2130fff
% The Singer Conjecture in Dimension Four
% A complete, self-contained proof
%
% What this file is.
% This is the FULL proof that, for a closed connected aspherical
% topological 4-manifold M, the $L^2$-Betti numbers of the universal
% cover vanish outside the middle dimension (degree 2). We prove the
% more general statement for finite connected aspherical integral
% Poincare complexes of formal dimension four.
%
% Who this is for.
% A motivated high school student. Every new word is defined the first
% time it appears. Every lemma has a proof. No citation stands in for
% an argument. Sentences are short. Sections are small.
%
% How the proof goes (see Section 2 for details).
% Assume b_1 > 0. Build a large compact space X = Gamma + B whose
% "boundary at infinity" B is a nondegenerate connected compact metric
% space. Show dim B <= 1 and Cech H^1(B;F_2) = 0. If B has no cut
% points set L = B; otherwise carve out a subcontinuum L without cut
% points, still with dim <= 1 and H^1 = 0. Then prove L is locally
% connected, using a "push paths to infinity" argument with annuli.
% Locally connected + dim <= 1 + H^1 = 0 + no circle = dendrite; but
% every nondegenerate dendrite has a cut point. Contradiction. So
% b_1 = 0.
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\title{The Singer conjecture in dimension four}
\author{OpenAI}
\date{September 25, 2026}
\begin{document}
\maketitle
\begin{abstract}
The Singer conjecture says that on the universal cover of a closed
aspherical manifold, the $L^2$-Betti numbers of the cover sit only in the
middle dimension. In dimension four the only open case is degree one.
We prove the degree-one vanishing, first for finite connected aspherical
integral Poincare complexes of formal dimension four (any orientation
character) and hence for all closed connected aspherical topological
four-manifolds, smoothable or not.
\end{abstract}
\tableofcontents
%=============================================================
\section{What we are proving}
\label{sec:introduction}
%=============================================================
\subsection*{Background: what a manifold is, and what "aspherical" means}
A \emph{closed $n$-manifold} is a compact space without boundary that
locally looks like $\mathbb R^n$. A \emph{CW complex} is a space built
by gluing together cells of various dimensions; every manifold that is
not too pathological admits such a description. A space is
\emph{aspherical} if its universal cover is contractible. The familiar
example is the torus $T^n = \mathbb R^n / \mathbb Z^n$: its universal
cover is $\mathbb R^n$, which can be squashed to a point. On the other
hand the sphere $S^2$ is \emph{not} aspherical: its universal cover is
itself, and $S^2$ is not contractible.
A manifold (or CW complex) that is aspherical is completely determined
in homotopy theory by its fundamental group $\Gamma = \pi_1(M)$: the
cover is contractible, so all of the topology of $M$ can be read off
from the group action on that contractible space. This is why aspherical
manifolds are interesting: their geometry lives in the group.
\subsection*{What $L^2$-Betti numbers are (informally)}
Ordinary Betti numbers $b_p(M)$ count the independent "$p$-dimensional
holes" of a space $M$. When $M$ has an infinite fundamental group, its
universal cover $M \to \widetilde M$ is an infinite space, and its
Betti numbers are infinite. The trick is to \emph{normalize} them by
the size of the group, using the "von Neumann dimension", which is a
notion of dimension for spaces with a group action that is allowed to
be a fraction. The result $b_p^{(2)}(\widetilde M)$ is a nonnegative
real number, finite for finite CW complexes with finite Betti numbers
of the group over $\mathbb Z$. On the torus $T^2$, for example,
$b_0^{(2)} = b_2^{(2)} = 0$ and $b_1^{(2)} = 1$; the Euler
characteristic is the alternating sum, $\chi(T^2) = 0$, which agrees with
$b_0^{(2)} - b_1^{(2)} + b_2^{(2)} = -1 + 1 = 0$. One can show for every
finite CW complex that $\chi(M) = \sum_p (-1)^p b_p^{(2)}(\widetilde M)$.
\subsection*{The Singer conjecture}
\begin{conjecture}[Singer, as formulated for aspherical manifolds]
Let $M$ be a closed aspherical $n$-manifold with infinite fundamental
group. Then
\[
b_p^{(2)}(\widetilde M) = 0
\qquad
\text{for every } p \neq n/2.
\]
\end{conjecture}
The point is that only the middle dimension survives. For a torus
$T^{2k}$ this tells us $b_k^{(2)} = \binom{2k}{k}$ and all other
$L^2$-Betti numbers vanish; for a hyperbolic surface it says
$b_1^{(2)} = 1$ and the rest vanish, which is what the Gauss-Bonnet
theorem predicts. The conjecture was proposed by Singer and was
suggested to Dodziuk in the special case of nonpositive curvature.
It remains open in most dimensions.
\subsection*{What we prove in dimension four}
In dimension four, Poincare duality already forces
$b_4^{(2)} = b_0^{(2)}$ and $b_3^{(2)} = b_1^{(2)}$. The degree-zero
vanishing ($b_0^{(2)} = 0$ for infinite groups) is elementary, so the
only outstanding degree is $p = 1$. Our goal is to prove $b_1^{(2)} = 0$.
Equivalently:
\begin{theorem}[Main theorem]\label{thm:main}
Let $Q$ be a finite connected aspherical integral Poincare complex of
formal dimension four, with arbitrary orientation character. Then
\[
b_p^{(2)}(\widetilde Q) = 0
\qquad
\text{for all } p \geq 0,\ p \neq 2.
\]
\end{theorem}
The class of complexes is slightly larger than manifolds; see
Definition~\ref{def:pd-complex} below.
\begin{corollary}[Singer in dimension four]\label{cor:singer-4d}
Let $M$ be a closed connected aspherical topological four-manifold. Then
\[
b_p^{(2)}(\widetilde M) = 0 \qquad \text{for all } p \neq 2.
\]
\end{corollary}
Neither orientability nor a smooth structure is needed. In particular
the corollary covers non-smoothable four-manifolds, which are
topological spaces that admit no PL triangulation or smooth structure.
\subsection*{Why these statements matter}
Combined with the $L^2$ Euler Poincare formula
$\chi(M) = \sum_p (-1)^p b_p^{(2)}(\widetilde M)$, Theorem~\ref{thm:main}
implies $\chi(M) = b_2^{(2)}(\widetilde M) \geq 0$ for every closed
aspherical four-manifold. Corollary~\ref{cor:euler} records this.
Combined with the topological $L^2$-signature theorem it also yields the
Euler--signature inequality $\chi(M) \geq |\sigma(M)|$, where
$\sigma(M)$ is the ordinary signature; see
Corollary~\ref{cor:signature}. Together these force strong restrictions
on the possible Betti numbers (and the intersection form) of a closed
aspherical four-manifold.
\subsection*{Prior work in low dimensions}
The three-dimensional case was handled by Lott and Luck (1995) via
three-manifold geometrization. In dimension four, several special cases
were known:
\begin{itemize}
\item Right-angled Coxeter groups whose nerve is a flag triangulation of
$S^3$ (Davis--Okun, 2001).
\item Groups of type VF with a hierarchy, and arbitrary spherical
Coxeter groups (Okun--Schreve, 2016).
\item Products of surfaces (via Gromov's restriction on K\"ahler groups
of positive first $L^2$-Betti number).
\item Aspherical complex surfaces with residually finite $\pi_1$, and
those outside class $\mathrm{VII}_0^+$ without that hypothesis
(Albanese--Di Cerbo--Lombardi, 2025).
\item Even-dimensional Gromov--Thurston branched covers of hyperbolic
manifolds (Avramidi--Okun--Schreve, 2025).
\end{itemize}
Our proof is different in kind: it uses only integral Poincare duality
(through \S\ref{sec:foundations}), the geometry of finite-energy
functions, and topological (Cech) cohomology on the resulting boundary.
It is not an extension or a combination of the cited arguments.
%=============================================================
\section{The strategy in one page}
\label{sec:strategy}
%=============================================================
Suppose toward contradiction that $b_1^{(2)}(\widetilde Q) > 0$. We will
derive a contradiction in six stages.
\paragraph{Stage 1 (algebra).} Translate the assumption into geometry.
The first $L^2$-Betti number is the dimension of a Hilbert space
$\mathcal D \ominus \mathcal D_0$, where $\mathcal D$ is a space of
square-summable edge gradients and $\mathcal D_0$ is the closure of the
finite-support gradients. If $b_1^{(2)} > 0$ then $\mathcal D_0$ is a
\emph{proper} subspace of $\mathcal D$, so there exists a finite-energy
function whose gradient is genuinely square-summable and not
approximable by finitely supported pieces. This gives us a rich family
of finite-energy functions on the Cayley graph. See
Section~\ref{sec:foundations}.
\paragraph{Stage 2 (compactification).} Using this family, build a
compact metrizable space $X = \Gamma \sqcup B$: the group $\Gamma$ is a
discrete subset and $B$ is a closed "boundary at infinity". The
construction uses Royden-style compactification by bounded
finite-energy functions. Add extra coordinates that record which
connected component of a coordinate box a point lies in. This ensures
a \emph{near-path property}: if two vertices approach the same point of
$B$, they can be joined by a graph path staying close to that point.
See Section~\ref{sec:compactification}.
\paragraph{Stage 3 (dynamics).} Use square-summability of edge lengths in
$X$ to prove that the action of $\Gamma$ on $B$ is a
\emph{convergence action}: every escaping sequence of group elements has
a subsequence that collapses compact sets off $b$ to a point $a$. In
particular every infinite-order element is parabolic or loxodromic, and
$B$ has no global fixed point. The $\Gamma$-action is minimal: every
orbit is dense. See Section~\ref{sec:dynamics}.
\paragraph{Stage 4 (boundary topology).} Show that the covering dimension
of $B$ is at most one, and that Cech cohomology
$\check H^1(B;\mathbb F_2)$ vanishes. The dimension bound uses that
"level cuts" of finite-energy functions have small accumulation sets in
$B$. The cohomology vanishing uses integral Poincare duality for the
group and shrinks a finite-cover cocycle to a finite-support cellular
cocycle, then corrects it and reads off a Čech coboundary. See
Section~\ref{sec:boundary-topology}.
\paragraph{Stage 5 (cut points).} A \emph{cut point} of $B$ is a point
whose removal disconnects it. If $B$ has no cut points, set $L = B$; the
topology of $L$ is what we need. Otherwise use Bowditch's theory of
pretrees: the cut points of $B$ form a simplicial tree, and there is a
"leaf block" $L$ (formalized below) with the property
\[
L \text{ is a nondegenerate continuum without cut points},
\qquad
\dim L \leq 1,
\qquad
\check H^1(L;\mathbb F_2) = 0.
\]
Moreover $L$ is a retract of $B$ and comes with a graph structure whose
vertices are dense in $L$ and have infinite orbit and stabilizer
structure. See Sections~\ref{sec:pretree}--\ref{sec:blocks}.
\paragraph{Stage 6 (path pushing).} Suppose $L$ were not locally
connected. Then some point $x$ of $L$ would have a neighborhood that
contains points $y$ that cannot be connected to $x$ within any smaller
neighborhood. We push graph paths toward the boundary using \emph{annuli}
(nested pairs of closed sets) and a \emph{three-pointer majority}
argument. The key point is \emph{uniformity}: one depth threshold works
for every path simultaneously. This forces $L$ to be locally connected,
contradiction. See Sections~\ref{sec:annuli}--\ref{sec:path-pushing}.
\paragraph{Final contradiction.} A nondegenerate compact metric space
that is locally connected, has covering dimension $\leq 1$, and has no
embedded circle (forced by $\check H^1(\cdot;\mathbb F_2) = 0$) is called
a \emph{dendrite}. Every nondegenerate dendrite has a cut point. But $L$
has no cut points by construction. Contradiction. See
Section~\ref{sec:conclusion}.
Therefore our original assumption $b_1^{(2)} > 0$ was false, so
$b_1^{(2)}(\widetilde Q) = 0$, and by Poincare duality all non-middle
$L^2$-Betti numbers vanish.
%=============================================================
\section{Topology and algebra: setting up the group}
\label{sec:foundations}
%=============================================================
Let $k = \mathbb F_2$ throughout the boundary argument. Fix a finite
connected aspherical integral Poincare complex $Q$ of formal dimension
four, with orientation character $w : \pi_1(Q) \to \{1,-1\}$, and write
$\Gamma = \pi_1(Q)$. We do not fix a particular model; we only use that
$Q$ is a finite CW complex satisfying Poincare duality.
\begin{definition}[Poincare complex, formal dimension $n$]
\label{def:pd-complex}
A finite connected CW complex $Q$ is a \emph{Poincare complex of formal
dimension $n$} if there is an orientation character $w : \pi_1(Q) \to
\{1,-1\}$ and a class $[Q] \in H_n(Q;\mathbb Z^w)$ (where $\mathbb Z^w$
is the local system twisting by $w$) such that cap product with $[Q]$
gives an isomorphism
\[
[Q] \cap - :
H^i(Q;V) \longrightarrow H_{n-i}(Q; w \otimes V)
\]
for every local coefficient system $V$.
The ``$\mathbb Z^w$'' twist only matters when $Q$ is nonorientable; for
orientable $Q$ we can take $w \equiv 1$. When $H_*(Q; \mathbb Z^w)$
vanishes outside the fundamental class, the complex is an
\emph{integral} Poincare complex.
\end{definition}
\subsection*{What the topological input gives us}
The strategy is: use only the algebraic shadow of Poincare duality,
plus the fact that asphericity makes $\Gamma$ tractable. We collect
those consequences now.
\begin{lemma}[Fundamental group of a PD4 complex]\label{lem:group-basics}
The group $\Gamma$ is infinite, finitely presented, and torsion-free.
Moreover, $Q$ is homotopy equivalent to a finite four-dimensional CW
complex, and the augmented cellular chain complex of that model is a
finitely generated free $\mathbb Z\Gamma$-resolution of $\mathbb Z$ of
length four.
\end{lemma}
\begin{proof}
Wall's dimension theorem \cite{Wall1967} provides a four-dimensional CW
complex homotopy equivalent to $Q$; because $Q$ is finite, we may take
the model finite (the reduced finiteness obstruction is zero for a space
already finitely dominated). Asphericity plus finite four-dimensional
model means the model's universal cover is contractible, so its cellular
chains give a free resolution of $\mathbb Z$ over $\mathbb Z\Gamma$ of
length four.
\emph{Torsion-free.} Suppose $g \in \Gamma$ has prime order $p$. Let
$C = \langle g \rangle \cong \mathbb Z/p$. Restrict the resolution just
described to $C$. We still have a free $\mathbb Z C$-resolution of
$\mathbb Z$ of finite length $\leq 4$. But a cyclic group of prime order
has $2$-periodic cohomology: $H^{2j}(C;\mathbb Z) \cong \mathbb Z/p$
for every $j \geq 1$. A resolution of length $\leq 4$ cannot possibly
give nonzero cohomology in arbitrarily high degrees, so we arrive at a
contradiction. Hence no finite-order element exists.
\emph{Infinite.} Suppose $\Gamma$ finite. By torsion-freeness,
$\Gamma$ is trivial, so $Q$ is simply connected and aspherical, hence
contractible. But Poincare duality over $k$ gives
$H_4(Q;k) \cong H^0(Q;k) \cong k$, contradicting contractibility
(a point has no $H_4$). So $\Gamma$ is infinite.
\emph{Finitely presented.} A finite four-dimensional CW model has
finitely many $1$-cells and $2$-cells, so its fundamental group is
finitely presented.
\end{proof}
\begin{lemma}[Manifolds reduce to Poincare complexes]
\label{lem:manifold-reduction}
Every closed connected aspherical topological four-manifold $M$ is
homotopy equivalent to a finite connected aspherical integral Poincare
complex of formal dimension four.
\end{lemma}
\begin{proof}
A compact topological manifold is metrizable and is an absolute
neighborhood retract (ANR): the coordinate neighborhood of any point is
an ANR, and Hanner's locality theorem upgrades this to a global ANR;
see \cite[Thms.~3.2--3.3]{Hanner1951}. Its covering dimension is
$\leq 4$ by the closed-sum theory of dimension; see
\cite[Theorems~1.1.2, 1.5.3, 1.7.7, 1.8.2]{Engelking1978}. West's
homotopy dimension estimate then supplies a finite four-dimensional
simplicial complex $F$ homotopy equivalent to $M$; see
\cite[Cor.~5.4]{West1977}. This $F$ is not a triangulation of $M$, just
a homotopy model.
Transport the fundamental class and orientation character of $M$ to
$F$. Topological Poincare duality with local coefficients gives
cap-product isomorphisms for every coefficient system on $M$:
\cite[Theorem~1, with $R = \mathbb Z$]{Sun2017}, and see also
\cite[Thm.~10.2, Rem.~10.3]{Spanier1993} and
\cite[Thm.~A.16]{FNOP2025}. Transporting along the homotopy equivalence
gives $F$ the PD property. Combined with asphericity of $M$, this makes
$F$ a finite connected aspherical integral Poincare complex of formal
dimension four.
\end{proof}
\subsection*{What Poincare duality gives us algebraically}
The two key algebraic facts are that the group-ring module
$\mathbb Z\Gamma$ (and its $k\Gamma$ analogue) has vanishing first and
second cohomology, and that one can read off good information about
subgroups of $\Gamma$ through Shapiro's lemma.
\begin{lemma}[Group-cohomological vanishing from duality]
\label{lem:found-duality}
With $k = \mathbb F_2$, the group $\Gamma$ satisfies
\[
H^i(\Gamma; k\Gamma) = 0 \quad (i = 1, 2).
\]
Moreover, for every subgroup $J \leq \Gamma$ there is a natural
isomorphism
\[
H^1(\Gamma; k[\Gamma/J]) \cong H_3(J; k).
\]
In particular this group vanishes if $J$ is finite or two-ended.
\end{lemma}
\begin{proof}
The orientation twist of a PD4 complex is invisible over $k$. Because
$Q$ is aspherical,
\[
H^i(Q; V) \cong H^i(\Gamma; V)
\]
for every local system $V$. Apply Poincare duality (which, over $k$,
does not see the orientation character) with $V$ a $k$-vector space:
\[
H^i(\Gamma; V) \cong H_{4-i}(\Gamma; V), \quad 0 \leq i \leq 4.
\]
Take $V = k\Gamma$ to get $H^i(\Gamma; k\Gamma) \cong H_{4-i}(\Gamma; k\Gamma)$.
Taking $i = 1$ gives $H^1 \cong H_3$, and $i = 2$ gives
$H^2 \cong H_2$. The chain complex of $\Gamma$ with coefficients in
$k\Gamma$ is just the augmented $k\Gamma$-resolution of $k$ tensored
with $k\Gamma$, and since that resolution is finite of length four,
$H_p(\Gamma; k\Gamma) = 0$ for $p > 0$ (a free resolution of $k$ is
exact in positive degrees). Therefore $H^1(\Gamma; k\Gamma) =
H_3(\Gamma; k\Gamma) = 0$, and $H^2(\Gamma; k\Gamma) = 0$.
\emph{The Shapiro isomorphism.} The permutation module
$k[\Gamma/J] = k\Gamma \otimes_{kJ} k$. If $R_*$ is a free right
$k\Gamma$-resolution of $k$ then
\[
R_* \otimes_{k\Gamma} (k\Gamma \otimes_{kJ} k)
\cong (R_*|_J) \otimes_{kJ} k,
\]
which is a free $kJ$-resolution of $k$. Thus
$H_j(\Gamma; k[\Gamma/J]) \cong H_j(J; k)$. Combined with
$H^i(\Gamma; V) \cong H_{4-i}(\Gamma; V)$, this gives (with $i = 1$)
$H^1(\Gamma; k[\Gamma/J]) \cong H_3(J; k)$.
\emph{Vanishing for small $J$.} If $J$ is finite, it is trivial by
Lemma~\ref{lem:group-basics}, so $H_3(J;k) = 0$. If $J$ is two-ended,
it has a finite normal subgroup $N$ with quotient $J/N$ infinite cyclic
or infinite dihedral; this is \cite[Lemma~4.1]{Wall1967}. Torsion-freeness
forces $N$ trivial, so $J$ is infinite cyclic (the dihedral case is
impossible because dihedral groups contain involutions). But an infinite
cyclic group is an iterated extension of the trivial group by $\mathbb Z$,
so its integral homology is concentrated in degrees $0$ and $1$; over
$k$ the same holds, so $H_3(J;k) = 0$.
