point at distance at most 2r from q, so has diameter at least r. Every negative side contains 1, so has diameter at least d_0. Thus at most N_+=N(d_0,r) positive sides fail to lie in V; these form an initial segment. If t>N_++1, remove this initial segment and one further annulus. The retained final chain has at least t-N_+-1 members, and its first negative side contains Y\setminus V.

Since U\cap V=\varnothing, the last positive side of the first retained chain and the first negative side of the second cover Y. The defining order relation therefore concatenates the two chains. Their total length is at least \ell+t-N_--N_+-2. If t\le N_++1, the first retained chain alone has at least this length. This proves (9.1), including a constant and threshold independent of b.

For the final assertion choose r with 4r<\rho(q,z). Eventually \rho(b_n,q)\ge3r, so the inequality applies. For any prescribed integer N, Lemma 8.3 supplies a chain of length N from \{1\} to a neighborhood of z. Eventually this neighborhood contains b_n, proving D(\{1\},b_n)\to\infty.□

Proof. For m=0 the first assertion follows from I_0=I\setminus A. Suppose m\ge1. Suppose that distinct u_n\in I\setminus I_m tend to x\in L\setminus A. Eventually u_n\notin A, so choose a_n\in C(u_n) with D(A,a_n)<m. Since there are only finitely many orbit representatives and the u_n are distinct, a_n escapes every finite subset of H. The comparison \rho(gp_i,g)\to0 in Proposition 7.4, applied to the finite list of representatives, gives a_n\to x. A chain from A to a neighborhood of x of length m, supplied by Lemma 8.3, contradicts D(A,a_n)<m for large n. The assertions about Q_m and approximation follow from Lemma 8.2.

Finally, for each u\in E\setminus A choose a single a_u\in C(u). The depth D(A,a_u) is finite. Taking m larger than all these finitely many depths excludes every vertex of E from I_m.□

Proof. If no such finite set and index exist, choose indices tending to infinity and pairwise distinct vertices; relabel them so that

Compactness gives a further subsequence w_n\to z\in E. The limit lies in L: this is automatic when I\subset L, and follows from isolatedness of group vertices when I=H. Since F_n tends to q\notin E, eventually w_n\notin F_n. The deletion of g_nw_n therefore gives a representative g_nb_n\in C(g_nw_n) such that

After restricting to one orbit representative we have w_n=b_np_i. Distinctness of the w_n forces b_n to escape finite subsets of H, and the comparison with group vertices gives b_n\to z\ne q. Lemma 9.2 now yields

for large n, a contradiction.□

Proof. Choose an open neighborhood O of z in Y whose closure misses q. Apply Lemma 9.4 to the compact set E=\overline O. There are a finite set S\subset I and an index n_0 such that

The deleted vertices I\setminus T have no distinct accumulation at any point of O\cap L. Lemma 8.2 therefore gives a neighborhood V_z\subset O\cap L on which the component of K[T] approached by vertices is constant; denote it by C. Shrink V_z if needed. Since F_n\to\{q\}, all g_nw for w\in V_z lie outside A once n is large.

Fix such an n. Every w\in V_z is approached by vertices of C. Their images approach g_nw and lie in the connected subgraph g_nC of K[I_{m_n+1}]. The defining property of Q_{m_n+1} therefore identifies \ell_n(w) with the component containing g_nC, for every w\in V_z. This gives the asserted neighborhood and threshold, simultaneously for all its points.

On L\setminus\{q\}, declare z\sim z' if their labels agree for all sufficiently large n. Eventual equality is an equivalence relation, even though the label set can change with n. The first part shows that every equivalence class is open. Since q is not a cut point, L\setminus\{q\} is connected, so this partition has one class.□

Proof. If the first assertion fails at arbitrarily large depths, choose m_n\to\infty and g_n violating it with D(A,g_n)\ge m_n. Pass to a subsequence with g_n^{-1}A\to\{q\}. Two fixed points of Z avoid q. By Lemma 9.5, their images under g_n eventually lie outside A and have the same Q_{m_n+1} label, a contradiction. This also proves uniqueness of the majority label, since two different labels cannot each occur at two points of a three-point set.

After choosing a threshold for the first assertion, suppose that the second fails at arbitrarily large m. Choose a violating sequence and, since S is finite, restrict to one fixed s\in S. Again pass to a subsequence with g_n^{-1}A\to\{q\}. Choose two points of Z\setminus\{q\} and two points of sZ\setminus\{q\}. Lemma 9.5 makes the labels of all four images under g_n equal for all sufficiently large n. They determine the majority labels of both g_nZ and g_nsZ, contradicting the chosen inequality.□

Proof. First suppose I\subset L. We claim that for every v\in I_m and every g\in C(v),

Let v=gp_i. For all j\ge0, the element gh_i^j also belongs to C(v). The minimum in (9.2) gives D(A,gh_i^j)\ge m for every j. The move h_i belongs to S, so Lemma 9.6 gives

Here g and m are fixed. As j\to\infty, the three anchors gh_i^jZ tend to v. Since v\in L\setminus A and Q_{m+1} is locally constant there, all three anchors eventually have label Q_{m+1}(v). This proves (9.4). The label at v is defined even when v\notin I_{m+1}; it records the component approached by allowed vertices near v.