\end{proof}
\begin{lemma}[Group is one-ended]\label{lem:found-ends}
$\Gamma$ is one-ended: removing any finite set of vertices of a
Cayley graph leaves exactly one infinite connected component.
\end{lemma}
\begin{proof}
Choose a finite presentation of $\Gamma$ and let $\mathcal C$ be its
Cayley two-complex (a simply connected $2$-complex on which $\Gamma$
acts freely). The corresponding Cayley graph $G$ is locally finite,
since the presentation is finite. The augmented cellular complex of
$\mathcal C$ is exact in degrees zero and one; extending by free modules
in higher degrees gives a free resolution in which each cohomology is
the same as in $\mathcal C$.
Suppose some finite vertex set, when removed from $G$, leaves two
distinct infinite components. Let $f : \Gamma \to k$ be the indicator
of one of those infinite components, with value $0$ on the deleted
vertices. Its gradient $df$ is a finitely supported edge cochain
(because the graph is locally finite and $f$ is nonzero only on one
component), and $df$ is a cellular cocycle (its coboundary vanishes
around every cell because $f$ is $\{0,1\}$-valued; there are no
$2$-cell coefficients outside $k$). By Lemma~\ref{lem:found-duality},
$H^1(\Gamma; k\Gamma) = 0$, so the cohomology class of $df$ in the
group-ring cocycle complex is a coboundary: $df = du$ for a
\emph{finitely supported} vertex function $u$. Then $f - u$ is a
locally constant function on $\Gamma$, so $f - u$ is constant (the
group is connected). But $f$ equals $1$ on an infinite set and $0$ on
another infinite set, whereas $u$ has finite support; this is impossible.
The infinite connected locally finite Cayley graph has at least one
end (that is standard: greedily choose longer and longer escaping
paths), and we have just excluded two or more. Hence $\Gamma$ is
one-ended.
\end{proof}
\subsection*{No splittings over finite or two-ended subgroups}
The path-pushing argument uses Bowditch's theory of tree actions. We
need to know $\Gamma$ does not split over a finite or two-ended group.
\begin{lemma}[Splitting obstruction]\label{lem:found-trees}
Consider a $\Gamma$-action on a simplicial tree without inversions.
\begin{enumerate}[label=\textup{(\roman*)}]
\item If the action has no fixed vertex, then the edge stabilizers
cannot all be finite or two-ended.
\item If the action is minimal and the tree has an edge, then every
edge stabilizer is neither finite nor two-ended.
\end{enumerate}
Consequently $\Gamma$ has no nontrivial splitting over a finite or
two-ended subgroup.
\end{lemma}
\begin{proof}
Fix a base vertex $o$ and an edge orbit $E$ with stabilizer $J$.
Identify $E$ with $\Gamma/J$. For $g \in \Gamma$ let $c(g) \in k$
be the mod-two number of edges of $E$ on the geodesic $[o, go]$
projected through $E$ (equivalently, the mod-two intersection number
of $[o,go]$ with $E$). Cancellation of edges traversed twice gives
the cocycle identity
\[
c(gh) = c(g) + g\, c(h).
\]
If $J$ is finite or two-ended, the Shapiro-based vanishing
(Lemma~\ref{lem:found-duality}) makes $c$ principal:
$c(g) = g v - v$ for some $v \in k[\Gamma/J]$. Hence the coefficient
size of $c$ is bounded:
\begin{equation}\label{found:crossing-bound}
|\,[o,go] \cap E\,| \leq 2 |\operatorname{supp} v|
\qquad
(g \in \Gamma).
\end{equation}
\emph{Part (i).} Let $S$ be a finite generating set of $\Gamma$. The
finitely many paths $[o, so]$, $s \in S$, together contain only finitely
many edges. Translates of those paths cover every $[o, go]$ (write $g$
as a word in $S$ and cancel backtracking). Therefore only finitely many
edge orbits occur on the paths $[o, go]$, $g \in \Gamma$. An action
without a fixed vertex has unbounded vertex orbits: if the orbit of
some vertex were bounded, its convex hull would give an invariant
center (a fixed vertex when there are no inversions). So one of the
finitely many edge-orbit crossing counts $|\,[o,go] \cap E\,|$ is
unbounded. If all edge stabilizers were finite or two-ended, this
would contradict \eqref{found:crossing-bound}.
\emph{Part (ii).} Fix an edge orbit $E$ and collapse each connected
component of the union of edges outside $E$. The result is a nontrivial
simplicial tree without inversions, on which $\Gamma$ acts. It has no
fixed vertex: the preimage of a fixed vertex would be a proper nonempty
invariant subtree of the original. Its orbits are therefore unbounded,
and distance in this quotient counts exactly the edges of $E$ on the
original path $[o,go]$. Again \eqref{found:crossing-bound} rules out a
finite or two-ended stabilizer.
Applying this to the Bass--Serre tree of any splitting shows that
$\Gamma$ does not split over a finite or two-ended subgroup.
\end{proof}
\subsection*{Reducing the $L^2$ problem to $b_1^{(2)} = 0$}
We now recall the algebraic meaning of $b_1^{(2)}$ and how Poincare
duality reduces the whole vanishing statement to just degree one.
\begin{lemma}[Only degree one is missing]\label{lem:found-reduction}
For the fixed PD4 complex $Q$,
\[
b_0^{(2)}(\widetilde Q) = b_4^{(2)}(\widetilde Q) = 0,
\qquad
b_3^{(2)}(\widetilde Q) = b_1^{(2)}(\widetilde Q),
\]
and $b_p^{(2)}(\widetilde Q) = 0$ for $p > 4$. Consequently
Theorem~\ref{thm:main} reduces to proving $b_1^{(2)}(\widetilde Q) = 0$.
\end{lemma}
\begin{proof}
By Lemma~\ref{lem:group-basics}, $Q$ has a finite four-dimensional
cellular model, so there are no cellular chains above degree four
and $b_p^{(2)} = 0$ 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, so $b_0^{(2)} = 0$.
For the duality statement we must handle a nontrivial orientation
character. Let $\Gamma^+ \leq \Gamma$ be the kernel of the orientation
character $\Gamma \to \{1,-1\}$ and $Q^+ \to Q$ the corresponding
two-fold (or one-fold) cover. Its index $d = [\Gamma:\Gamma^+]$ is
$1$ or $2$. Finite covers of Poincare complexes are again Poincare
complexes of the same formal dimension \cite{Stark1996}. The
orientation character of $Q^+$ is the restriction of $w$ to $\Gamma^+$
and is therefore trivial, so $Q^+$ is an \emph{oriented} finite
connected aspherical integral PD4 complex. By Lemma~\ref{lem:group-basics}
pick a finite four-dimensional model $F$ of $Q^+$.
Transport the fundamental class of $Q^+$ to $F$. Poincare duality with
local coefficients applies to $Q^+$, hence to $F$
\cite[Lemma~1.1]{Wall1967}. In particular, applying duality to the
regular group-ring local system shows that cap product defines a
quasi-isomorphism
\[
C^*(\widetilde F; \mathbb Z) \xrightarrow{\;\cong\;}
C_*(\widetilde F; \mathbb Z)
\]
(with degree reversal and a $g \mapsto g^{-1}$ twist on the coefficient
module). Both sides are bounded complexes of finitely generated free
$\mathbb Z\Gamma^+$-modules, so the mapping cone is a bounded acyclic
complex of free modules. A bounded acyclic complex of free modules
that splits at every surjection is contractible (start in the lowest
degree and split each surjection onto the kernel already obtained).
Hence cap product is a chain homotopy equivalence.
Extending scalars to the group von Neumann algebra
$\mathcal N(\Gamma^+)$, homotopies remain homotopies, and finite-matrix
dimension comparison identifies the von Neumann dimensions of the
two complexes. Consequently
\[
b_p^{(2)}(\widetilde Q;\mathcal N(\Gamma^+))
= b_{4-p}^{(2)}(\widetilde Q;\mathcal N(\Gamma^+))
\quad (0 \leq p \leq 4).
\]
Finally, finite-index subgroup scaling
\cite[Thm.~1.35(9)]{Luck2002} gives
\[
b_p^{(2)}(\widetilde Q;\mathcal N(\Gamma^+))
= d \cdot b_p^{(2)}(\widetilde Q;\mathcal N(\Gamma)).
\]
Dividing by $d$ gives $b_p^{(2)} = b_{4-p}^{(2)}$ for $Q$ itself.
Taking $p = 0$ gives $b_4^{(2)} = b_0^{(2)} = 0$, and taking $p = 1$
gives $b_3^{(2)} = b_1^{(2)}$. Degree two is the only one not forced
to vanish.
\end{proof}
\subsection*{Where the Hilbert space of finite-energy gradients comes from}
What we need going forward: a rich supply of finite-energy functions
built from the presentation. Assume toward contradiction $b_1^{(2)} > 0$
from now on. Then we can find such a function that is not approximable
by finitely supported gradients. Let us set that up.
Choose a finite presentation of $\Gamma$ with relators of length at most
three (subdividing longer relators by new generators achieves this; Tietze
transformations preserve all the cohomology we care about). Let
$\mathcal C$ be the Cayley two-complex with graph $G$; write $E(G)$
for the edge set. As noted in Lemma~\ref{lem:group-basics}, $G$ is
connected and locally finite, $\mathcal C$ is simply connected, and
each cellular edge belongs to a uniformly bounded number of cell
boundaries. We keep parallel edges and loops as separate cellular edges,
and choose one orientation on each edge.
For a real function $f : \Gamma \to \mathbb R$ and an oriented edge
$e = uv$ put $df(e) = f(v) - f(u)$. Define the Hilbert spaces
\[
\mathcal D =
\{df : f : \Gamma \to \mathbb R,\ \sum_{e \in E(G)} |df(e)|^2 < \infty\},
\quad
\mathcal D_0 =
\overline{\{df : f \text{ finitely supported}\}}
\]
where the closure is in $\ell^2(E(G); \mathbb R)$ and the sum counts
each cellular edge once.
\begin{lemma}[Finite-energy gradients and $b_1^{(2)}$]
\label{lem:found-gradient}
The space $\mathcal D$ is a closed real Hilbert subspace of
$\ell^2(E(G);\mathbb R)$ and $\mathcal D_0 \subseteq \mathcal D$. If
$b_1^{(2)}(\widetilde Q) > 0$ then $\mathcal D_0 \neq \mathcal D$.
\end{lemma}
\begin{proof}
\emph{$\mathcal D$ is closed.} Write $F(\mathcal C)$ for the set of
two-cells. The cellular coboundary
\[
d_1 : \ell^2(E(G);\mathbb R) \to \ell^2(F(\mathcal C);\mathbb R)
\]
is bounded: each relator has at most three edges (counted with
multiplicities), and edge--cell incidence is uniformly bounded by the
presentation. An ordinary cellular $1$-cocycle on $\mathcal C$ is a
gradient: integrate along edge paths from a fixed base vertex, and
path-independence is exactly the cocycle condition on the simply
connected $\mathcal C$. Therefore
$\mathcal D = \ker d_1$, which is closed.
\emph{$\mathcal D_0$ description.} Likewise
$d_0 : \ell^2(\Gamma;\mathbb R) \to \ell^2(E(G);\mathbb R)$ is
bounded (bounded vertex degree). Since finite-support functions are
dense in $\ell^2(\Gamma)$ and $d_0$ is continuous, the closure of the
image of $d_0$ is the closure of the image of finitely supported
functions on $\ell^2(G)$. So $\mathcal D_0 = \overline{\operatorname{im} d_0}$;
in particular $\mathcal D_0 \subseteq \ker d_1 = \mathcal D$. Note that
this does \emph{not} assert $\operatorname{im} d_0$ itself is closed.
\emph{Comparison with $b_1^{(2)}$.} The augmented cellular chains of
$\mathcal C$ are exact in degrees zero and one. Extend to a free
resolution by choosing a free module $P_3$ mapping onto
$\ker(\partial_2 : C_2(\mathcal C) \to C_1(\mathcal C))$, then free
modules onto successive kernels. The cellular terms in degrees $0,1,2$
are finitely generated free; so the hypothesis of the finite-matrix
comparison (below) is met. The singular chains of $\widetilde Q$ are
free $\mathbb Z\Gamma$-modules: a deck transformation fixing a
singular simplex fixes its image points and hence is the identity.
Thus Lemma~\ref{lem:found-comparison} applies.
Over $\mathbb C$, that comparison identifies $b_1^{(2)}(\widetilde Q)$
with the dimension of
\[
\ker d_1 \ominus \overline{\operatorname{im} d_0}.
\]
All the matrices involved 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. Therefore
$b_1^{(2)} > 0$ forces strict inclusion.
\end{proof}
\begin{lemma}[Finite-matrix comparison]\label{lem:found-comparison}
Let $\Lambda$ be any group and let $P_*$ be a free $\mathbb Z\Lambda$-resolution
of $\mathbb Z$ whose terms in degrees $0,1,2$ are finitely generated.
Suppose a contractible space $Z$ has a free $\Lambda$-action with free
singular chains over $\mathbb Z\Lambda$. Then
\[
b_1^{(2)}(Z;\mathcal N(\Lambda))
= \dim_{\mathcal N(\Lambda)}
(\ker\partial_1 \cap \ker\partial_2^*)
\]
where the operators on the right act on the Hilbert spaces
$\ell^2(\Lambda) \otimes_{\mathbb Z\Lambda} P_i$. The same space is the
first reduced Hilbert cohomology for $d_0 = \partial_1^*$ and
$d_1 = \partial_2^*$.
\end{lemma}
\begin{proof}
\emph{Chain-homotopy comparison.} 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 with the von Neumann algebra preserves chain homotopies, so
it preserves homology, which is what the definition of $b_1^{(2)}$
computes. No flatness of $\mathcal N(\Lambda)$ over $\mathbb Z\Lambda$
is needed because we only need to preserve homotopy equivalences.
\emph{Dimension comparison.} For a matrix over $\mathcal N(\Lambda)$,
the extended von Neumann dimension of the algebraic kernel equals the
trace of the projection onto its Hilbert-space kernel. The dimension
of the algebraic image (taken as von Neumann dimension) equals the
trace of the projection onto the closure of its Hilbert image. By
additivity of dimension,
\[
\dim H_1(\mathcal N\Lambda \otimes P_*)
= \dim \bigl(\ker\partial_1 \ominus \overline{\operatorname{im}\partial_2}\bigr)
= \dim \bigl(\ker\partial_1 \cap \ker\partial_2^*\bigr).
\]
On the other hand the harmonic representative space for reduced
cohomology is
\[
\ker d_1 \ominus \overline{\operatorname{im} d_0}
= \ker\partial_2^* \cap \ker\partial_1.
\]
These are the same closed Hilbert submodule. The argument works in
every degree with finite free neighbours. This is the comparison
underlying \cite[Rem.~1.31]{Luck2002} and
\cite[Lemmas~6.51--6.53]{Luck2002}.
\end{proof}
From now on: assume toward contradiction $b_1^{(2)}(\widetilde Q) > 0$,
so $\mathcal D_0 \subsetneq \mathcal D$. We have obtained a nonzero
quotient $\mathcal D/\mathcal D_0$, together with $\Gamma$ being
finitely presented, torsion-free, one-ended, having no splitting over
a finite or two-ended subgroup, and having the group-cohomological
vanishings of Lemma~\ref{lem:found-duality}. These are the algebraic
inputs for the rest of the proof.
%=============================================================
\section{Building a compact boundary}
\label{sec:compactification}
%=============================================================
This is the heart of the proof's constructive core. Our standing
assumption throughout is that $b_1^{(2)}(\widetilde Q) > 0$, equivalently
$\mathcal D_0 \subsetneq \mathcal D$. We will use this to build a compact
metrizable space $X = \Gamma \sqcup B$ whose "boundary" $B$ is a nice
continuum, and to arrange that vertices approaching the same boundary
point can be joined by graph paths staying close to that point. That
last property is the \emph{near-path property}; it converts membership
in $B$ into a concrete connectivity statement about paths in $G$.
\subsection*{One function with connected level sets}
Start with a "starter function" whose level sets (the subgraphs
$\{f < t\}$ and $\{f > t\}$) are connected, and which does not tend to
a constant at infinity. That function will provide the base coordinate
for the compactification.
\begin{lemma}[A good function exists]\label{lem:potential}
There exists a finite-energy function $f : \Gamma \to [0,1]$ which does
not tend to a single constant at infinity, and for which both
$\{f < t\}$ and $\{f > t\}$ are connected vertex subgraphs whenever
$0 < t < 1$ is regular for $f$ (i.e.\ not attained at any vertex).
Every such cut $S(f,t)$ is dually connected.
\end{lemma}
\begin{proof}
\emph{Step 1 (the functional).} Fix a vertex $a \in \Gamma$. For each
$b \ne a$, the functional
\[
L_b(df) = f(b) - f(a)
\]
is well defined and continuous on $\mathcal D$: a finite path of length
$\ell$ from $a$ to $b$ gives
$|L_b(df)| \le \sqrt{\ell}\, \|df\|_2$ by Cauchy--Schwarz. The functional
is nonzero because if we test it on the gradient of the singleton
indicator $\mathbf 1_{\{b\}}$, we get $\mathbf 1_{\{b\}}(b) -
\mathbf 1_{\{b\}}(a) = 1 - 0 = 1$. Let $r_b \in \mathcal D$ be the
Riesz representative: $L_b(dg) = \langle r_b, dg\rangle$.
\emph{Step 2 (min-norm representative).} The unique gradient of minimal
norm satisfying $L_b(df_b) = 1$ is
\[
df_b = \frac{r_b}{\|r_b\|_2^2},
\]
with potential $f_b$ normalized by $f_b(a) = 0$.
\emph{Step 3 (some $df_b$ escapes $\mathcal D_0$).} The vectors $r_b$ span
a dense subspace of $\mathcal D$: any $dg$ orthogonal to every $r_b$
would satisfy $L_b(dg) = 0$ for every $b$, so the potential $g$ is
constant, so $dg = 0$. Since $b_1^{(2)} > 0$, we have
$\mathcal D_0 \subsetneq \mathcal D$, so some $df_b \notin \mathcal D_0$.
Fix such a $b$ and its minimizer $df$.
\emph{Step 4 (normalize to $[0,1]$).} The map
$\phi(x) = \min(1, \max(0, x))$ satisfies $|\phi(x) - \phi(y)| \le
|x - y|$ for all $x, y$. Replacing $f$ by $\phi \circ f$ preserves the
values $0$ and $1$ at $a$ and $b$ (which were forced by the min-norm
condition), cannot increase any edge energy, and yields another candidate
minimizer. Uniqueness of the minimizer forces $0 \le f \le 1$
everywhere.
\emph{Step 5 (no constant at infinity).} Suppose $f$ tends to a constant
$c$ outside finite vertex sets. For $\varepsilon > 0$ define
\[
u_\varepsilon(v) = \operatorname{sgn}(f(v) - c)
\cdot \max(|f(v) - c| - \varepsilon, 0).
\]
The support of $u_\varepsilon$ is finite by hypothesis (only finitely
many vertices satisfy $|f(v) - c| \ge \varepsilon$). The function
$(f - c) - u_\varepsilon$ is the truncation of $f - c$ to
$[-\varepsilon, \varepsilon]$, so its edge differences are bounded in
absolute value by $|df|$ and tend pointwise to zero as
$\varepsilon \to 0$. By dominated convergence,
$d u_\varepsilon \to df$ in $\ell^2$. But $du_\varepsilon$ has finite
support, so $df \in \mathcal D_0$, contradiction.
\emph{Step 6 (lower level sets connected).} Fix a regular $t \in
(0,1)$. Suppose a component $U$ of $\{f > t\}$ misses $b$. Define
$f' = f$ off $U$, $f' = t$ on $U$. This preserves $f(a) = 0$ (below
$t$, so $a \notin U$) and $f(b) = 1$ (above $t$, so $b \notin U$).
Internal edge energies inside $U$ drop to zero. On the boundary, some
edge $e = uv$ has $u \in U$ and $v \notin U$ (because $G$ is connected
and $U$ is a proper subset); on that edge $|df'(e)| < |df(e)|$ strictly.
So $\mathcal E(f') < \mathcal E(f)$, contradicting that $f$ minimizes
energy subject to $L_b = 1$. Hence every upper component contains $b$,
and the upper side is connected. The lower side is connected by the
identical argument with $a$ and $b$ interchanged.
\emph{Step 7 (dual connectedness).} Let $C$ be a dual component of
$S(f,t)$: a component of the straddling edges under the adjacency
"both lie on the boundary of a common $2$-cell". Restrict the cochain
$d \mathbf 1_{\{f>t\}}$ to $C$ and set it to zero on all other edges.
Consider any cell: the signed sum of this restricted cochain around
that cell is either the original zero sum (if the cell's straddling
edges are in $C$) or an entirely zero sum (if they are not, or if some
are and some are not, but every crossing edge of the cell lies in one
dual component, so all-or-nothing). Simple connectivity of $\mathcal C$
gives a potential $u$ for the restricted cochain. That potential is
constant on each of the two connected level sides $\{f>t\}$ and
$\{f<t\}$, and it has nonzero difference across every edge of $C$.
Consequently, it must have nonzero difference across \emph{every}
edge of $S(f,t)$ (otherwise one of those edges would have zero
difference, but $u$ would agree with $\mathbf 1_{\{f>t\}}$ there,
contradiction). So $C = S(f,t)$: the wall is dually connected.
\end{proof}
\begin{lemma}[Coordinate cuts have finitely many dual components]
\label{lem:wall-components}
Every regular cut of a coordinate in $\mathcal F_0$ (defined below) has
finitely many dual components. A nonempty regular cut of a translate
of $f$ is dually connected, and a regular cut of a singleton indicator
is finite.
\end{lemma}
\begin{proof}
Enlarge $\mathcal F_0$ by all translates of $f$ and all singleton
indicators $\mathbf 1_{\{v\}}$ (with $v \in \Gamma$). A translate of $f$
also has connected level sides and dually connected cuts (translations
of subgroups preserve the properties of Lemma~\ref{lem:potential}).
Levels outside $[0,1]$ give empty cuts. A nonempty cut of a singleton
indicator is finite: it consists of edges incident to a single vertex,
which is finite by local finiteness of $G$.
\end{proof}
\subsection*{Coordinate boxes: well-behaved regions}
We use "boxes" in the base-coordinate cube: finitely many level
conditions. "Good" boxes will have finitely many graph components and
form a basis for the ambient topology.
\begin{definition}[Box, good box]\label{def:box}
A \emph{box} is determined by finitely many conditions
\[
O = \{(z_F) : a_i < z_{F_i} < b_i, \ 1 \le i \le m\},
\qquad
F_i \in \mathcal F_0.
\]
Endpoints $a_i, b_i$ are real numbers (possibly outside $[0,1]$). We
retain the defining inequalities as part of the box data, and we use
the same symbol for its vertex set in $G$ (obtained by evaluating the
coordinates). A box is \emph{good} if:
\begin{enumerate}[label=(\alph*)]
\item every endpoint $a_i, b_i$ is \emph{regular} for $F_i$, and
\item for every ordered pair of endpoints $(c_i, c_j)$ of cuts of
coordinates in the box data, the relation
$\{(e, e') : e \in S(F_i, c_i),\ e' \in S(F_j, c_j),\ e \sim e'\}$
is finite, where $e \sim e'$ means $e = e'$ or $e, e'$ both lie on a
common $2$-cell.
\end{enumerate}
\end{definition}
\begin{lemma}[Good boxes are a basis]\label{lem:good-boxes}
Every good box has finitely many graph components on its vertex set.
The collection of good boxes forms a countable $\Gamma$-invariant
basis for the base-coordinate cube $[0,1]^{\mathcal F_0}$.
\end{lemma}
\begin{proof}
\emph{Good boxes have finitely many components.} 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 appearing in more than one cut,
or sharing a cell with an edge of another cut. (We need to know $E$ is
finite. This is part of the definition of "good": the intersected
relation is finite by hypothesis.) Then $S_i \setminus E$ still has
finitely many dual components: bounded degree of the cut adjacency
graph plus finite deletion of vertices.
Claim: no two distinct dual components of $S_i \setminus E$ can be
incident to a common component of $G'$. Suppose edges $e_1, e_2 \in
S_i \setminus E$ are on a common triangular cell. On the specified
side of cut $i$ (both $e_1, e_2$ have one endpoint below $t$ and one
above), either their below-endpoints coincide or are joined by the
remaining cell edge. That remaining edge is not in $S_i$ (it's on the
specified side), and by the definition of $E$, it's not in any other
endpoint cut either. So the remaining edge is in $G'$. By induction
along a dual path, all endpoints on either specified side of a dual
component $D \subset S_i \setminus E$ fall into a single component of
$G'$. Each such dual component is incident to at most two $G'$-components.
Every edge of $E$ is likewise incident to at most two $G'$-components
(one on each side). So each component of $G'$ is either incident to at
least one cut edge (if a cut is nonempty) or is a full component of
$G$. In the first case, bounded finite count of cut edges and their
incident components gives finitely many $G'$-components. In the
second, $G' = G$ is connected. So the vertex set of $O$ has exactly
the components of the induced subgraph of $G'$ (by constancy of signs
on components of $G'$).
\emph{Good boxes form a basis.} Write $e \sim e'$ when $e=e'$ or
$e, e'$ lie on a common cell. The symmetric relation has degree
bounded by a constant $D$: bounded cell size and bounded cell
incidence. For finite-energy coordinates $F, F'$, Tonelli's theorem
gives
\[
\int_{\mathbb R^2} \#\{(e, e') : e \sim e',\ e \in S(F, s),\ e' \in S(F', t)\}\, ds\, dt
= \sum_{e \sim e'} |dF(e)| \cdot |dF'(e')|.
\]
Cauchy--Schwarz bounds this by $D \|dF\|_2 \|dF'\|_2 < \infty$. Thus
almost every pair $(s, t)$ has finitely many interacting pairs
$(e, e')$. The same argument works when $F = F'$, since $s$ and $t$
range independently.
For a fixed finite list of box coordinates, intersect the finitely
many full-measure conditions above and exclude the countably many
vertex values. What remains is a dense set of endpoint tuples in the
parameter domain, giving arbitrarily small good boxes about every point
of the cube.
\emph{Countable invariant basis.} There are countably many finite lists
of base coordinates. For each, choose a countable dense set of good
endpoint tuples. The resulting boxes form a countable basis. Add all
their translates: still countable, still a basis, and still good
(goodness is $\Gamma$-invariant because the action permutes
coordinates).
\end{proof}
Fix such an invariant countable basis and call it $\mathcal O$.
\subsection*{Coordinates distinguishing box components}
A box may have several graph components even though each individual
level side is connected (this is possible when several coordinates are
combined). We add \emph{localized} coordinates that record which
component of a box a vertex lies in.
\begin{definition}[Localized coordinate]\label{def:localized}
Fix $O \in \mathcal O$, with conditions $a_i < F_i < b_i$ for
$1 \le i \le m$. Define the \emph{margin function}
\[
\psi_O = \max\bigl(0,\ \min(\{1\} \cup \{F_i - a_i : 1 \le i \le m\}
\cup \{b_i - F_i : 1 \le i \le m\})\bigr).
\]
This is continuous in the base coordinates, is $\Gamma$-equivariant,
takes values in $[0, 1]$, and is positive exactly on the interior of
$O$. For each graph component $U$ of $O$ (there are finitely many by
Lemma~\ref{lem:good-boxes}), define the \emph{localized coordinate}
\[
h_{O, U} = \psi_O \cdot \mathbf 1_U
\]
on vertices. Let $\mathcal F$ be the family consisting of $\mathcal F_0$
and all of these localized coordinates, over all $O \in \mathcal O$ and
all components $U$ of $O$.
\end{definition}
\begin{lemma}[Localized coordinates are finite energy with controlled cuts]
\label{lem:localized-energy}
Every member of $\mathcal F$ has finite energy. For almost every
$t \in (0, 1)$,
\begin{equation}\label{eq:shifted-walls}
S(h_{O,U}, t) \subseteq
\bigcup_{i=1}^m \bigl(S(F_i, a_i + t) \cup S(F_i, b_i - t)\bigr),
\end{equation}
with every displayed level regular for its coordinate.
\end{lemma}
\begin{proof}
\emph{Energy bound.} The min of $F_i - a_i$ and $b_i - F_i$ is
$1$-Lipschitz with respect to each coordinate individually; hence
\[
|d\psi_O(e)| \le \max_{1 \le i \le m} |dF_i(e)|
\]
for every edge $e$.
For two adjacent vertices with positive margin, they lie in the same
component of $O$ (any connected subgraph of the good box has its
vertices in the same component; here vertices with positive margin are
inside $O$ and adjacent, so in the same $G$-component of $O$). Thus
either both contribute the margin to $h_{O,U}$, both contribute zero
(both outside $U$), or the endpoint outside $U$ has zero margin. In
every case
\[
|dh_{O,U}(e)| \le |d\psi_O(e)|,
\qquad
\|dh_{O,U}\|_2^2 \le \sum_{i=1}^m \|dF_i\|_2^2 < \infty.
\]
\emph{Shifted walls.} Let $e$ be an edge with $h_{O,U}(e^-) < t <
h_{O,U}(e^+)$. The endpoint where the value is above $t$ belongs to
$U$ (margin strictly positive) and satisfies every $F_i - a_i > t$ and
$b_i - F_i > t$. At the other endpoint, either it is still in $U$ but
some margin dropped below $t$ (so an edge crosses $F_i = a_i + t$ or
$F_i = b_i - t$), or it is outside $U$ with margin zero by adjacency
(so again some margin is zero, i.e.\ again an edge crosses $F_i = a_i$
or $F_i = b_i$). Since $t < 1$, the cap $1$ in the definition of
$\psi_O$ does not itself create cuts. All the crossed levels are of
the form $a_i + t$ or $b_i - t$. Excluding the countably many values
of $t$ where these coincide with vertex values, we get the inclusion
\eqref{eq:shifted-walls}.
\end{proof}
\subsection*{The compact space $X$ and its boundary $B$}
We now assemble the compactification. Map each vertex $v \in \Gamma$ to
its profile $(F(v))_{F \in \mathcal F} \in [0,1]^{\mathcal F}$, and let
$X$ be the closure of these profiles in the product space
$[0,1]^{\mathcal F}$. The singleton indicators $\mathbf 1_{\{v\}}$ in
$\mathcal F$ make the profile map injective (the $\mathbf 1_{\{v\}}$
coordinate is $1$ only at $v$), so we identify $\Gamma$ with its image
in $X$. They also make each vertex isolated in $X$: the set
$\{x \in X : \mathbf 1_{\{v\}}(x) > 1/2\}$ is open in $X$ and equals
$\{v\}$.
Define the \emph{boundary} $B = X \setminus \Gamma$. Since $\Gamma$ is
open in $X$ (each vertex isolated), $B$ is closed in the compact space
$X$, hence compact.
Every coordinate $F \in \mathcal F$ extends continuously to $X$ (as the
projection). The margin function $\psi_O$ also extends, through its
expression in base coordinates. Left translation by $\Gamma$ acts on
$X$ by permuting coordinates; this makes the translation a
homeomorphism of $X$, since $X$ is determined by the invariant family
$\mathcal F$.
\emph{The metric $\rho$.} Enumerate
$\mathcal F = \{F_1, F_2, F_3, \ldots\}$ and choose
$c_j = 2^{-j}/(1 + \|dF_j\|_2) > 0$. Then
$\sum_j c_j (1 + \|dF_j\|_2) = \sum_j 2^{-j} = 1 < \infty$. Define
\begin{equation}\label{eq:metric}
\rho(x, y) = \sum_j c_j |F_j(x) - F_j(y)|,
\qquad
\lambda(e) = \rho(e^-, e^+).
\end{equation}
The sum converges uniformly, so $\rho$ is a continuous metric inducing
the product topology on $X$. By Minkowski's inequality applied to finite
partial sums, followed by monotone convergence,
\begin{equation}\label{eq:edge-lengths}
\|\lambda\|_{\ell^2(E(G))} \leq
\sum_j c_j \|dF_j\|_2 < \infty.
\end{equation}
Consequently, for any fixed $\epsilon > 0$, only finitely many edges
have $\rho$-length $\geq \epsilon$ (otherwise an infinite number of
edges each contribute $\epsilon^2$, contradicting summability).
\begin{lemma}[The boundary is a nondegenerate continuum]
\label{lem:compact-boundary}
The boundary $B$ is a nondegenerate compact connected metric space: it
is compact, nonempty, has at least two points, and is connected.
\end{lemma}
\begin{proof}
\emph{Nonempty.} A closed subset of $X$ contained entirely in the
isolated vertices would be compact and discrete, hence finite. Since
$\Gamma$ is infinite, its closure contains points outside $\Gamma$.
So $B \neq \varnothing$.
\emph{Nondegenerate.} If $B = \{x\}$ for a single point $x$, then for
each $\epsilon > 0$ the vertex set $\{v : |f(v) - f(x)| \ge \epsilon\}$
would have to be finite (its closure in $X$ would otherwise meet
$B = \{x\}$ at a point other than $x$, or fail to include $x$ but be
nonempty, and any accumulation would be in $B$). This would make $f$
tend to the constant $f(x)$ at infinity, contradicting
Lemma~\ref{lem:potential}. So $B$ has at least two points.
\emph{Connected.} Suppose $B$ is disconnected, so $B = B_1 \sqcup B_2$
with $B_1, B_2$ nonempty compact disjoint. The distance between them
is some $\delta > 0$ (compactness). Choose neighborhoods $N_1, N_2$ of
$B_1, B_2$ in $X$ whose closures have distance $> \delta/2$ (so they
are disjoint) and cover $B$ (since $B = B_1 \sqcup B_2$). The
complement $X \setminus (N_1 \cup N_2)$ is a closed subset of $X$
disjoint from $B$, hence a finite vertex set. Call it $F$.
Each $N_i$ contains infinitely many vertices (else its intersection
with $B$ would be at most finite, contradicting that $B_i$ is nonempty
and has no isolated points, since it is a closed subset of the compact
metric space $X$; more concretely, any point of $B_i$ is a limit of
vertices of $\Gamma$). So infinitely many vertices lie in $N_1$ and
infinitely many in $N_2$.
Only finitely many graph edges join a vertex of $N_1$ to a vertex of
$N_2$: if infinitely many did, then infinitely many such edges would
have length $\geq \delta/2$, contradicting \eqref{eq:edge-lengths}.
Delete their vertices together with the finite set $F$. The remaining
graph has no path between the two sides: any path from $N_1$ to $N_2$
would either stay in $X \setminus (N_1 \cup N_2)$ (impossible: that
set is finite and all its vertices are deleted), or cross via one of
the finitely many edges between $N_1$ and $N_2$ (all their endpoints
deleted). So the remaining graph splits into at least two infinite
components. A connected locally finite graph minus finitely many
vertices has only finitely many components, so we really get infinite
components on each side, contradicting one-endedness
(Lemma~\ref{lem:found-ends}).
\end{proof}
\subsection*{The near-path property}
Connectedness of $B$ is not enough to control paths near a boundary
point. We now show that vertices approaching the same boundary point
can be joined by graph paths whose vertices also approach that point.
This is where the localized coordinates are essential.
\begin{lemma}[The near-path property]\label{lem:compact-near-path}
For every $x \in B$ and every open neighborhood $W$ of $x$ in $X$, there
is an open neighborhood $V \subseteq W$ of $x$ such that any two
vertices in $V$ can be joined by a finite graph path whose vertices all
lie in $W$.
Equivalently: if $u_n, v_n \in \Gamma$ both converge to the same $x \in
B$, then there exist joining graph paths $P_n$ with
$\sup\{\rho(w, x) : w \text{ is a vertex of } P_n\} \to 0$.
\end{lemma}
\begin{proof}
\emph{Component identities.} For a good box $O \in \mathcal O$ with its
finite component list $U_1, \ldots, U_r$, the following identities hold
on vertices:
\[
\sum_{j=1}^r h_{O, U_j} = \psi_O,
\qquad
h_{O, U_i} h_{O, U_j} = 0 \quad (i \neq j).
\]
By continuity they extend to all of $X$. Consequently, if
$\psi_O(x) > 0$, exactly one component $U_O(x)$ satisfies
\[
h_{O, U_O(x)}(x) = \psi_O(x) > 0,
\]
and all vertices sufficiently near $x$ also fall in this component (by
continuity of the localized coordinates and the fact that the $h_{O,U}$
are nonnegative, sum to $\psi_O$).
\emph{Choice of a small feature.} Shrink $W$ to a neighborhood defined
by finitely many coordinate conditions, each with positive tolerance
around its value at $x$. The base coordinates give a continuous
projection $X \to [0,1]^{\mathcal F_0}$. Choose a good basis box $O'$
about the base profile of $x$ small enough to have three properties,
stated verbatim below as (P1)--(P3):
\begin{enumerate}[label=(P\arabic*)]
\item every tested base coordinate satisfies its prescribed tolerance
throughout $O'$;
\item for every tested localized coordinate $h_{O,U}$ with
$\psi_O(x) > 0$, we have $O' \subset O$, and the variation of
$\psi_O$ over $O'$ is $< $ the prescribed tolerance;
\item if a tested box has $\psi_O(x) = 0$, its margin is
$< $ the prescribed tolerance throughout $O'$.
\end{enumerate}
These are finitely many open conditions on the base profile of $x$, so
Lemma~\ref{lem:good-boxes} supplies such a good basis box $O'$.
\emph{Selection of the relevant component.} We have $\psi_{O'}(x) > 0$
(the box $O'$ contains the base profile of $x$). Let $U' =
U_{O'}(x)$ be the component selected by $h_{O',U'}(x) = \psi_{O'}(x)$.
For any tested box $O$ with $\psi_O(x) > 0$: since $O' \subset O$ and
$U'$ is connected inside $O$, we have $U' \subseteq$ a single component
of $O$. That component must be $U_O(x)$ because, by density of vertices
near $x$ and continuity of coordinates, there is a vertex near $x$
where both $h_{O', U'}$ and $h_{O, U_O(x)}$ are positive; and their
product identities force these two positive-mass coordinates to be the
same one.
\emph{Conclusion for $U'$.} Now consider any tested localized coordinate
$h_{O,U}$, evaluated on $U'$:
\begin{itemize}
\item If $\psi_O(x) > 0$ and $U = U_O(x)$, then $h_{O,U}|_{U'} =
\psi_O|_{U'}$ (since $U'$ is inside the selected component and
therefore the indicator $\mathbf 1_U$ equals $1$ there). Its variation
is controlled by (P2).
\item If $\psi_O(x) > 0$ but $U \ne U_O(x)$, then $h_{O,U} = 0$
identically on $U'$ and at $x$.
\item Finally, if $\psi_O(x) = 0$, then $0 \le h_{O,U} \le \psi_O$
everywhere (since $h_{O,U} = \psi_O \mathbf 1_U$), so the value is
controlled by (P3).
\end{itemize}
Thus all vertices of $U'$ lie in $W$.
\emph{The neighborhood $V$.} Set
\[
V = W \cap \{y \in X : h_{O', U'}(y) > \tfrac12 \psi_{O'}(x)\}.
\]
This is open in $X$, contains $x$ (because $h_{O', U'}(x) = \psi_{O'}(x)
> \frac12 \psi_{O'}(x)$), and every vertex in $V$ belongs to $U'$.
Since $U'$ is a connected component of a subgraph, any two vertices of
$U'$ can be joined by a path in $U'$, hence in $W$. That proves the
neighborhood formulation.
\emph{Sequential formulation.} Apply the neighborhood formulation to
successively smaller metric balls $W_k = B_\rho(x, 1/k)$ about $x$,
obtaining $V_k$ with the stated property, and choose radii slowly
enough that any prescribed pair of converging sequences
$u_n, v_n \to x$ eventually lie inside the same $V_k$. Conversely,
failure of the sequential statement for a fixed $W$ would give two
sequences $u_n, v_n \to x$ with no joining paths in $W$, contradicting
the neighborhood formulation.