If v,w\in I_m are adjacent, choose their representatives g,gs using the finite edge list. Both depths are at least m by the minimum convention. Lemma 9.6 and (9.4) give Q_{m+1}(v)=Q_{m+1}(w). Thus Q_{m+1} is constant along every path in K[I_m]. If Q_m(x)=Q_m(y), choose vertices in this component sufficiently near x and y that their Q_{m+1} labels equal those of x and y. Such vertices exist by Lemma 9.3 and local constancy of Q_{m+1}. A path in their common component proves (9.3).

Now suppose I=H. There is a single representative p_1=1, and C(v)=\{v\}. The finite edge moves and Lemma 9.6 imply that J_m is constant along paths of K[I_m]. For a fixed x\in L\setminus A and fixed m\ge M, we claim that

These vertices belong to I_m when sufficiently near x, by Lemma 9.3. If the claim fails, take a sequence g_n\to x with the wrong label, and pass to a subsequence g_n^{-1}\to b\in L. The convergence property in Proposition 7.4 sends at least two fixed points of Z\setminus\{b\} to x under g_n. Local constancy of Q_{m+1} near x gives the asserted majority label, a contradiction. Approximating x and y inside their common Q_m component and using constancy of J_m along a connecting path now proves (9.3) in this case as well.□

Proof. Fix x\in L and a relatively open neighborhood U\subset L of x. Choose an open neighborhood O of x in Y with 1\notin O and \overline O\cap L\subset U, and apply the preceding construction to A=Y\setminus O. Let M be the threshold in Lemma 9.7. By local constancy of Q_M, there is a relative neighborhood V\subset L\setminus A of x such that Q_M(y)=Q_M(x) for all y\in V. Induction using Lemma 9.7 gives

Fix y\in V. For each m\ge M choose vertices a_m,b_m\in I_m in this common component with \rho(a_m,x)<1/m and \rho(b_m,y)<1/m, and choose a finite path P_m in K[I_m] joining them. Every vertex of P_m lies in O. We next show that the maximum distance between consecutive vertices of these paths tends to zero.

Use the finite list of edge representatives (u_e,v_e). The comparison with group vertices gives

uniformly over this finite list. For each \epsilon>0, choose a finite F\subset H outside which every one of these edge distances is less than \epsilon. There are only finitely many endpoints of the edges (gu_e,gv_e) with g\in F. By Lemma 9.3, none of these endpoints belongs to I_m once m is large. Consequently every edge in K[I_m] then has endpoint distance less than \epsilon. This proves the mesh assertion without any local-finiteness assumption on K.

Regard P_m as its finite vertex set. The space of nonempty closed subsets of the compact metric space Y is compact in the Hausdorff metric, so pass to a subsequence with P_m\to C. The limit contains x,y and lies in \overline O. It also lies in L: this is immediate if I\subset L; if I=H, Lemma 9.3 excludes each fixed group vertex from P_m for large m, and isolatedness of that vertex excludes it from the Hausdorff limit.

The limit C is connected. Otherwise write it as a union of disjoint nonempty compact sets C_0,C_1, whose distance is some \delta>0. For large m, the set P_m lies in the union of their disjoint \delta/3-neighborhoods and meets both. A path with these vertices must have two consecutive vertices in different neighborhoods, at distance at least \delta/3. This contradicts the mesh assertion. Thus C is a connected subset of \overline O\cap L\subset U containing x and y.

Every y\in V therefore belongs to the component of x in U. Repeating the argument at every point of every open subset of L shows that its components are open. This is local connectedness of L.□

Proof. Represent a class on A by a cocycle on the nerve of a finite relative open cover \{V_i\} of A. Choose open sets W_i\subset Z with W_i\cap A=V_i. The cover consisting of the W_i and Z\setminus A has a finite open refinement \mathcal U of multiplicity at most two. Its nerve is a graph. Retain the indices of the cover members when restricting to A, even if two restrictions coincide. After empty restrictions are discarded, the nerve of \mathcal U|_A is a subgraph of this graph. A refinement map pulls the original cocycle back to that subgraph. Extend the resulting 1-cochain by zero on all remaining edges of N(\mathcal U). Since there are no 2-simplices, the extension is a cocycle. Its Čech class restricts to the prescribed class on A.□

Proof. Every connected open set V\subset L is covered by connected open sets W whose closures lie in V and whose diameters are at most any prescribed \varepsilon>0. Indeed, choose a sufficiently small metric ball about each point with closure in V, then a connected open neighborhood inside a still smaller ball. The intersection graph of this cover is connected: otherwise the unions belonging to its graph components would separate V. Thus any two points of V can be joined by a finite chain W_1,\ldots,W_m from this cover, with the first point in W_1, the last in W_m, and W_i\cap W_{i+1}\ne\varnothing. We may arrange m\ge2 by repeating a member if necessary.