\end{proof}
We summarize the properties of $X$, $B$, and the metric $\rho$ obtained
so far:
\begin{itemize}
\item[(i)] $X = \Gamma \sqcup B$ is compact metrizable; $\Gamma$ is
discrete (its points are isolated); $B$ is a closed nonempty
connected compact subset with at least two points.
\item[(ii)] The edge-length function $\lambda(e) = \rho(e^-, e^+)$ is
square-summable: $\|\lambda\|_{\ell^2} < \infty$.
\item[(iii)] (near-path) For every $x \in B$ and every neighborhood $W$
of $x$ in $X$, there is a neighborhood $V \subseteq W$ of $x$ such
that any two vertices of $V$ can be joined by a graph path inside $W$.
\end{itemize}
These three properties are the only role of the compactification until
Section~\ref{sec:boundary-topology}.
%=============================================================
\section{Convergence dynamics on the boundary}
\label{sec:dynamics}
%=============================================================
Now we bring the action of $\Gamma$ into play. Recall the metric
$\rho$ on $X$ and the edge-length function $\lambda(e) =
\rho(e^-, e^+)$. From \eqref{eq:edge-lengths} we know
$\lambda \in \ell^2(E(G))$.
Standing notations: for a function $F : X \to \mathbb R$ and a regular
level $t$, $S(F, t)$ is the set of edges whose endpoints lie on
opposite sides of $t$. Two edges are \emph{near} if they are the same
edge or occur together on the boundary of a common $2$-cell.
\subsection*{Escape to infinity and the two-sided move}
First a gentle observation that drives everything else.
\begin{lemma}[Moving a bounded distance moves us a little]\label{lem:dyn-right-move}
For every fixed $s \in \Gamma$,
\begin{equation}\label{eq:dyn-right-move}
\rho(u, us) \longrightarrow 0
\qquad
(u \longrightarrow \infty \text{ in } \Gamma).
\end{equation}
Here ``$u \to \infty$'' means $u$ escapes every finite subset of $\Gamma$.
The convergence is uniform over any fixed finite set of choices of $s$.
\end{lemma}
\begin{proof}
Fix a finite edge path from $1$ to $s$ and translate it by $u$. Each
translated edge escapes every finite edge set as $u \to \infty$. Because
$\lambda \in \ell^2$, and because $\ell^2$ functions tend to zero outside
finite sets, each summand $\lambda(ue)$ tends to zero. Summing the
finitely many edges of the path gives
$\rho(u, us) \le \sum_{e} \lambda(ue) \to 0$. The same argument works
for a fixed finite list of choices of $s$ (the corresponding paths give
finitely many edges, and each summand still tends to zero).
\end{proof}
\subsection*{Lengths of translated level sets}
The square-summability controls how the level sets of any base
coordinate behave after translation.
\begin{lemma}[Translated cut lengths]\label{lem:dyn-length}
For every base coordinate $F \in \mathcal F_0$ and every $g \in \Gamma$,
\begin{equation}\label{eq:dyn-length}
\int_0^1 \sum_{e \in S(F, t)} \ \sum_{e' \text{ near } e}
\lambda(ge') \, dt
= \sum_e |dF(e)| \sum_{e' \text{ near } e} \lambda(ge').
\end{equation}
This quantity is finite and tends to $0$ as $g \to \infty$. There is a
constant $C$, independent of $g, F, t$, such that for every dually
connected set $D$ of edges,
\begin{equation}\label{eq:dyn-dual-diameter}
\operatorname{diam}_\rho \{gv : v \text{ is an endpoint of an edge of } D\}
\leq C \sum_{e \in D} \sum_{e' \text{ near } e} \lambda(ge').
\end{equation}
Moreover the sum of these diameter bounds over distinct dual components
is bounded by the same expression with their union in place of $D$.
\end{lemma}
\begin{proof}
\emph{Equality \eqref{eq:dyn-length}.} For each edge, the set of levels
$t$ at which it straddles is an interval of length $|dF(e)|$.
Substituting $\sum_{e \in S(F,t)} X_e$ for a summable family $(X_e)$
and applying Tonelli's theorem exchanges the level integral with the
edge sum, giving the equality.
\emph{Finiteness and vanishing.} Define
$a(e') = \sum_{e : e' \text{ near } e} |dF(e)|$. Bounded incidence of
``near'' (uniformly in either variable, because each cell has bounded
size and bounded cell incidence) makes $a \in \ell^2(E(G))$ (Cauchy-Schwarz
against the $\ell^2$ function $e \mapsto |dF(e)|$ over the uniformly
bounded neighborhoods). The right side of \eqref{eq:dyn-length} is
$\sum_{e'} a(e')\lambda(ge')$, finite by Cauchy-Schwarz. To see it
tends to zero as $g \to \infty$, approximate $a$ and $\lambda$ in
$\ell^2$ by finitely supported functions: for finitely supported
$a_0, \lambda_0$, the pairing $\sum_{e'} a_0(e')\lambda_0(ge')$
vanishes for large $g$ (the supports of $a_0$ and $g \cdot \operatorname{supp}(\lambda_0)$
are eventually disjoint), and Cauchy-Schwarz controls the error.
\emph{Diameter bound \eqref{eq:dyn-dual-diameter}.} Fix a dually
connected set $D$ and any two edges $e_1, e_2 \in D$. Join them by a
simple dual path: a sequence of edges $e_1 = E_0, E_1, \ldots, E_n =
e_2$ where consecutive $E_i, E_{i+1}$ lie on a common cell. For each
consecutive pair, traverse the boundary of a common cell. This uses
each consecutive pair exactly once, and the remaining edges used in the
traversal are ``near'' those in $D$. Bounded cell size and bounded cell
incidence give a uniform bound $C'$ on the number of times any near
edge is used. Then by triangle inequality
\[
\rho(gu, gv) \le C' \sum_{e \in D} \sum_{e' \text{ near } e} \lambda(ge'),
\]
where $u, v$ are any two endpoints of edges in $D$. Summing over all
pairs (or taking the diameter directly) yields \eqref{eq:dyn-dual-diameter}
with $C = C'$. Summing over dual components gives the last claim.
\end{proof}
\begin{lemma}[Collapse of translated cuts]\label{lem:dyn-walls}
Suppose $g_n \to a \in B$. Given finitely many coordinates from $\mathcal F$,
there is a subsequence such that, for almost every level of each
coordinate, all endpoints of the $g_n$-translated straddling set tend
uniformly to $a$. If also $g_n^{-1} \to b \in B$, the subsequence can
be chosen so that the corresponding inverse-translated straddling sets
tend uniformly to $b$.
\end{lemma}
\begin{proof}
Fix a finite list $F_1, \ldots, F_r$ of base functions.
Lemma~\ref{lem:dyn-length} applied to each yields sums
$\sum_n \int_0^1 \sum_{e \in S(F_i, t)} \sum_{e' \text{ near } e}
\lambda(g_n e') \, dt$. Along a subsequence these are summable in $n$
(any nonnegative summable-in-$n$-sequence argument works: pick a
subsequence making the sums finite). For almost every level, the
nonnegative inner sums then tend to zero as $n \to \infty$.
Fix a regular base cut $S(F, t)$ at such a level. It has finitely many
dual components (Lemma~\ref{lem:wall-components}). Choose one endpoint
$v_C$ in each nonempty dual component. The translates $g_n v_C$ tend to
$a$ by Lemma~\ref{lem:dyn-right-move}. The diameter estimate
\eqref{eq:dyn-dual-diameter} with our chosen level makes every endpoint
of the translated cut tend to $a$ too (the sum tends to zero).
For a localized coordinate $h_{O,U}$: its straddling set is contained
in a finite union of base straddling sets at levels of the form
$a_i + t$ or $b_i - t$ (Lemma~\ref{lem:localized-energy}). Include
those base coordinates in the finite list. Translations and reflections
of the level parameter preserve measure zero sets, so almost every
$t$ has the required estimates for all the covering cuts. Levels outside
$[0,1]$ contribute empty cuts.
To handle both directions, choose the summable-in-$\ell^2$-over-$n$
subsequence simultaneously for $g_n$ and $g_n^{-1}$.
\end{proof}
\subsection*{The convergence action}
\begin{proposition}[Convergence action]\label{prop:dyn-convergence}
If $g_n \to a \in B$ and $g_n^{-1} \to b \in B$, then
\begin{equation}\label{eq:dyn-convergence}
g_n x \longrightarrow a
\quad
\text{uniformly for $x$ in each compact subset of } X \setminus \{b\}.
\end{equation}
Consequently $\Gamma$ acts on $B$ by a convergence action; that is, on
the space of distinct triples of $B$, the action is properly
discontinuous.
\end{proposition}
\begin{proof}
\emph{Compact-uniform convergence.} Suppose \eqref{eq:dyn-convergence}
fails. After taking a subsequence, there is $x_n \to r \ne b$ with
$g_n x_n \to s \ne a$. Since vertices are dense in $X$, choose
$v_n \in \Gamma$ with $\rho(v_n, x_n) < 1/n$ and
$\rho(g_n v_n, g_n x_n) < 1/n$. Then $v_n \to r$ and
$g_n v_n \to s$.
Choose a coordinate $F_1$ separating $r$ from $b$ (so $F_1(r) \ne
F_1(b)$) and a coordinate $F_2$ separating $s$ from $a$. Apply
Lemma~\ref{lem:dyn-walls} to the pair $(F_1, F_2)$ and to a level
strictly between the respective values that avoids the null exceptional
sets; call these the first and second cuts.
After the subsequence, the first translated cut has all endpoints
uniformly near $a$, and the second inverse-translated cut has all
endpoints uniformly near $b$.
Let $C_n$ be the graph component of the $r$-side of the first cut
containing $v_n$ (the $r$-side being the side of the cut where
$F_1 > F_1(r)$, say). Note $F_1(b) < t < F_1(r)$, so a small neighborhood
of $b$ misses the whole $r$-side (all its vertices satisfy $F_1 > t$).
Both sides have vertices (the cut straddles).
A graph path from $v_n$ to the other side of the cut must contain a
first straddling edge; its endpoint $w_n$ on the $r$-side of that edge
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 of the first
cut for large $n$. Since $w_n$ is on the $r$-side and near $a$ (uniformly),
$g_n w_n$ lies on the $s$-side of the second cut ($s \ne a$ and the second
cut separates $s$ from $a$). But then $g_n$ of a path in $C_n$ from
$v_n$ to $w_n$ must include an edge of the second cut (since $g_n v_n$
and $g_n w_n$ sit on opposite sides of it), a contradiction: the
inverse-translated second cut has no endpoints on the $r$-side of the
first cut for large $n$.
\emph{Proper discontinuity on triples.} Suppose, for contradiction,
that infinitely many distinct group elements take some point of one
compact set $K_1$ of distinct triples to some compact set $K_2$ of
distinct triples. Pass to converging subsequences: source triple
$(x_1, x_2, x_3) \in K_1$ and target triple $(y_1, y_2, y_3) \in K_2$
and group elements $g_n$. Passing further, $g_n \to a \in B$ and
$g_n^{-1} \to b \in B$ (limits exist by compactness of $X$; they lie in
$B$ since group vertices are isolated). At least two coordinates of the
source triple differ from $b$ (only one coordinate can equal $b$).
By \eqref{eq:dyn-convergence}, both of those coordinates map to $a$ in
the limit, contradicting that the target triple has distinct entries.
\end{proof}
\subsection*{Parabolic elements and minimality}
In a convergence action on a compact metric space with at least three
points, every infinite-order element is either \emph{parabolic} (one
fixed point, and iterates converge to that point everywhere) or
\emph{loxodromic} (two fixed points, one attracting and one repelling).
Moreover, the stabilizer of a point that contains a loxodromic element
is virtually cyclic. These classical facts are
\cite[Lemma~6.2 and Prop.~6.4]{Bowditch1999Treelike}; they apply
because $B$ is a nondegenerate continuum.
\begin{lemma}[Parabolic comparison]\label{lem:dyn-parabolic}
Let $h \in \Gamma$ be parabolic with unique fixed point $p \in B$. Then
$h^n \to p$ and $h^{-n} \to p$ in $X$ as $n \to +\infty$; both sequences
of maps converge to the constant map $p$ uniformly on compact subsets of
$X \setminus \{p\}$ and pointwise on $B$. Furthermore,
\begin{equation}\label{eq:dyn-parabolic-comparison}
\rho(g, gp) \longrightarrow 0
\qquad
(g \to \infty \text{ in } \Gamma).
\end{equation}
\end{lemma}
\begin{proof}
\emph{Convergence of powers.} The powers $h^n$ escape every finite
subset of $\Gamma$ as $n \to \pm\infty$ (a finite set contains finitely
many powers of the infinite-order $h$). If a subsequence of $h^n$
converges to $z \in B$, then by Lemma~\ref{lem:dyn-right-move}
$\rho(h^n, h^{n+1}) \to 0$ along that subsequence, while continuity of
left-multiplication by $h$ on $X$ gives $h^{n+1} \to h z$. So $hz = z$,
and uniqueness of the parabolic fixed point forces $z = p$.
Compactness of $X$ and the argument for both signs give convergence of
the entire sequences.
Let $x \in X \setminus \{p\}$ and let $W$ be a small ball about $p$
disjoint from a compact neighborhood of $x$. Then $h^{-N} W$ misses $x$
for large $N$ (else $x = h^{-N} y$ with $y \in W$, contradicting that
$h^N x = y \to p$ requires $x = p$). So $h^n x \to p$ for any such
$x$, and compact-uniformity follows from compactness.
\emph{Comparison \eqref{eq:dyn-parabolic-comparison}.} Suppose the assertion
fails for a sequence $g_n \to \infty$. Passing to a subsequence, we
can arrange $g_n \to a \in B$ and $g_n p \to c \ne a$.
First, the cosets $g_n \langle h \rangle$ escape every finite subset
of $\Gamma$. If some $w$ were in infinitely many of them, then
$g_n = w h^{j_n}$ along a subsequence with $|j_n| \to \infty$, and both
$g_n \to w p$ and $g_n p \to w p$ (since $h^{\pm j_n}$ acts almost
constantly on points near $p$ for large $|j_n|$), contradiction.
For fixed $n$, the sequence $u_{n,j} = g_n h^j$, $j \ge 0$, starts at
$g_n$ and tends to $g_n p$. By the whole-coset escape and Lemma~\ref{lem:dyn-right-move},
\[
\sup_{j \ge 0} \rho(u_{n,j}, u_{n,j+1}) \longrightarrow 0
\quad \text{as } n \to \infty.
\]
Choose three distinct numbers strictly between $0$ and $\rho(a, c)$.
For each number, choose the first index $j^{(i)}_n$ at which
$\rho(a, u_{n, j^{(i)}_n})$ reaches it. Vanishing of the step size plus
compactness gives (after a common subsequence) three selected vertex
sequences with distinct limits $z_1, z_2, z_3 \in B$.
The conjugate $k_n = g_n h g_n^{-1}$ sends each selected vertex to its
successor, so asymptotically it fixes the three points $z_i$. If $k_n$
escapes finite sets, pass to a subsequence with $k_n \to a'$ and
$k_n^{-1} \to b'$; at least two of the $z_i$ avoid $b'$, and
Proposition~\ref{prop:dyn-convergence} sends both to $a'$, contradicting
distinctness. If $k_n$ stays in a finite set (after passing to a
subsequence), the $k_n$ converge (as group elements) to a fixed
conjugate of $h$, and by continuity that element fixes all three $z_i$;
but $h$ is parabolic, so its conjugate has only one fixed point.
Contradiction. This proves \eqref{eq:dyn-parabolic-comparison}.
\end{proof}
\begin{proposition}[Minimality]\label{prop:dyn-minimal}
The action of $\Gamma$ on $B$ is minimal: every orbit is dense.
\end{proposition}
\begin{proof}
\emph{Step 1.} Every invariant subset with at least two points is dense.
Given $a \in B$, choose group vertices $g_n \to a$ and pass to a
subsequence with $g_n^{-1} \to b \in B$. An invariant subset with at
least two points contains some $z \ne b$; by
Proposition~\ref{prop:dyn-convergence}, $g_n z \to a$. Since
$\{g_n z\} \subset$ the invariant subset, its closure contains $a$. Since
$a$ was arbitrary, the invariant subset is dense.
\emph{Step 2 (no global fixed point).} Suppose some $c \in B$ is fixed
by every element of $\Gamma$. Choose $a \ne c$ and vertices $g_n \to a$.
Passing to a subsequence, Proposition~\ref{prop:dyn-convergence} plus
$g_n c = c$ forces $g_n^{-1} \to c$ (because the source triple involving
$c$ must go to some target with the same entry $c$, which is the
repelling limit).
Choose disjoint nonempty closed neighborhoods $A, C$ of $a, c$ in $B$.
For large $n$, $g_n A \subset A$ and $g_n^{-1} C \subset C$ (by
compact-uniform convergence).
Fix one such nonidentity $g_n$. It has infinite order (torsion-free).
It cannot be parabolic: Lemma~\ref{lem:dyn-parabolic} would make the
forward orbit of a point of $A$ and the backward orbit of a point of $C$
both tend to the same fixed point, which would then belong to both
disjoint closed sets. So it is loxodromic.
Since $\Gamma$ fixes $c$, the stabilizer of $c$ is all of $\Gamma$ and
contains a loxodromic element; the classical fact quoted above then
makes $\Gamma$ virtually cyclic. An infinite virtually cyclic group has
two ends, contradicting Lemma~\ref{lem:found-ends}. So no global fixed
point exists.
\end{proof}
%=============================================================
\section{Dimension and cohomology of the boundary}
\label{sec:boundary-topology}
%=============================================================
Two topological facts about $B$ will drive the contradiction: its
covering dimension is at most one, and its first Cech cohomology
vanishes. Neither requires local connectivity. The dimension bound
comes from small accumulation sets of coordinate cuts; the cohomology
vanishing transfers a finite-cover cocycle to the presentation complex
and cancels its finite coboundary there.
\subsection*{Thin walls and covering dimension}
For a coordinate $F \in \mathcal F$ and a regular level $t$, let
$P(F,t)$ be the set of endpoints of edges in $S(F,t)$, and put
\[
E(F,t) = \overline{P(F,t)} \cap B.
\]
In words: $E(F,t)$ is the part of $B$ near which the cut $S(F,t)$
accumulates. All closures and diameters use the metric $\rho$ on $X$.
\begin{lemma}[Walls have measure zero]\label{lem:topo-thin-walls}
For every coordinate $F \in \mathcal F$, the compact set $E(F,t)$ has
$\mathcal H^1_\rho$-measure zero for almost every $t \in (0,1)$; here
$\mathcal H^1_\rho$ is one-dimensional Hausdorff measure with respect
to $\rho$.
\end{lemma}
\begin{proof}
\emph{Base coordinates.} Let $F$ be a base coordinate. The
level-length estimate of Lemma~\ref{lem:dyn-length}, with $g$
the identity, gives
\[
\sum_{e \in S(F,t)} \sum_{e' \text{ near } e} \lambda(e') < \infty
\]
for almost every $t$.
Fix such a level. Traversing cell boundaries along a simple dual path
shows that the diameter of the endpoint set of a dual component $D$ is
at most
\[
C \sum_{e \in D} \sum_{e' \text{ near } e} \lambda(e'),
\]
where $C$ depends only on the finite presentation. Bounded local
incidence bounds the multiplicity with which a nearby edge is counted.
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, because deleting finitely many vertices from a
connected locally finite graph creates only finitely many components.
The endpoint-set closures of these components cover $E(F,t)$: deleting
finitely many isolated graph vertices does not change boundary
accumulation. The sum of their diameters is at most
\[
C \sum_{e \in S(F,t) \setminus D_n} \sum_{e' \text{ near } e} \lambda(e')
\longrightarrow 0.
\]
So the maximum diameter tends to zero, and these finite covers prove
$\mathcal H^1_\rho(E(F,t)) = 0$.
\emph{Localized coordinates.} A localized coordinate's straddling set
is contained in a finite union of base-coordinate walls at the shifted
levels appearing in Lemma~\ref{lem:localized-energy}. The 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 a level parameter
preserve null sets. The result follows.
\end{proof}
\begin{proposition}[Covering dimension at most one]
\label{prop:topo-dimension}
The boundary $B$ has covering dimension $\dim B \leq 1$: it admits
arbitrarily fine finite open covers of multiplicity at most two.
\end{proposition}
\begin{proof}
Choose a finite grid of levels in finitely many coordinates, each level
avoiding vertex values and satisfying Lemma~\ref{lem:topo-thin-walls}.
Let $E = \bigcup E(F_i, t_i)$ be the union of the corresponding wall
accumulation sets. It is compact and has $\mathcal H^1_\rho$-measure
zero.
\emph{$E$ has small clopen partitions.} Fix $x \in E$. The distance map
$y \mapsto \rho(x,y)$ is Lipschitz, so its image on $E$ has Lebesgue
measure zero. Hence there are arbitrarily small positive radii $r$
with the sphere of radius $r$ about $x$ disjoint from $E$. The sets
$E \cap B_\rho(x,r)$ are then clopen (both closed and open) in $E$.
Compactness gives a finite small clopen cover, which becomes a disjoint
partition by successively subtracting earlier members.
\emph{Sign tuples on $B \setminus E$.} 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 $X$
containing no endpoint of any straddling edge. Lemma~\ref{lem:compact-near-path}
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; that sign also applies
at all sufficiently nearby boundary points, so it is locally constant
on $B$. This definition is valid even when $F(x) = t$: it uses the
signs on graph vertices approaching $x$.
\emph{Small diameter.} The finitely many sign tuples partition
$B \setminus E$ into finitely many disjoint sets open in $B$. Within
any one of them, 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, then
choose the grids sufficiently fine. Every sign-tuple set consequently
has arbitrarily small diameter.
\emph{Assembling the cover.} Partition $E$ into finitely many small
clopen sets. They are compact, disjoint 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 of these pieces occurs (the
sign-tuple sets are disjoint from each other and from a neighborhood
of $E$ near each point of $E$): so the cover 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 exactly
$\dim B \leq 1$.
\end{proof}
\subsection*{Cech cohomology vanishes}
The vanishing argument uses only the near-path property, the shrinking
diameters of cell boundaries at infinity, and the group-cohomological
vanishing from Lemma~\ref{lem:found-duality}. It does \emph{not}
replace Cech cohomology by singular cohomology on the possibly
nonlocally-connected boundary.
\begin{proposition}[Cech cohomology vanishes]\label{prop:topo-cech}
With coefficients $k = \mathbb F_2$ one has
$\check H^1(B; k) = 0$.
\end{proposition}
\begin{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 is a coboundary after refinement.
\emph{Step 1 (labeling graph vertices).} Choose $\delta > 0$ so that
every ball of radius $3\delta$ in $B$ is contained in some $U_i$
(Lebesgue number). For each vertex $u \in \Gamma$, choose a nearest
point $b(u) \in B$ and a label $i(u) \in \{1, \ldots, N\}$ with
$B_B(b(u), 3\delta) \subset U_{i(u)}$. If $u \to x \in B$, then
$\rho(b(u), x) \le 2\rho(u,x) \to 0$, so all labels occurring
sufficiently near $x$ contain a fixed neighborhood of $x$ in $B$.
\emph{Step 2 (cellular cochain).} Define a cellular $1$-cochain on
$\mathcal C$ by
\[
\alpha(uv) = d_{i(u) i(v)}
\]
when that pair occurs in the cover nerve, and by $0$ otherwise.
Claim: the coboundary $\omega = d\alpha$ has finite support. Suppose
not. Then infinitely many distinct cells have nonzero coboundary.
Local finiteness of $G$ forces their vertices to escape every finite
set. Each cell boundary has at most three edges, whose lengths tend
to zero (since $\lambda \in \ell^2$), so a subsequence of the vertex
sets converges to one point $x \in B$. All labels on every sufficiently
late cell contain $x$, i.e.\ contain a fixed neighborhood of $x$ in
some $U_i$. The nerve cocycle identity then gives zero around the
cell boundary, contradiction. Repeated labels or boundary edges cause
no difficulty: the identity is a cellular sum, and $d_{ii} = 0$.
\emph{Step 3 (group-ring cocycle).} We explain why $\omega$ 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 a free module mapping
onto $\ker \partial_2$, then free modules onto successive kernels. For
any finite-support scalar cellular cochain $\omega$, the formula
\[
\Phi_\omega(c) = \sum_{g \in \Gamma} \omega(g^{-1} c)\, g
\]
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
this way.
For every finite cellular $2$-cycle $z$ and every $g \in \Gamma$,
\[
\omega(g^{-1} z) = \alpha(\partial(g^{-1} z)) = 0
\]
(since $\partial$ of a cycle is zero). So $\Phi_\omega$ vanishes on
$\ker \partial_2$ and is a cocycle on the extended free resolution.
Lemma~\ref{lem:found-duality} gives $H^2(\Gamma; k\Gamma) = 0$, so
there is a module $1$-cochain whose coboundary is $\Phi_\omega$. Taking
identity coefficients gives a finite-support scalar $1$-cochain $\eta$
with $d\eta = d\alpha$. Ordinary simple connectivity of $\mathcal C$
then supplies a scalar function $l$ on its vertices with
\[
\alpha - \eta = dl.
\]
\emph{Step 4 (back 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 give
\[
l(u) + d_{i, i(u)} = l(v) + d_{i, i(v)}.
\]
By Lemma~\ref{lem:compact-near-path}, this expression has one constant
value on all vertices sufficiently near $x$; call it $l_i(x)$. It is
independent of the chosen neighborhood, since graph vertices approach
every boundary point.
\emph{Step 5 (locally constant).} The function $l_i$ is locally
constant on $U_i$: 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
\[
l_i + l_j = d_{ij}.
\]
Refine the cover by the open sets $U_i \cap l_i^{-1}(a)$ for
$a \in k$, omitting empty ones. On this finite refinement the original
cocycle is the coboundary of the assignment $a$ to the corresponding
cover member. Its Cech class vanishes.
\end{proof}
We have shown $\dim B \leq 1$ and $\check H^1(B; k) = 0$. These alone
do not ensure local connectivity. The next steps isolate a continuum
$L \subset B$ without cut points carrying the same properties, and then
prove local connectivity by pushing graph paths toward the boundary.
%=============================================================
\section{The tree of global cut points}
\label{sec:pretree}
%=============================================================
We now isolate the argument that uses cut points of $B$. Throughout
this section we only use:
\begin{itemize}
\item[$(\Gamma 0)$] $B$ is a nondegenerate compact metric continuum;
\item[$(\Gamma 1)$] $\Gamma$ is finitely presented, torsion-free, one-ended;
\item[$(\Gamma 2)$] $\Gamma$ acts on $B$ minimally by convergence;
\item[$(\Gamma 3)$] $\Gamma$ has no nontrivial splitting over a finite or
two-ended subgroup.
\end{itemize}
All these are established (Sections~\ref{sec:foundations},
\ref{sec:dynamics}). We will not use the construction of $B$ again
except through these properties.
\subsection*{Cut points and the pretree}
A point $p \in B$ is a \emph{global cut point} if $B \setminus \{p\}$
is disconnected. Let $P$ be the set of global cut points. For distinct
$a, p, b \in P$, write $apb$ when $a$ and $b$ lie on opposite sides of
an open separation of $B \setminus \{p\}$.
The relation $apb$ is a \emph{pretree}: a strict betweenness relation
satisfying
\[
\neg(aba), \qquad abc \iff cba, \qquad \neg(abc \wedge acb),
\qquad abc,\ d \ne b \implies abd \text{ or } cbd.
\]
Write $(a,b) = \{p : apb\}$ and $[a,b] = (a,b) \cup \{a,b\}$, with
$[a,a] = \{a\}$. A subset is \emph{full} if it contains $[a,b]$
whenever it contains $a$ and $b$. A pretree is \emph{median} if every
triple has a unique point in the three pairwise closed intervals
(called the \emph{median}). It is \emph{complete} if every nonempty
full linearly ordered subset is an interval, allowed to be open or
half-open.
Bowditch's theory of pretrees applies: there is a complete median
pretree $\Phi$ containing $P$ equivariantly with the same betweenness,
together with a specific map and various structural results. We use
the following precise statement of that theory; we cite the specific
locations in \cite{Bowditch1999Treelike} and prove the one additional
fact that is essential for the all-parabolic case.
\begin{proposition}[Bowditch's pretree package]\label{prop:pret-package}
Let a group act minimally by convergence on a nondegenerate compact
metric continuum $B$, and let $P \ne \varnothing$ be its set of global
cut points. Then there is a complete median pretree $\Phi$ containing
$P$ equivariantly, with the same betweenness on $P$, satisfying:
\begin{enumerate}[label=\textup{(\roman*)}]
\item \emph{Adjacency.} Distinct $a,b \in \Phi$ are adjacent if $(a,b) =
\varnothing$. Every adjacent pair in $\Phi$ has one point in $P$ and
one outside. A full subset whose intervals are finite is the vertex
set of a simplicial tree, with these adjacent pairs as edges.
\item \emph{Support.} There is an equivariant map $\phi : B \to \Phi$,
equal to the identity on $P$, and for every $q \in \Phi$ an
equivariantly assigned nonempty compact subset $R_B(q) \subset B$.
For $q \in P$, $R_B(q) = \{q\}$; for $q \notin P$,
\[
R_B(q) = \phi^{-1}(q) \cup \Lambda(q), \qquad
\Lambda(q) = \{p \in P : p \text{ adjacent to } q\}.
\]
No point of $\Phi$ is fixed by all of $\Gamma$.
\item \emph{Dendritic quotient.} Beginning with equality on $\Phi$,
successively collapse finite-interval classes (two points lie in one
class if their closed interval in the current quotient is finite),
and at limit ordinals take the union of the preceding equivalence
relations. This process stabilizes. Its final quotient is a dendrite
$D(B)$; there is a continuous equivariant surjection $B \to D(B)$,
and the induced action is minimal and convergence.
\item \emph{Fixed-point restrictions.} If a loxodromic element fixes
$q, r \in \Phi$ with $[q,r]$ infinite, then $q, r$ are terminal (no
point lies strictly between and neither is between two others). If a
parabolic element has its unique fixed point in $B \setminus P$, then
it has exactly one fixed point in $\Phi$.
\end{enumerate}
\end{proposition}
\begin{proof}[Reference]
The complete median pretree $\Phi$ is the flow completion of $P$ from
\cite[Thm.~3.19]{Bowditch1999Treelike}. Adjacency and the simplicial-tree
description are \cite[Lemmas~3.28 and 3.34]{Bowditch1999Treelike}.
The compact supports are constructed in \cite[Sec.~5]{Bowditch1999Treelike};
their formula is \cite[Lemma~6.8]{Bowditch1999Treelike}, and absence of
a global fixed point is \cite[Lemma~6.11]{Bowditch1999Treelike}.
Full equivalence relations and their successive finite-interval
collapses are in \cite[Lemmas~4.2--4.4]{Bowditch1999Treelike}, and the
dendritic quotient identification is in \cite[Sec.~5 and Thm.~5.23]{Bowditch1999Treelike}.
The induced action is \cite[Lemma~6.5]{Bowditch1999Treelike}; minimality
follows from equivariance and surjectivity. Fixed-point assertions are
\cite[Lemmas~6.9 and 6.10]{Bowditch1999Treelike}. All these constructions
apply to a continuum before local connectivity is known.
\end{proof}
\subsection*{All cut points are parabolic}
The next result uses the splitting obstruction to show that the
parabolic case is the only one that can occur, and that the dendritic
quotient is a point.
\begin{lemma}[Cut-point stabilizers are parabolic]
\label{lem:pret-parabolic-cuts}
Under the assumptions $(\Gamma 0)$--$(\Gamma 3)$, the dendritic
quotient $D(B)$ is a single point. Every $p \in P$ has infinite
stabilizer containing no loxodromic element, and hence contains a
parabolic element of infinite order fixing $p$.
\end{lemma}
\begin{proof}
Bowditch's dendrite obstruction \cite[Lemma~1.4]{Bowditch1999Connected}
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, and torsion-freeness implies the torsion
condition. So $D(B)$ is a point.
By \cite[Thm.~6.1]{Bowditch1999Treelike}, 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
every point of $P$ is parabolic in the sense that its stabilizer is
infinite and contains no loxodromic element. Any nonidentity element
of this stabilizer has infinite order because $\Gamma$ is torsion-free;
by the convergence action classification it is therefore parabolic.
\end{proof}
The cited Theorem~6.1 does not itself produce an invariant
finite-interval class when all cut points are parabolic. We now prove
that, adapting the limit and higher-successor steps of Bowditch's proof
following his Lemma~6.13. The parabolic fixed-point observation below
permits those steps.
\begin{lemma}[Parabolic fixed points in $\Phi$]
\label{lem:pret-parabolic-star}
Let $\gamma$ be parabolic with unique fixed point $a \in P$. Every
point of $\Phi$ fixed by $\gamma$ is either $a$ or adjacent to $a$.
Consequently, if an infinite-order element fixes two points of $\Phi$
whose interval is infinite, the element is loxodromic and both points
are terminal.
\end{lemma}
\begin{proof}
Every nonempty compact $\gamma$-invariant subset of $B$ contains $a$:
the iterates of any other point converge to $a$
(Lemma~\ref{lem:dyn-parabolic}). If $q \in \Phi$ is fixed by $\gamma$,
then $R_B(q) \subset B$ is such a compact invariant subset, so contains
$a$. For $q \in P$ this gives $q = a$ (unique fixed point in $B$).
For $q \notin P$: $a \in R_B(q)$ and $\phi(a) = a \ne q$, so by the
formula of Proposition~\ref{prop:pret-package}(ii) we must have
$a \in \Lambda(q)$, i.e.\ $q$ is adjacent to $a$.
Any two points equal to or adjacent to $a$ have a finite interval
between them, contained in the union of their intervals to $a$. So a
parabolic element fixing a point of $P$ cannot have two fixed points
with an infinite interval between them. A parabolic element whose
fixed point lies outside $P$ cannot have them either by
Proposition~\ref{prop:pret-package}(iv). The infinite-order element
that has such a pair of fixed points is therefore loxodromic, and by
Proposition~\ref{prop:pret-package}(iv) both fixed points are terminal.
\end{proof}
\subsection*{An invariant finite-interval class}
The final quotient $D(B)$ is a point. We now show an invariant class
already appears at the first finite-interval collapse.
An equivalence relation on $\Phi$ is \emph{full} if all its classes are
full. For a full relation $\sim$, the quotient pretree has the induced
betweenness: a class $V$ lies between distinct classes $U, W$ when some
$v \in V$ lies between every $u \in U$ and every $w \in W$. Write
$\sim_0$ for equality and define
\begin{equation}\label{eq:pret-relations}
x \sim_{\beta + 1} y
\iff [x,y] \text{ meets finitely many } \sim_\beta \text{-classes},
\qquad
\sim_\lambda = \bigcup_{\beta < \lambda} \sim_\beta
\ (\lambda \text{ a limit ordinal}).
\end{equation}
These are the increasing invariant full relations from
Proposition~\ref{prop:pret-package}(iii); quotients remain complete
median pretrees. In particular $x \sim_1 y$ means $[x,y]$ is finite.
Goal: a fixed point of $\Phi / \sim_1$.
We record a fact from completeness that will control the direction of an
edge between collapsed classes. Let $U, V$ be disjoint nonempty full
subsets of a complete median pretree with full union. A \emph{facing
point of $V$ toward $U$} is a point $p \in V$ such that
\begin{equation}\label{eq:pret-facing}
[u,v] \cap V = [p,v]
\qquad
(u \in U, v \in V).
\end{equation}
\begin{lemma}[Facing point exists and is unique]\label{lem:pret-facing}
There is at most one facing point of $V$ toward $U$ on each side, and
at least one of $U, V$ admits one.
\end{lemma}
\begin{proof}
Choose $u \in U$, $v \in V$. Partition $[u,v]$ by the two full sets.
The initial part $A = [u,v] \cap U$ has, by completeness, the form
$[u,t]$ or $[u,t)$.
\emph{Closed case.} $t \in U$. Then $t$ is the "last point of $U$"
on $[u,v]$, and the point immediately adjacent (or the endpoint of the
adjacent open interval, if $V$ extends into $[u,v]$) witnesses the
facing point either in $U$ or in $V$.
\emph{Half-open case.} Put $m = \mathrm{med}(u,v,t)$. If $m \ne t$
then $A = [u,m]$, reducing to the closed case with $m$ in place of $t$.
Otherwise $t \in [u,v]$, $t \notin A$, so $t \in V$. Then $t$ is the
facing point of $V$.
This proves existence. Uniqueness: if $p, p' \in V$ both work, then
$\mathrm{med}(u,p,p')$ must equal both $p$ and $p'$ (via
\eqref{eq:pret-facing}), forcing $p = p'$. Interchanging $U$ and $V$
shows the symmetric statement. This is also
\cite[Lemma~6.13]{Bowditch1999Treelike}.
\end{proof}
\begin{lemma}[An invariant finite-interval class]\label{lem:pret-invariant-class}
Under $(\Gamma 0)$--$(\Gamma 3)$, $\Gamma$ preserves a finite-interval
class of $\Phi$.
\end{lemma}
\begin{proof}
Lemma~\ref{lem:pret-parabolic-cuts} says the terminal quotient in
\eqref{eq:pret-relations} is a point. Let $\alpha$ be the least ordinal
for which $\Phi/\sim_\alpha$ has a point fixed by $\Gamma$.
Proposition~\ref{prop:pret-package}(ii) gives $\alpha \ne 0$. We rule
out limits and successors greater than one.
\emph{Limit case.} Suppose $\alpha$ is a limit ordinal. Choose $x$ in
an invariant $\sim_\alpha$-class. For each member $s$ of a finite
generating set of $\Gamma$, the relation $x \sim_\alpha sx$ holds at
some stage $\beta_s < \alpha$. Let $\beta = \max_s \beta_s < \alpha$.
Then invariance and transitivity of $\sim_\beta$ give $x \sim_\beta gx$
for every word $g$ in the generators and their inverses, i.e.\ the
$\sim_\beta$-class of $x$ is invariant, contradicting minimality of
$\alpha$.
\emph{Higher successor case.} Suppose $\alpha = \beta + 1$ with
$\beta \geq 1$. 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 minimality of $\alpha$.
For adjacent classes $U, V \in \Sigma$, their union is full. So by
Lemma~\ref{lem:pret-facing} there is a facing 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, contradicting $U \ne V$. So exactly one side
has a facing point, and an automorphism cannot interchange $U, V$.
Thus the action on $\Sigma$ has no edge inversions.
\emph{Every edge with infinite stabilizer is a leaf edge.} Let $UV$ be
an edge of $\Sigma$ with infinite stabilizer, with $p \in V$ its facing
point (say; all direction labels below are interchangeable). Choose a
nonidentity stabilizing element $\gamma$; it has infinite order (torsion-free).
$\gamma$ preserves $U$ and $V$ and fixes $p$.
Claim: $\gamma$ fixes a second point outside $V$. For $y \in U$ put
$x = \mathrm{med}(p, y, \gamma y)$. Fullness gives $x \in [y,\gamma y]
\subset U$; both $x$ and $\gamma x$ lie in the linearly ordered interval
$[p,\gamma y]$. If $x = \gamma x$, done. Otherwise, replacing
$(\gamma, x)$ by $(\gamma^{-1}, \gamma x)$ if necessary, we get
$x \in (p, \gamma x)$. The intervals $[p, \gamma^n x]$, $n \geq 0$, are
then strictly increasing. Their union is a full directed arc with
first point $p$ and no last point. Completeness gives an endpoint $z$
with
\begin{equation}\label{eq:pret-endpoint}
\bigcup_{n \geq 0} [p, \gamma^n x] = [p, z).
\end{equation}
The endpoint is unique in a median pretree
\cite[Lemma~3.13]{Bowditch1999Treelike}. The union in
\eqref{eq:pret-endpoint} is $\gamma$-invariant, so $z$ is fixed.
Since $x \in (p,z)$ but $x \notin V$, fullness of $V$ forces $z \notin V$.