Starting with [0,1], the set U, and endpoint values a,b, construct successive finite interval partitions \mathcal P_n as follows. Associate to every interval I\in\mathcal P_n a connected open set V_I containing its two already assigned endpoint values. For n\ge1 require \operatorname{diam}V_I\le2^{-n}. Inside V_I, choose the finite chain above with closures in V_I and diameters at most 2^{-(n+1)}. Subdivide I into m equal intervals and associate these to the chain members. At each new division point assign a point of the corresponding consecutive intersection. Keep the old endpoint values. These assignments agree at shared endpoints, and both endpoints of every child interval lie in its associated open set.

Every subdivision has at least two children, so \operatorname{mesh}(\mathcal P_n)\le2^{-n}. The set D of all partition endpoints is therefore dense in [0,1], and the compatible assignments define f:D\to L. Nesting ensures that f(D\cap I)\subset V_I for every I\in\mathcal P_n. For fixed n\ge1, let \delta_n>0 be the minimum length of a cell of \mathcal P_n. If s,t\in D and |s-t|<\delta_n, they lie in the same or adjacent cells. In the adjacent case use the assigned value at their common endpoint. In either case,

Thus f is uniformly continuous and extends continuously to [0,1] by completeness of L. For each cell I\in\mathcal P_1, its extended image lies in \overline{V_I}\subset U. Hence the entire path lies in U, and its endpoints are a,b.□

Proof. The continuum contains no embedded circle. Such a circle C would be closed, and Lemma 10.1 would give a surjection from \check H^1(L;\mathbb F_2)=0 onto \check H^1(C;\mathbb F_2)\cong\mathbb F_2.

A locally connected compact metric continuum is a Peano continuum. Lemma 10.2 and Moore’s theorem give an arc between any two distinct points of a connected open subset of L. Choose an arc between two distinct points a,b of L and an interior point p of that arc. If p were not a cut point, the connected open set L\setminus\{p\} would contain a second arc between a and b. The component of the first arc minus the second that contains p is an open subarc with both endpoints on the second arc. Its closure, together with the subarc of the second joining those endpoints, is an embedded circle. This contradiction shows that p is a cut point.□

Proof of Theorem 1.1. Suppose that b_1^{(2)}(\widetilde Q)>0. The finite-energy construction, the boundary topology proved in Propositions 5.2 and 5.3, and the reduction of Proposition 7.4 give a nondegenerate compact metric continuum L without cut points, with \dim L\le1 and \check H^1(L;\mathbb F_2)=0. Proposition 9.8 makes L locally connected. Lemma 10.3 is a contradiction. Hence b_1^{(2)}(\widetilde Q)=0.

Lemma 2.3 gives b_0^{(2)}=b_4^{(2)}=0, b_3^{(2)}=b_1^{(2)}, and vanishing in degrees above four. Its duality argument includes nontrivial orientation characters. Consequently b_p^{(2)}(\widetilde Q;\mathcal N(\Gamma))=0 for every p\ge0 with p\ne2, as asserted.□

Proof. For a finite CW complex the L^2 Euler–Poincaré formula is

[21, Theorem 1.35(2)]. Theorem 1.1 leaves only the degree-two term, whose von Neumann dimension is nonnegative. The characterization of zero follows from the same theorem. For a topological manifold, pass to the finite homotopy model of Lemma 2.1; both invariants are preserved.□

Proof. The topological L^2-signature theorem gives \sigma(M)=\sigma^{(2)}(\widetilde M) [23, Theorem 0.2]. The latter is the difference of the positive and negative von Neumann dimensions of the lifted middle-dimensional intersection form, so |\sigma^{(2)}(\widetilde M)|\le b_2^{(2)}(\widetilde M). Corollary 11.1 now proves the inequality.□

Proof. Lück’s approximation theorem [20, Theorem 0.1] identifies the limit with b_p^{(2)}(\widetilde Q). Apply Theorem 1.1 and Corollary 11.1. A finite homotopy model of a topological manifold lifts to a homotopy equivalence on each corresponding finite cover, giving the last assertion.□