So $\gamma$ fixes $p$ and a second point $q \notin V$. The points $p,q$
belong to $q$ in different $\sim_\beta$-classes (since $\beta \ge 1$,
$\sim_\beta$-inequality implies infinite original interval). By
Lemma~\ref{lem:pret-parabolic-star}, $p$ is terminal in $\Phi$. Now for
$v \in V$ and $u \in U$, by facing we have $p \in [u,v]$. Since
$p \ne u$, terminality forces $v = p$. So $V = \{p\}$, i.e.\ $V$ is a
leaf vertex of $\Sigma$ (if it had two neighbors, $p$ would lie
strictly between representatives of them in $\Phi$, contradicting
terminality).
\emph{Delete leaf edges and get an invariant subtree.} Delete every
edge with infinite stabilizer, together with its leaf endpoint. The
remaining vertices form a nonempty invariant subtree (removing selected
leaves preserves paths between remaining vertices). If nothing
remained, the original tree would have at most two vertices, so an
action without inversions would have a fixed vertex, already excluded.
All remaining edges have finite stabilizer. A fixed vertex of the
remaining tree would also be fixed in $\Sigma$.
But a finitely generated one-ended group acting without inversions on
a simplicial tree with finite edge stabilizers fixes a vertex
\cite[Lemma~6.6]{Bowditch1999Treelike}. Contradiction.
The least ordinal is therefore $\alpha = 1$, which is exactly the
assertion of the lemma.
\end{proof}
\subsection*{The simplicial tree}
\begin{proposition}[Tree of cut points]\label{prop:pret-tree}
Let $(\Gamma 0)$--$(\Gamma 3)$ hold and suppose $P \ne \varnothing$.
Then there is a nontrivial minimal simplicial $\Gamma$-tree
$\mathcal T$ with no edge inversions and finite quotient, such that:
\begin{enumerate}[label=\textup{(\roman*)}]
\item $P$ is an invariant subset of its vertices, and every edge has
one endpoint in $P$ and one outside;
\item $\mathcal T$ is spanned by $P$; its other vertices have degree
at least two;
\item every edge stabilizer is infinite;
\item if $a, b, q \in P$ and $q$ separates $a$ and $b$ in $B$, then
$q \in [a,b]_{\mathcal T}$.
\end{enumerate}
Every point of $P$ is fixed by an infinite-order parabolic element.
\end{proposition}
\begin{proof}
Let $C \subset \Phi$ be the invariant finite-interval class supplied by
Lemma~\ref{lem:pret-invariant-class}. It has more than one point
(else $\Phi$ would have a fixed point, excluded by
Proposition~\ref{prop:pret-package}(ii)). So its simplicial tree has
an edge, and Proposition~\ref{prop:pret-package}(i) places one endpoint
of that edge in $P$.
\emph{$P \subset C$.} Choose $p \in C \cap P$. The orbit $\Gamma p$ is
dense in $B$ (Proposition~\ref{prop:dyn-minimal}) and lies in $C$. Now
let $q \in P$. Take an open separation of $B \setminus \{q\}$. Both
sides are nonempty open subsets, so each meets $\Gamma p$. For orbit
points $a,b$ on opposite sides, separation gives $q \in [a,b]_\Phi$.
Since $C$ is full, $q \in C$.
\emph{$\mathcal T$ spanned by $P$.} Let $\mathcal T$ be the subtree of
$C$ spanned by $P$ (union of intervals between points). Its intervals
agree with those of $\Phi$, so it retains the separation compatibility
(iv). Adjacency in Proposition~\ref{prop:pret-package}(i) gives the
bipartition $P$, and forbids edge inversions. A vertex outside $P$
cannot be a leaf of a tree spanned by $P$.
\emph{Minimality.} Let $\mathcal S \subset \mathcal T$ nonempty
invariant subtree. A single vertex would be a fixed point of $\Phi$,
excluded. Otherwise $\mathcal S$ contains an edge, hence a point of $P$.
Its orbit is again dense; separation gives $P \subset \mathcal S$.
So $\mathcal S = \mathcal T$.
\emph{Finite quotient.} Choose a base vertex $o$ and a finite
generating set $S$ for $\Gamma$. The union of the finite paths
$[o,so]$, $s \in S$, is a finite subtree after adjoining $o$. Its
translates form a connected invariant subtree, hence cover
$\mathcal T$ by minimality. So the quotient has finitely many vertex
and edge orbits.
\emph{Infinite edge stabilizers.} If an edge had finite stabilizer,
collapsing all edges outside its orbit gives a nontrivial tree with one
edge orbit and a minimal action without inversions. Bass-Serre theory
yields a nontrivial splitting over that edge stabilizer, contrary to
$(\Gamma 3)$.
The parabolic assertion is Lemma~\ref{lem:pret-parabolic-cuts}.
\end{proof}
%=============================================================
\section{The subcontinuum without cut points}
\label{sec:blocks}
%=============================================================
We have seen that $B$ always has $\dim \leq 1$ and
$\check H^1 = 0$ (Propositions~\ref{prop:topo-dimension},
\ref{prop:topo-cech}). We now show that $B$ always contains a
\emph{nondegenerate subcontinuum $L$ without cut points} that retains
these two properties, is a retract of $B$, and comes equipped with the
graph-theoretic structure needed for path pushing.
If $B$ has no cut points, we just take $L = B$; the case is simple.
Otherwise we use the tree $\mathcal T$ from Section~\ref{sec:pretree}
to extract one of its "leaf blocks".
\subsection*{The setup}
Assume $P \ne \varnothing$ for the rest of this subsection. The tree
$\mathcal T$ from Proposition~\ref{prop:pret-tree} is bipartite between
$P$ and the remaining vertices; every edge has a $P$-endpoint and a
non-$P$-endpoint. Choose \emph{any} vertex $r \in \mathcal T \setminus P$
and set
\[
R = \{p \in P : p \text{ is adjacent to } r \text{ in } \mathcal T\},
\qquad
H = \Gamma(r) \leq \Gamma.
\]
Here $\Gamma(r)$ is the stabilizer of $r$.
Since $\mathcal T$ is spanned by $P$, a non-$P$ vertex cannot be a leaf,
so $|R| \geq 2$. For distinct $a,b \in R$, the tree interval is $a - r -
b$. By Proposition~\ref{prop:pret-tree}(iv), no cut point of $B$
separates $a$ and $b$.
Also $H$ is exactly the setwise stabilizer of $R$: any element
preserving $R$ preserves the unique midpoint $r$ of the interval
$a - r - b$ for any two distinct $a, b \in R$.
\subsection*{A graph on the cut set}
\begin{lemma}[Graph on $P$]\label{lem:block-graph}
There is a connected $\Gamma$-graph $K^*$ with vertex set $P$ and
finitely many edge orbits such that the endpoints of every edge have
tree-distance two in $\mathcal T$. The graph $K$ obtained by
restricting $K^*$ to $R$ is connected with finitely many $H$-orbits of
vertices and edges.
Define the projection $\pi : P \to R$ by: $\pi(p)$ is the neighbor of
$r$ on the tree interval from $r$ to $p$. Then an edge of $K^*$ with
both endpoints in $R$ is fixed by $\pi$, and every other edge is
collapsed to a vertex. Moreover the stabilizer in $H$ of every vertex
of $K$ is infinite.
\end{lemma}
\begin{proof}
\emph{Constructing $K^*$.} The finite quotient of $\mathcal T$
(Proposition~\ref{prop:pret-tree}) gives finitely many $\Gamma$-orbits
in $P$. Choose representatives $p_0, \ldots, p_k$ of $P$-orbits and a
finite symmetric generating set $S$ of $\Gamma$. Start with the graph
containing all translates of the edges $p_0 \to s p_0$ for $s \in S$,
and of $p_0 \to p_i$ for $1 \le i \le k$ (omit loops, i.e.\ when
$p_0 = s p_0$).
This graph is connected: the $s$-edges connect the whole
$p_0$-orbit, and the $p_0 \to p_i$-edges connect every other orbit to
$p_0$. It has finitely many edge orbits.
Replace each edge by the \emph{consecutive pairs} of $P$-vertices along
the tree interval between its endpoints. These tree intervals are
finite (they are between $P$-points in $\mathcal T$). The replacement
is equivariant. So the resulting $K^*$ is still connected, has finitely
many edge orbits, and the endpoints of every edge are at tree-distance
two.
\emph{Projection properties.} Let $a s b$ be an edge of $K^*$ with
midpoint $s$ (so $a, s, b$ are collinear in $\mathcal T$ and $s$ is the
midpoint of the interval $a$-$b$). Two cases:
\begin{itemize}
\item $s = r$: then $a, b \in R$ and $\pi(a) = a$, $\pi(b) = b$ — the
edge is fixed by $\pi$.
\item $s \ne r$: then $a$ and $b$ lie in the same component of
$\mathcal T \setminus \{r\}$, so they have the same neighbor on the
path to $r$, i.e.\ $\pi(a) = \pi(b)$ — the edge collapses.
\end{itemize}
This shows: projecting a path between two members of $R$ and deleting
stationary steps gives a path in $K$, so $K$ is connected.
\emph{Orbits and stabilizers.} If a group element carries one edge of
$K$ to another, it carries their unique tree midpoints (both equal to
$r$) to one another, so it lies in $H$. Consequently every
$\Gamma$-orbit of $K^*$-edges meets $K$ in at most one $H$-orbit, so
there are finitely many $H$-edge orbits in $K$. Since $K$ is connected
with at least two vertices, it has finitely many $H$-vertex orbits.
Finally, for $p \in R$: $H_p \subseteq H$ is the stabilizer of the tree
edge $rp$ in the whole group $\Gamma$, which is infinite by
Proposition~\ref{prop:pret-tree}(iii).
\end{proof}
The graph $K$ can have infinite valence and infinite vertex stabilizers.
This will cause no trouble: our arguments use orbit sets, not finite
valence.
\begin{lemma}[Paths in $K$ near $B$]\label{lem:block-near-path}
Suppose $a_n, b_n \in P$ both tend to $x \in B$. Then there are paths
in $K^*$ from $a_n$ to $b_n$ whose vertices all tend uniformly to $x$.
If a path in $K^*$ has distinct projected endpoint labels in $R$, then
after stationary steps are removed, its projection uses only vertices
of the original path.
\end{lemma}
\begin{proof}
By Lemma~\ref{lem:block-graph} (finitely many representatives), write
$a_n = g_n p_{i_n}$ and $b_n = h_n p_{j_n}$ with $i_n, j_n \in
\{0, \ldots, k\}$. We may choose $g_n, h_n$ to escape every finite
subset of $\Gamma$: each representative $p_{i}$ has infinite stabilizer
$H_{p_i}$ (since $p_i$ is a vertex of $P$ and each has infinite
stabilizer by Proposition~\ref{prop:pret-tree}), so we can multiply by
any element of that stabilizer.
By Proposition~\ref{prop:pret-tree} every $P$-point is a parabolic
fixed point; a parabolic fixed point has $\rho(g p_i, g) \to 0$ as
$g \to \infty$ (Lemma~\ref{lem:dyn-parabolic}). Applied to the finite
list of possible $p_i$, we get $g_n \to x$ and $h_n \to x$ in $X$.
Now by Lemma~\ref{lem:compact-near-path}, the Cayley paths from $g_n$
to $h_n$ (in $G$, the original graph) have all vertices tending
uniformly to $x$. Fix finite $K^*$-paths from $p_0$ to $sp_0$ for every
Cayley generator $s$, and from $p_0$ to each representative $p_i$.
Replace each Cayley step $g \to gs$ by the $g$-translate of the first
appropriate $K^*$-path, and similarly at the endpoints (using the
translated paths to $p_{i_n}$ and $p_{j_n}$). The result is a
$K^*$-path from $a_n$ to $b_n$.
The fixed $K^*$-path templates involve only finitely many points of $P$.
For each such point $t$, we have $\rho(gt, g) \to 0$ as $g$ escapes
finite sets (parabolic comparison). The Cayley vertices escape finite
sets uniformly because they tend to the boundary point $x$. So all
transferred path vertices tend uniformly to $x$.
Regarding the last assertion: by Lemma~\ref{lem:block-graph}, every
non-constant projected edge (i.e.\ an edge whose image under $\pi$ is
not a collapsed vertex) was already an edge of the original path. So
when the projected endpoints differ, every vertex of the projected
path lies on the projection of a path edge in the original path.
\end{proof}
\subsection*{Retraction onto the subcontinuum}
\begin{proposition}[Retraction and the subcontinuum $L$]
\label{prop:block-retraction}
Assume $P \ne \varnothing$. The map $\pi : P \to R$ extends to a
continuous retraction
\[
\mathfrak r : B \to L,
\qquad
L = \overline{R}^{\, B}.
\]
The map $\mathfrak r$ is locally constant on $B \setminus L$ and is
the identity on $L$. Moreover:
\[
L \text{ is a nondegenerate continuum without cut points},
\qquad
\dim L \leq 1,
\qquad
\check H^1(L; \mathbb F_2) = 0.
\]
\end{proposition}
\begin{proof}
\emph{Minimality gives density.} By Proposition~\ref{prop:dyn-minimal},
the invariant nonempty set $P$ is dense in $B$.
\emph{$\pi$ extends continuously on $B \setminus L$.} Let
$x \in B \setminus L$. Choose a neighborhood $O$ of $x$ in $B$ with
closure disjoint from $L$. Claim: $\pi$ is constant on $P \cap V$ for
some neighborhood $V \subseteq O$ of $x$. If not, we could choose
$a_n, b_n \in P \to x$ with $\pi(a_n) \ne \pi(b_n)$.
Lemma~\ref{lem:block-near-path} would join $a_n$ to $b_n$ by paths
whose vertices eventually lie entirely in $O$. Any such path with
distinct projected endpoints uses at least one vertex of $R$ lying in
$O$ (by the last part of that lemma). But $O \subseteq B \setminus L$
and $R \subseteq L$, so $O \cap R = \varnothing$ — contradiction.
\emph{$\pi$ extends continuously on $L$.} Now let $x \in L$ and
$a_n \in P \to x$. Choose $b_n \in R \to x$ (density of $R$ in $L$).
Join $a_n$ to $b_n$ by paths from Lemma~\ref{lem:block-near-path}.
Whenever $\pi(a_n) \ne b_n$, the same lemma places $\pi(a_n)$ on the
original path, so $\rho(\pi(a_n), b_n) \to 0$, hence
$\pi(a_n) \to x$. On the other hand, if $\pi(a_n) = b_n$, then
$\pi(a_n) = b_n \to x$ directly. So in all cases $\pi(a_n) \to x$.
\emph{Formally constructing $\mathfrak r$.} Take the closure of the
graph of $\pi$ in $B \times L$. By density of $P$ and compactness of
$L$, this closed set projects onto $B$. The two paragraphs above show
it is a graph: exactly one point of $L$ over each point of $B$. Its
projection onto $B$ is a continuous bijection from a compact space to a
Hausdorff space, hence a homeomorphism. The second coordinate defines
$\mathfrak r : B \to L$.
By the first paragraph, $\mathfrak r$ is locally constant on
$B \setminus L$. It's the identity on $L$ by construction (the graph
of $\pi$ contains $(q, q)$ for every $q \in R$, and by density and
continuity it contains $(q,q)$ for every $q \in L$).
\emph{Continuum properties.} $B$ is connected and $L$ is its
continuous image, so $L$ is connected. $R$ has at least two points,
so $L$ is nondegenerate. Since $L$ is closed in the compact metric
space $B$, it is compact metric, and $\dim L \le \dim B \leq 1$.
\emph{Čech cohomology of $L$.} Let $i : L \hookrightarrow B$ be the
inclusion. Then $\mathfrak r \circ i = \mathrm{id}_L$. Functoriality of
Čech cohomology gives $i^* \mathfrak r^* =
\mathrm{id}_{\check H^1(L;\mathbb F_2)}$. So $\mathfrak r^*$ injects
$\check H^1(L;\mathbb F_2)$ into $\check H^1(B;\mathbb F_2) = 0$.
Thus $\check H^1(L;\mathbb F_2) = 0$.
\emph{$L$ has no cut points.} Suppose toward contradiction that $q \in L$
is a cut point of $L$: $L \setminus \{q\} = U \sqcup V$ with $U, V$
nonempty open in $L$. By density, there is $a \in R \cap U$ and
$b \in R \cap V$. The extra fiber
\[
F = \mathfrak r^{-1}(\{q\}) \setminus \{q\}
\]
is closed in $B \setminus \{q\}$ by continuity of $\mathfrak r$. It is
also open in $B \setminus \{q\}$: it lies in $B \setminus L$, where
$\mathfrak r$ is locally constant, hence near any point of $F$ the map
$\mathfrak r$ is constant on a neighborhood; since $\mathfrak r$ takes
value $q$ at that point of $F$, the whole neighborhood is in $F$.
Therefore
\[
B \setminus \{q\}
= \bigl(\mathfrak r^{-1}(U) \cup F\bigr) \sqcup \mathfrak r^{-1}(V)
\]
is a separation: $\mathfrak r^{-1}(U)$, $\mathfrak r^{-1}(V)$ are open
(by continuity), $F$ is open (as above) and disjoint from both, and
$a \in \mathfrak r^{-1}(U)$, $b \in \mathfrak r^{-1}(V)$ lie on
opposite sides. This would make $q$ a cut point of $B$ separating two
members of $R$, contradicting the observation (from
Proposition~\ref{prop:pret-tree}(iv)) that no cut point of $B$ separates
two distinct members of $R$.
\end{proof}
\subsection*{The common setting for path pushing}
We now include the case $P = \varnothing$ and state the common setting
that the path-pushing argument uses. In the $P = \varnothing$ case we
just take $L = B$ and the original graph.
\begin{proposition}[Setting for path pushing]\label{prop:block-setting}
There are a subgroup $H \leq \Gamma$, a compact space
$Y = H \sqcup L$ (with the metric $\rho$ inherited from $X$), and a
connected $H$-graph $K$ with vertex set $I \subseteq Y$, such that:
\begin{enumerate}[label=\textup{(\arabic*)}]
\item $Y = \overline{H}^{\,X}$; the points of $H$ are isolated in $Y$;
$L$ is a nondegenerate continuum without cut points satisfying
$\dim L \leq 1$ and $\check H^1(L; \mathbb F_2) = 0$.
\item If $h_n \to a \in L$ and $h_n^{-1} \to b \in L$, then $h_n$
converges to the constant map $a$, uniformly on compact subsets of
$Y \setminus \{b\}$.
\item $K$ has finitely many $H$-orbits of vertices and edges; either
$I = H$ or $I \subseteq L$; every point of $L$ is the limit of a
sequence of distinct vertices of $I$.
\item For each fixed $w \in I$: $\rho(hw, h) \to 0$ as $h \to \infty$
in $H$.
\item For every $x \in L$ and every neighborhood $W$ of $x$ in $Y$,
there is a neighborhood $V$ of $x$ in $Y$ such that any two vertices
of $I \cap V$ can be joined by a finite path of $K$ with all vertices
in $W$.
\item If $I \subseteq L$, then for every $p \in I$ there is an
infinite-order element $h \in H_p$ with $h^j z \to p$ for every
$z \in L$ as $j \to +\infty$.
\end{enumerate}
No finite-generation of $H$ and no local finiteness of $K$ is assumed.
\end{proposition}
\begin{proof}
\emph{Case $P = \varnothing$.} Take $H = \Gamma$, $Y = X$, $L = B$,
$K$ = the Cayley graph $G$ with $I = H$. Continuum properties are
those of $B$ (Lemma~\ref{lem:compact-boundary}; absence of cut points
is the assumption of this case). (2) is
Proposition~\ref{prop:dyn-convergence}. (3), (4) are from the
construction of $X$, Proposition~\ref{prop:dyn-convergence},
Lemma~\ref{lem:compact-near-path}, and Lemma~\ref{lem:dyn-right-move}.
Every boundary point is approached by distinct group vertices because
isolated vertices accumulate on $B$ everywhere (dense in $X$). (5) is
Lemma~\ref{lem:compact-near-path}. (6) has no content: $I = H$ means
$I$ is disjoint from $L$.
\emph{Case $P \ne \varnothing$.} Take
\[
H = \Gamma(r), \qquad I = R, \qquad
L = \overline{R}^{\,B}, \qquad Y = H \sqcup L,
\]
and $K$ the graph of Lemma~\ref{lem:block-graph}. Property (1) is
Lemma~\ref{lem:compact-boundary}, Proposition~\ref{prop:block-retraction}.
For the remaining interfaces:
$Y = \overline{H}^{\,X}$: fix $p \in R$. If an escaping sequence
$h_n \in H$ converges to a boundary point, then parabolic comparison
$(\rho(h,hp) \to 0)$ and $h_n p \in R$ show that the limit lies in
$L$. Conversely, for any $p \in R$, the infinite stabilizer $H_p$
supplies distinct $h_n$ with $h_n = p$ (i.e.\ $h_n p = p$, but the
$h_n$ themselves are distinct and tend to $p$ by parabolic comparison).
So $R \subseteq \overline{H}^{\,X}$, and taking closures gives
$L \subseteq \overline{H}^{\,X}$. No point of $\Gamma \setminus H$ lies
in the closure (group vertices are isolated). This proves equality.
Property (2): restriction of Proposition~\ref{prop:dyn-convergence} to
$H$.
Property (3): finite orbit sets come from Lemma~\ref{lem:block-graph}.
$I = R \subseteq L$ (since $R \subseteq \overline R^{\,B} = L$). Every
point of $L$ is a limit of distinct members of $R$: $R$ is dense in $L$
and $L$ has no isolated points (else that point would be isolated in
the continuum $B$, which has no isolated points).
Property (4): parabolic comparison (Lemma~\ref{lem:dyn-parabolic}).
Property (5): near-path for $K$. Take two sequences $a_n, b_n \in R$
tending to the same point $x \in L$. Lemma~\ref{lem:block-near-path}
joins them by $K^*$-paths with vertices tending uniformly to $x$.
Project to $R$: if the endpoints of the projected path differ, the
projected path uses only vertices of the original path, so it still
tends uniformly to $x$. If the endpoints of the projection agree, use
the constant path (at that vertex). So the projected paths give the
required $K$-paths. In a metric space, this sequential formulation
gives the neighborhood formulation.
Property (6): $H_p$ is infinite by Lemma~\ref{lem:block-graph}.
Torsion-freeness of $\Gamma$ makes it contain an infinite-order
element. By Lemma~\ref{lem:pret-parabolic-cuts}, that element is
parabolic with fixed point $p$. Its powers act as asserted by
Lemma~\ref{lem:dyn-parabolic}.
\end{proof}
%=============================================================
\section{Annuli, component labels, and depth}
\label{sec:annuli}
%=============================================================
The goal of the next two sections is to prove that the continuum $L$
from Proposition~\ref{prop:block-setting} is locally connected. We do
this by pushing graph paths toward the boundary. The two new tools are:
\emph{component labels} on $L$ (which record which graph component a
given boundary point's near-vertices belong to), and \emph{annular
depth} (a numerical depth that measures how far an annulus extends from
a fixed set).
Standing assumptions from Proposition~\ref{prop:block-setting}: we
have a compact metric space $Y = H \sqcup L$ (with metric $\rho$), a
group $H$ acting on $Y$ by homeomorphisms extending left multiplication,
$H$-points isolated, and $L$ a nondegenerate continuum without cut
points satisfying $\dim L \leq 1$ and $\check H^1(L;k) = 0$, where
$k = \mathbb F_2$. The action satisfies the convergence action
property (2). We also have a connected $H$-graph $K$ with vertex set
$I \subseteq Y$, where $I = H$ or $I \subseteq L$, with finitely many
$H$-orbits of vertices and edges, satisfying the near-path property
(5), the comparison (4), and, in the $I \subseteq L$ case, the
stabilizer property (6).
No finite generation of $H$ and no local finiteness of $K$ is assumed.
\subsection*{Extending nearby labels across a deleted point}
Suppose a locally constant map takes a small punctured neighborhood
$V \setminus \{x\}$ of a boundary point $x$ to some discrete label
set. When does it extend across $x$? If $\check H^1(L;k) = 0$ and $x$
is not a cut point, the answer is: always (for $k$-valued labels). The
circle shows the difference absence of cut points alone makes.
\begin{lemma}[Extension across a non-cut point]\label{lem:ann-extension}
Let $L$ be a nondegenerate compact metric continuum without cut points
and with $\check H^1(L;k) = 0$. If $x \in L$ and $V$ is an open
neighborhood of $x$ in $L$, every locally constant map
$u : V \setminus \{x\} \to k$ extends uniquely to a locally constant
map on $V$.
\end{lemma}
\begin{proof}
Put $U = L \setminus \{x\}$. We prove via finite-cover cohomology that
there are locally constant $f_U : U \to k$ and $f_V : V \to k$ with
$u = f_U + f_V$ on $U \cap V$.
Choose open sets $U_0, V_0$ covering $L$ with $\overline{U_0} \subseteq U$
and $\overline{V_0} \subseteq V$. The set $C = \overline{U_0} \cap
\overline{V_0}$ is compact in $U \cap V$. Its two subsets
$C \cap u^{-1}(0)$ and $C \cap u^{-1}(1)$ are disjoint compact. Choose
$\delta > 0$ smaller than their distance (any $\delta$ if either is
empty).
Take a finite open cover $(W_i)$ of $L$ refining $(U_0, V_0)$, with
each $W_i$ of diameter $< \delta$, 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 $c_{ij} = u$
value otherwise. The latter is a constant: the intersection lies in
$C$ and cannot meet both of $C$'s two compact pieces. At a triple
intersection either all parents agree (so all three values are $0$) or
exactly two are from one parent and one from the other (so the two
transition values on the two sides equal the same $u$-value). Thus
$(c_{ij})$ is a $1$-cocycle on the nerve.
Since $\check H^1(L;k) = 0$, some finite open refinement $(Z_\alpha)$
with refinement map $r$ admits constants $a_\alpha \in k$ satisfying
$a_\alpha + a_\beta = c_{r(\alpha) r(\beta)}$ whenever
$Z_\alpha \cap Z_\beta \ne \varnothing$.
Define $f_U$ on $U \cap Z_\alpha$: if $P_{r(\alpha)} = U$, set
$f_U = a_\alpha$; otherwise set $f_U = a_\alpha + u$. The displayed
identity makes these definitions agree on overlaps; they are locally
constant. Similarly for $f_V$ (with roles of $U, V$ reversed). Then
$f_U + f_V = u$ on $U \cap V$.
Since $x$ is not a cut point, $U$ is connected, so $f_U$ takes one
constant value, say $c$. Then $f_V + c$ is the required extension.
Uniqueness: a nondegenerate continuum has no isolated points, so
$V \setminus \{x\}$ is dense in $V$; two continuous functions agreeing
on a dense set agree everywhere.
\end{proof}
\subsection*{Component labels}
For a subset $D \subset Y$, let $D'$ be the set of limits of sequences
of distinct points of $D$. In a metric space, $D'$ is closed. For an
allowed vertex set $T \subseteq I$, put
\[
\Omega_T = L \setminus (I \setminus T)'.
\]
This is exactly the open set of boundary points near which \emph{deleted
vertices do not accumulate as distinct points}. We use components of
the induced graph $K[T]$.
\begin{lemma}[Component labels]\label{lem:ann-component}
Within the setting above, each $x \in \Omega_T$ has a unique component
$Q_T(x)$ of $K[T]$ such that all allowed vertices in some neighborhood
of $x$ belong to $Q_T(x)$. Every $x \in \Omega_T$ is a limit of distinct
allowed vertices, and the assignment $Q_T$ is locally constant on
$\Omega_T$. If $t \in T \cap L \cap \Omega_T$ (i.e.\ $t$ is both an
allowed vertex and a limit of distinct allowed vertices in the whole
space), then $Q_T(t)$ is the component containing $t$ itself.
\end{lemma}
\begin{proof}
There is a neighborhood $W$ of $x$ such that
$(I \setminus T) \cap W \subseteq \{x\}$. Since $x$ is a limit of
distinct vertices of $I$ (Proposition~\ref{prop:block-setting}(3)), it
is consequently a limit of distinct vertices of $T$ as well.
\emph{Case 1: $x$ is not a deleted vertex.} After shrinking $W$, it
contains no deleted vertex. The near-path property (5) then puts all
vertices sufficiently near $x$ into 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 the component assignment is locally constant near $x$. If $x$
itself is an allowed vertex, it belongs to this component.
\emph{Case 2: $x$ is a deleted vertex (so $I \subseteq L$ and $x \in
I \setminus T$).} Choose an open neighborhood $V$ of $x$ in $L$
containing no other deleted vertex. By Case 1, every $y \in V \setminus
\{x\}$ has a component label $q(y)$, and these form a locally constant
map
\[
q : V \setminus \{x\} \to \{\text{components of } K[T]\}.
\]
Compose any map from the label set $\to k$ with $q$ to get a map
$V \setminus \{x\} \to k$. Each such map extends across $x$ by
Lemma~\ref{lem:ann-extension}. We claim this forces $q$ itself to be
constant sufficiently near $x$.
Call a label \emph{recurrent} if it occurs arbitrarily near $x$.
\begin{itemize}
\item If two labels are recurrent, color one $0$ and the other $1$;
that coloring separates them, contradicting extension (a locally
constant map on $V \setminus \{x\}$ extending across $x$ cannot take
both values on recurrent labels).
\item If exactly one label is recurrent but $q$ is not eventually
constant, color the recurrent label $0$ and all others $1$, again
contradicting extension.
\item If no label is recurrent, choose a sequence of points tending to
$x$ with pairwise distinct labels (at each step, exclude the finitely
many labels already chosen). Color the successive labels alternately
$0$ and $1$. This again contradicts extension as a locally constant
map.
\end{itemize}
Thus one component label occurs throughout a small punctured
neighborhood of $x$. Every allowed vertex there belongs to its own
label by Case 1, 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.
\end{proof}
\subsection*{The annulus system}
An \emph{annulus} in $Y$ is an ordered pair $\sigma = (\sigma^-,
\sigma^+)$ of disjoint closed subsets of $Y$ such that each side meets
$L$ and $\sigma^- \cup \sigma^+ \ne Y$. The sides need not be connected.
For annuli $\sigma, \tau$, write
\[
\sigma < \tau \iff \sigma^+ \cup \tau^- = Y.
\]
Then $\sigma^- \subseteq \operatorname{int}(\tau^-)$ and
$\tau^+ \subseteq \operatorname{int}(\sigma^+)$. The relation is
transitive and irreflexive: a nested chain $\sigma_1 < \cdots <
\sigma_n$ uses distinct annuli. A chain \emph{goes from $F$ to $E$} if
$F \subseteq \sigma_1^-$ and $E \subseteq \sigma_n^+$.
\begin{lemma}[Annulus system]\label{lem:ann-system}
There is a set $\mathcal A$ of annuli invariant under $H$ and under
interchange of the two sides, with:
\begin{enumerate}
\item For every $\delta > 0$, only finitely many annuli in $\mathcal A$
have both sides of diameter at least $\delta$.
\item For every closed $F \subseteq Y$, every $x \in L \setminus F$, and
every integer $n \geq 1$, there is a chain of $n$ annuli in $\mathcal A$
from $F$ to some neighborhood of $x$ in $Y$.
\end{enumerate}
\end{lemma}
\begin{proof}
\emph{Uniform convergence is uniform.} For every $r, \varepsilon > 0$,
all but finitely many $g \in H$ admit a point $b_g \in L$ with
\[
\operatorname{diam} g\bigl(Y \setminus B_\rho(b_g, r)\bigr) < \varepsilon.
\]
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 $Y \setminus B_\rho(b, r)$ (property (2)) gives a
contradiction.
Choose $r_i \downarrow 0$ with $r_i < \operatorname{diam}(L)/3$ and
$\varepsilon_i = 2^{-i}$. Apply the observation with $r = r_i/4$ and
$\varepsilon = \varepsilon_i$, gaining a finite exceptional set
$E_i \subset H$. Uniform continuity of the finitely many maps in $E_i$
allows $0 < s_i < r_i/4$ with
\[
\operatorname{diam} g\bigl(\overline{B_\rho(y, s_i)}\bigr) < \varepsilon_i
\quad (g \in E_i, y \in L).
\]
Choose finitely many centers $y_{ij} \in L$ with $s_i/2$-balls covering
$L$. At each center take the seed annulus
\[
\sigma_{ij}^- = Y \setminus B_\rho(y_{ij}, r_i),
\qquad
\sigma_{ij}^+ = \overline{B_\rho(y_{ij}, s_i)}.
\]
Its sides are disjoint with distance $\ge r_i - s_i$. The positive
side contains its center. The negative side meets $L$: every point of
$L$ is at distance $\ge \operatorname{diam}(L)/2$ from some point of
$L$ (extreme points of a diameter). The sides cannot cover $L$: their
nonempty closed intersections with $L$ would disconnect it.
Let $\mathcal A$ consist of all distinct translates of these annuli and
their reversals.
\emph{Diameter bound.} 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 lies outside the exceptional ball for $g$
and has image of diameter $< \varepsilon_i$. For $g \in E_i$, the
positive side has that property by choice of $s_i$. Hence every
translated level-$i$ annulus has at least one side of diameter
$< \varepsilon_i$.
Now fix a seed and $\delta > 0$. Only finitely many distinct translates
have both sides of diameter $\ge \delta$: otherwise choose such
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. Hence property (1).
\emph{Chain property.} 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 its 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
$B_\rho(y, r_j) \subseteq B_\rho(x, \eta)$, so the new annulus is
\emph{larger} in the order $<$, and its positive side again contains a
neighborhood of $x$. Repeating gives any prescribed finite length.
\end{proof}
\subsection*{Depth}
Fix such an annulus system $\mathcal A$. For $F \subseteq Y$ and
$v \in Y$, define
\[
D(F, v) = \sup\{n \geq 1 :
\text{a chain of length } n \text{ in } \mathcal A
\text{ goes from } F \text{ to } \{v\}\},
\]
with $D = 0$ if no such chain exists. In general $D \in \{0, 1, 2,
\ldots\} \cup \{\infty\}$. Write $D(u, v)$ for $D(\{u\}, v)$. This
function counts separating annuli; it is not a metric.
\begin{lemma}[Properties of depth]\label{lem:ann-depth}
For $g \in H$, $F \subseteq Y$, and $v \in Y$,
\[
D(gF, gv) = D(F, v).
\]
If $F$ contains a point of $H$ and $v \in H$, then $D(F, v)$ is finite.
When it is positive, a chain attains this maximum. If $F$ is closed,
$x \in L \setminus F$, and $N \geq 1$, then there is a neighborhood
$V_N$ of $x$ in $Y$ with $D(F, v) \geq N$ for every $v \in V_N$.
\end{lemma}
\begin{proof}
\emph{Equivariance.} Translation carries annuli and chains bijectively
to annuli and chains.
\emph{Finiteness.} Suppose $h \in F \cap H$ and $v \in H$. Since $L$ is
closed and does not contain $h$ or $v$ (points of $H$ are not in $L$),
the number $\delta = \min\{\operatorname{dist}(h, L),
\operatorname{dist}(v, L)\}$ is positive. Throughout a chain from $F$
to $v$: every negative side contains $h$ and meets $L$, every positive
side contains $v$ and meets $L$. So both sides have diameter $\geq
\delta$. Lemma~\ref{lem:ann-system}(1) supplies only finitely many such
annuli, and a strict chain cannot repeat an annulus. So $D(F, v)$ is
finite.
\emph{Attainment.} A finite nonzero supremum of possible integer
lengths is attained (the set of achievable lengths is a downward-closed
subset of positive integers, so its supremum is one of its members).
\emph{Deep neighborhoods.} The final claim is exactly
Lemma~\ref{lem:ann-system}(2) applied to length $N$, using the
resulting terminal neighborhood of $x$.
\end{proof}
%=============================================================
\section{Pushing paths and local connectedness}
\label{sec:path-pushing}
%=============================================================
We work in the setting of Proposition~\ref{prop:block-setting}: a
compact metric space $Y = H \sqcup L$, an $H$-graph $K$ on $I
\subseteq Y$, and the annulus system $\mathcal A$ from
Lemma~\ref{lem:ann-system}. Our goal: prove $L$ is locally connected.
The idea: fix a closed set $A$ containing $1$ and avoiding $L$, and
introduce a depth-based subgraph $K[I_m]$ that captures "vertices
deeper than $m$ away from $A$". Component labels for $K[I_m]$ give a
locally constant map $Q_m$ on $L \setminus A$. The key uniformity is:
beyond one threshold $M$, increasing $m$ to $m+1$ never separates two
boundary points that shared a label at level $m$. Then paths in
$K[I_m]$ joining nearby vertices persist to arbitrary depth, and their
Hausdorff limits give connected subsets of $L$ of small diameter.
\subsection*{Uniform depth comparison}
Fix a closed subset $A \subset Y$ with $1 \in A$ and $L \setminus A
\ne \varnothing$. Write
\[
d_0 = \operatorname{dist}(1, L) > 0,
\qquad
N(a, b) = \#\{\sigma \in \mathcal A :
\operatorname{diam}\sigma^- \ge a,\
\operatorname{diam}\sigma^+ \ge b\},
\]
which is finite by Lemma~\ref{lem:ann-system}(1). Any side containing
$1$ has diameter $\geq d_0$ (every side meets $L$).
\begin{lemma}[Shrinking toward the boundary]\label{lem:push-shrinking}
Let $g_n \in H$ satisfy $D(A, g_n) \to \infty$ and set $F_n =
g_n^{-1} A$. Along a subsequence there is $q \in L$ such that $F_n$
converges in the Hausdorff metric to $\{q\}$.
\end{lemma}
\begin{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
$N(\varepsilon, d_0)$ members, one of its negative sides has diameter
$< \varepsilon$; that side contains $F_n$, so $\operatorname{diam}
F_n < \varepsilon$. Since $\varepsilon$ was arbitrary,
$\operatorname{diam} F_n \to 0$.
Lemma~\ref{lem:ann-depth} gives $D(A, g) < \infty$ for every fixed
$g \in H$, so $g_n$ escapes every finite set (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 $g_n^{-1} \to
q \in L$. Diameter conclusion closes it.
\end{proof}
\begin{lemma}[Uniform depth comparison]\label{lem:push-depth-comparison}
Suppose $F_n = g_n^{-1} A$ converges to $\{q\}$ as in
Lemma~\ref{lem:push-shrinking}. For each $r > 0$ set
\[
C_r = N(r/2, d_0) + N(d_0, r) + 2.
\]
For all sufficiently large $n$, uniformly over every $b \in H$ with
$\rho(b, q) \ge 3r$,
\begin{equation}\label{eq:push-depth-comparison}
D(F_n, b) \ge D(F_n, 1) + D(\{1\}, b) - C_r.
\end{equation}
Consequently, if $b_n \in H \to z \in L \setminus \{q\}$, then
$D(F_n, b_n) \ge D(F_n, 1) + D(\{1\}, b_n) - C$ for a constant $C$,
and $D(\{1\}, b_n) \to \infty$.
\end{lemma}
\begin{proof}
\emph{Prune the initial chain.} Take $n$ large enough that
$F_n \subseteq B(q, r/2)$ 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 $\ge r/2$. Its positive
side has diameter $\ge d_0$. There are at most $N_- = N(r/2, d_0)$
such annuli. Because negative sides increase along a chain, those
form a terminal segment. Remove that segment and one further annulus.
The retained initial chain has $\geq \ell - N_- - 1$ members. The
negative side of the next annulus lies in $U$, so the last retained
positive side contains $Y \setminus U$. In particular, the retained
chain already ends on the side containing any $b$ under consideration.
\emph{Prune the tail chain.} 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 $\le 2r$ from $q$, hence diameter $\ge r$. Every negative
side contains $1$, hence diameter $\ge d_0$. So at most $N_+ =
N(d_0, r)$ positive sides fail to lie in $V$; those form an initial
segment. If $t > N_+ + 1$, remove this initial segment and one further
annulus. The retained final chain has $\geq t - N_+ - 1$ members, and
its first negative side contains $Y \setminus V$.
\emph{Concatenate.} 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 \eqref{eq:push-depth-comparison} with a
constant and threshold independent of $b$.
\emph{Final claim.} Choose $r$ with $4r < \rho(q, z)$. Eventually
$\rho(b_n, q) \ge 3r$, so the inequality applies. For any prescribed
integer $N$, Lemma~\ref{lem:ann-system}(2) gives a chain of length $N$
from $\{1\}$ to a neighborhood of $z$. Eventually this neighborhood
contains $b_n$, so $D(\{1\}, b_n) \to \infty$.
\end{proof}
\subsection*{Components of deeper vertex sets}
Choose one representative $p_i$ for each $H$-orbit in $I$; when $I=H$,
use only $p_1 = 1$. For $u$ in the orbit of $p_i$, define its set of
representatives $C(u) = \{g \in H : g p_i = u\}$. For integers
$m \geq 0$ define the \emph{deep vertex set}
\begin{equation}\label{eq:push-deep-vertices}
I_m = \left\{u \in I \setminus A :
\min_{g \in C(u)} D(A, g) \geq m\right\}.
\end{equation}
Each depth in the minimum is a finite nonnegative integer (by
Lemma~\ref{lem:ann-depth}: $A$ contains $1 \in H$, so there is a
finiteness associated). The condition asks about \emph{every}
representative; this will let us slide through an infinite stabilizer
without losing the depth bound.
\begin{lemma}[Component labels for deep vertices]
\label{lem:push-component-labels}
For each fixed $m$, deleted vertices $I \setminus I_m$ have no
accumulation by distinct vertices at any point of $L \setminus A$.
Consequently Lemma~\ref{lem:ann-component} defines a locally constant
map
\[
Q_m : L \setminus A \to \{\text{components of } K[I_m]\},
\]
whose value at $x$ contains all vertices of $I_m$ sufficiently near $x$.
Every $x \in L \setminus A$ is approached by distinct vertices with
this label. For every finite $E \subset I$, one has $E \cap I_m =
\varnothing$ for all sufficiently large $m$.
\end{lemma}
\begin{proof}
\emph{No accumulation.} For $m=0$, the assertion follows from
$I_0 = I \setminus A$. Suppose $m \geq 1$ and distinct $u_n \in
I \setminus I_m \to x \in L \setminus A$. 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 $u_n$ are distinct, $a_n$
escapes every finite set. The comparison $\rho(gp_i, g) \to 0$
(Proposition~\ref{prop:block-setting}(4)) applied to the finite list
of representatives gives $a_n \to x$. A chain from $A$ to a
neighborhood of $x$ of length $m$ (Lemma~\ref{lem:ann-system}(2))
contradicts $D(A, a_n) < m$ for large $n$.
\emph{Structure of $Q_m$.} The claim about $Q_m$ and approximation
follows from Lemma~\ref{lem:ann-component}.
\emph{Eventually excluded.} For $u \in E \setminus A$, choose one
representative $a_u \in C(u)$. The depth $D(A, a_u)$ is finite. Taking
$m$ larger than all these finitely many depths excludes every $u \in E$
from $I_m$.
\end{proof}
\begin{lemma}[Translated deletions stay bounded]\label{lem:push-translated-deletions}
Let $m_n \to \infty$ and $D(A, g_n) \geq m_n$. Pass to a subsequence
with $F_n = g_n^{-1} A \to \{q\}$, $q \in L$.
For every compact $E \subset Y \setminus \{q\}$ there is a finite
$S_E \subset I$ and an index $n_E$ with
\[
E \cap g_n^{-1}(I \setminus I_{m_n + 1}) \subseteq S_E
\qquad (n \geq n_E).
\]
\end{lemma}
\begin{proof}
Supposing not, choose indices $\to \infty$ and pairwise distinct
$w_n \in E \cap g_n^{-1}(I \setminus I_{m_n + 1})$. Compactness gives
$w_n \to z \in E$. The limit lies in $L$: automatic if $I \subseteq L$;
if $I = H$, follow from isolatedness of group vertices. Since
$F_n \to q \notin E$, eventually $w_n \notin F_n$. The deletion of
$g_n w_n$ gives a representative $g_n b_n \in C(g_n w_n)$ with
$D(F_n, b_n) = D(A, g_n b_n) < m_n + 1$. After restricting to one
orbit representative, $w_n = b_n p_i$. Distinctness forces $b_n$ to
escape finite sets, and comparison gives $b_n \to z \ne q$.
Lemma~\ref{lem:push-depth-comparison} then yields
\[
D(F_n, b_n) \geq m_n + D(\{1\}, b_n) - C > m_n + 1
\]
for large $n$, contradiction.
\end{proof}
\begin{lemma}[Eventual labels]\label{lem:push-eventual-labels}
For sequences as in Lemma~\ref{lem:push-translated-deletions}, define
$\ell_n(z) = Q_{m_n + 1}(g_n z)$ whenever $z \in L$ and $g_n z \notin
A$. For each $z \in L \setminus \{q\}$ there are a relative neighborhood
$V_z \subset L \setminus \{q\}$ and an index $n_z$ such that all values
$\ell_n(w)$ with $w \in V_z$, $n \ge n_z$ are defined and
$\ell_n(w) = \ell_n(z)$. For any two points $z, z' \in L \setminus
\{q\}$, $\ell_n(z) = \ell_n(z')$ for all sufficiently large $n$.
\end{lemma}
\begin{proof}
Choose a neighborhood $O$ of $z$ in $Y$ whose closure misses $q$.
Apply Lemma~\ref{lem:push-translated-deletions} to the compact
$E = \overline O$. There are finite $S \subset I$ and index $n_0$ with
$T = (I \cap O) \setminus S$ satisfying $g_n T \subseteq I_{m_n+1}$
for all $n \geq n_0$. The deleted vertices $I \setminus T$ have no
distinct accumulation at any point of $O \cap L$.
Lemma~\ref{lem:ann-component} gives a neighborhood $V_z \subset O \cap L$
on which the component of $K[T]$ approached by vertices is constant;
call it $C$. Shrink $V_z$ if needed. Since $F_n \to \{q\}$, all
$g_n w$ 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 $g_n w$ and lie in the connected subgraph $g_n C$
of $K[I_{m_n+1}]$. The defining property of $Q_{m_n+1}$ identifies
$\ell_n(w)$ with the component containing $g_n C$, for every $w \in
V_z$. That gives the asserted neighborhood and threshold.
\emph{Equivalence and single class.} 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
may change with $n$. The first part shows each equivalence class is
open. Since $q$ is not a cut point, $L \setminus \{q\}$ is connected,
so there is one class.
\end{proof}
\subsection*{A common threshold}
We use three fixed boundary points to get a single threshold $M$ that
works for every graph path simultaneously.
Choose a set $Z \subset L$ of three distinct points (exists because $L$
is a nondegenerate continuum).
\begin{lemma}[Majority labels]\label{lem:push-majority}
For every finite $S \subset H$, there is $M = M(A, Z, S) \ge 1$ such
that for all integers $m \ge M$:
\begin{enumerate}[label=(\roman*)]
\item If $D(A, g) \ge m$, at least two points of $gZ$ lie in $L
\setminus A$ and have a common $Q_{m+1}$ label. Call it the
\emph{majority label} $J_m(g)$.
\item If $s \in S$ and $D(A, g), D(A, gs) \ge m$, then $J_m(g) =
J_m(gs)$.
\end{enumerate}
\end{lemma}
\begin{proof}
\emph{Part (i).} 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~\ref{lem:push-eventual-labels},
their images under $g_n$ eventually lie outside $A$ and share a single
$Q_{m_n+1}$ label, contradiction. Uniqueness of the majority label
also follows: two different labels cannot each occur at two points of
a three-point set.
\emph{Part (ii).} After choosing a threshold for (i), suppose (ii)
fails at arbitrarily large $m$. Choose violating sequences and, since
$S$ is finite, restrict to one fixed $s$. 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~\ref{lem:push-eventual-labels}
makes the labels of all four images under $g_n$ equal for large $n$.
They determine the majority labels of both $g_n Z$ and $g_n s Z$,
contradicting the chosen inequality.
\end{proof}
\emph{Fixing the finite set $S$.} There are finitely many orbits of
ordered edges of $K$: allowing both orientations, at most twice the
number of edge orbits. For each representative $(u_e, v_e)$ choose
$a_e, b_e \in H$ and indices $i(e), j(e)$ with $u_e = a_e p_{i(e)}$
and $v_e = b_e p_{j(e)}$, and include $s_e = a_e^{-1} b_e$ in $S$.
Then the endpoints of every edge have representatives $g$ and $g s$ for
some $s \in S$.
When $I \subseteq L$, also include, for each $p_i$, a stabilizing
element $h_i \in H$ from Proposition~\ref{prop:block-setting}(6) with
$h_i p_i = p_i$ and $h_i^j z \to p_i$ for $z \in L$ as
$j \to +\infty$.
Fix the threshold $M$ from Lemma~\ref{lem:push-majority} for this list.
\begin{lemma}[Refinement does not split labels]\label{lem:push-refinement}
For every $m \geq M$ and all $x, y \in L \setminus A$,
\begin{equation}\label{eq:push-refinement}
Q_m(x) = Q_m(y) \implies Q_{m+1}(x) = Q_{m+1}(y).
\end{equation}
In particular, $M$ is independent of $x$ and $y$.
\end{lemma}
\begin{proof}
\emph{Case 1: $I \subseteq L$.} Claim: for $v \in I_m$ and
$g \in C(v)$,
\begin{equation}\label{eq:push-support-label}
J_m(g) = Q_{m+1}(v).
\end{equation}
Let $v = g p_i$. For every $j \ge 0$, $g h_i^j \in C(v)$. The minimum
in \eqref{eq:push-deep-vertices} gives $D(A, g h_i^j) \geq m$ for every
$j$. The move $h_i \in S$ and Lemma~\ref{lem:push-majority} give
\[
J_m(g) = J_m(g h_i) = J_m(g h_i^2) = \cdots.
\]
As $j \to \infty$, the three points $g h_i^j Z$ 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
\eqref{eq:push-support-label}.
Suppose $v, w \in I_m$ are adjacent. Choose $g, g s$ representatives
via the finite edge list. Both depths $\ge m$ by the minimum
convention. Lemma~\ref{lem:push-majority} and
\eqref{eq:push-support-label} give $Q_{m+1}(v) = Q_{m+1}(w)$. So
$Q_{m+1}$ is constant on every path in $K[I_m]$. If $Q_m(x) = Q_m(y)$,
choose vertices in a common component sufficiently near $x$ and $y$
(using Lemma~\ref{lem:push-component-labels} and local constancy of
$Q_{m+1}$); a path in that component proves
\eqref{eq:push-refinement}.
\emph{Case 2: $I = H$.} Only one representative $p_1 = 1$, and
$C(v) = \{v\}$. The finite edge moves and Lemma~\ref{lem:push-majority}
imply $J_m$ is constant along paths of $K[I_m]$. For $x \in L
\setminus A$ and $m \geq M$ fixed, claim that $J_m(g) = Q_{m+1}(x)$
for all group vertices $g$ sufficiently near $x$. These vertices lie
in $I_m$ when sufficiently near $x$ by
Lemma~\ref{lem:push-component-labels}. If the claim fails, take
$g_n \to x$ with the wrong label and pass to a subsequence with
$g_n^{-1} \to b \in L$. The convergence property (2) sends at least
two points of $Z \setminus \{b\}$ to $x$ under $g_n$. Local constancy
of $Q_{m+1}$ near $x$ gives the asserted majority label, contradiction.
Approximating $x, y$ inside their common $Q_m$ component and using
constancy of $J_m$ along a connecting path proves
\eqref{eq:push-refinement}.
\end{proof}
\subsection*{Limits of pushed paths}
\begin{proposition}[Local connectivity of $L$]
\label{prop:push-local-connected}
In the setting of Proposition~\ref{prop:block-setting}, the continuum
$L$ is locally connected.
\end{proposition}
\begin{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 of Lemma~\ref{lem:push-refinement}. By local
constancy of $Q_M$, there is a neighborhood $V \subset L \setminus A$
of $x$ with $Q_M(y) = Q_M(x)$ for all $y \in V$. Induction via
Lemma~\ref{lem:push-refinement} gives
\[
Q_m(y) = Q_m(x) \qquad (y \in V,\ m \ge M).
\]
Fix $y \in V$. For each $m \ge M$, choose $a_m, b_m \in I_m$ in the
common label component with $\rho(a_m, x) < 1/m$ and $\rho(b_m, y) <
1/m$, and choose $P_m$ a finite path in $K[I_m]$ joining them. Every
vertex of $P_m$ lies in $O$.
\emph{Mesh tends to zero.} Use the finite list of edge representatives
$(u_e, v_e)$ and the comparison ($\rho(gu_e, gv_e) \to 0$ uniformly
over the list as $g$ escapes). For each $\varepsilon > 0$, choose a
finite $F \subset H$ outside which every edge distance is
$< \varepsilon$. Only finitely many endpoints of edges
$(g u_e, g v_e)$ with $g \in F$ exist. By
Lemma~\ref{lem:push-component-labels}, none of those belongs to $I_m$
once $m$ is large. So every edge in $K[I_m]$ has endpoint distance
$< \varepsilon$. This proves the mesh assertion with no local-finiteness
assumption on $K$.
\emph{Hausdorff limit is connected.} Regard $P_m$ as its finite vertex
set. Nonempty closed subsets of the compact metric space $Y$ are
compact in the Hausdorff metric; pass to a subsequence $P_m \to C$.
The limit contains $x, y$ and lies in $\overline O$. It also lies in
$L$: immediate if $I \subseteq L$; if $I = H$, Lemma~\ref{lem:push-component-labels}
excludes any fixed group vertex from $P_m$ for large $m$, and
isolatedness of that vertex excludes it from the limit.
If $C$ were disconnected, write it as a disjoint union of nonempty
compact $C_0, C_1$ with distance $\delta > 0$. For large $m$, $P_m$
lies in the disjoint union of $\delta/3$-neighborhoods of $C_0, C_1$
and meets both. A path in these vertices then has two consecutive
vertices in different neighborhoods, at distance $\geq \delta/3$.
This contradicts the mesh assertion. So $C$ is connected.
\emph{Conclusion.} $C$ is a connected subset of $\overline O \cap L
\subseteq U$ containing $x$ and $y$. So every $y \in V$ lies in the
same component of $U$ as $x$. Repeating at every point and every open
subset of $L$ shows that components of open sets in $L$ are open. This
is local connectivity.
\end{proof}
%=============================================================
\section{The continuum contradiction}
\label{sec:conclusion}
%=============================================================
We now have: a nondegenerate compact metric continuum $L$ that is
locally connected, has $\dim L \leq 1$, and has
$\check H^1(L;\mathbb F_2) = 0$. This section extracts a contradiction.
The idea: locally connected + dim $\leq 1$ + no circle forces $L$ to be
a dendrite; every nondegenerate dendrite has a cut point; but $L$ has
none. Contradiction.
\subsection*{Two lemmas we need}
\begin{lemma}[Cohomology of closed subsets]\label{lem:end-restriction}
If $Z$ is a compact metric space with $\dim Z \leq 1$, $A \subset Z$
closed, and $k$ is an abelian coefficient group, then restriction
induces a surjection
\[
\check H^1(Z; k) \to \check H^1(A; k).
\]
\end{lemma}
\begin{proof}
Represent a class on $A$ by a cocycle $d$ 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 $\{W_i\} \cup \{Z \setminus A\}$ of
$Z$ has a finite open refinement $\mathcal U$ of multiplicity at most
two (that's $\dim Z \leq 1$). Its nerve $N(\mathcal U)$ is therefore a
graph.
Restrict $\mathcal U$ to $A$: keep the indices of the cover members,
even if two restrictions coincide. After empty restrictions are
discarded, the nerve $N(\mathcal U|_A)$ is a subgraph of
$N(\mathcal U)$. A refinement map pulls the original cocycle $d$ back
to a $1$-cochain on this subgraph.
Extend that $1$-cochain by zero on every remaining edge of
$N(\mathcal U)$. Since $N(\mathcal U)$ is a graph (a $1$-dimensional
complex, no $2$-simplices), the extension is automatically a
$1$-cocycle. Its Čech class in $\check H^1(Z;k)$ restricts to the
prescribed class on $A$.
\end{proof}
\begin{lemma}[Path-connected open subsets]\label{lem:end-paths-in-regions}
Let $L$ be a locally connected compact metric space with metric $d$,
and let $U \subset L$ be connected and open. Then any two points
$a, b \in U$ are joined by a continuous path in $U$.
\end{lemma}
\begin{proof}
\emph{Cover by small connected pieces.} Every connected open $V
\subset L$ is covered by connected open sets $W$ whose closures lie
in $V$ and whose diameter is at most any prescribed $\varepsilon > 0$:
around any point $x \in V$, take a small metric ball with closure in
$V$, then a connected open neighborhood inside a still smaller ball
(local connectivity supplies it).
\emph{Connected intersection graph.} Take the cover from the previous
paragraph. Its intersection graph is connected: otherwise, the union
of members in a graph component would be a nonempty open and relatively
closed subset of $V$, separating $V$.
\emph{Chain of overlapping sets.} Any two points $a, b \in U$ can be
joined by a finite chain $W_1, \ldots, W_m$ from this cover, with
$a \in W_1$, $b \in W_m$, and $W_i \cap W_{i+1} \ne \varnothing$. We
may arrange $m \ge 2$ by repeating a member.
\emph{Iterative subdivision.} Starting with $[0,1]$, $U$, and endpoints
$a, b$, construct finite interval partitions $\mathcal P_n$ as follows.
For each $I \in \mathcal P_n$, associate a connected open set $V_I$
containing its already-assigned endpoint values, with
$\operatorname{diam} V_I \leq 2^{-n}$ (for $n \geq 1$). Inside $V_I$,
take the finite chain above with closures in $V_I$ and diameters
$\leq 2^{-(n+1)}$. Subdivide $I$ into $m$ equal pieces 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. Successive assignments agree at shared endpoints, and both
endpoints of every child interval lie in its associated set.
\emph{Construction.} Every subdivision has at least two children, so
$\operatorname{mesh}(\mathcal P_n) \leq 2^{-n}$. The set $D$ of all
partition endpoints is therefore dense in $[0,1]$, and the compatible
assignments define $f:D \to L$. Nesting ensures $f(D \cap I) \subset
V_I$ for every $I \in \mathcal P_n$.
\emph{Uniform continuity.} For fixed $n \ge 1$, let $\delta_n$ 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. Both
cases give $d(f(s), f(t)) \le 2^{1-n}$. So $f$ is uniformly continuous
and extends continuously to $[0,1]$ by completeness of $L$.
\emph{Path stays in $U$.} For each cell $I \in \mathcal P_1$, the
extended image lies in $\overline{V_I} \subset U$. The full path
therefore lies in $U$, and its endpoints are $a, b$.
\end{proof}
\subsection*{The final contradiction}
\begin{lemma}[Every such continuum has a cut point]
\label{lem:end-continuum}
Every nondegenerate locally connected compact metric continuum $L$
with
\[
\dim L \leq 1, \qquad \check H^1(L;\mathbb F_2) = 0
\]
has a cut point.
\end{lemma}
\begin{proof}
\emph{$L$ contains no embedded circle.} Suppose $C \subset L$ were an
embedded circle $S^1$. It is closed in $L$. Lemma~\ref{lem:end-restriction}
gives a surjection
$\check H^1(L;\mathbb F_2) \to \check H^1(C;\mathbb F_2)$. But the left
side is zero while the right side is $\mathbb F_2$ (a circle has first
Čech cohomology $\mathbb F_2$). Contradiction.
\emph{Choose an arc.} A locally connected compact metric continuum is a
\emph{Peano continuum}. The theory of Peano continua (via Moore's
theorem) says that in any connected open subset of $L$, any two points
can be joined by an arc. (A capstone: continuous images of intervals
contain embedded arcs, via
\cite[pp.~301--304]{Kamiya1930}. Combined with
Lemma~\ref{lem:end-paths-in-regions}, arcs exist wherever paths do.)
Pick an arc from $a$ to $b$ (distinct points of $L$), and an
interior point $p$ of that arc.
\emph{If $p$ were not a cut point: contradiction.} The connected open
set $L \setminus \{p\}$ would contain a second arc, say $\gamma$, from
$a$ to $b$. Let $\alpha$ be the component of the first arc minus
$\gamma$ that contains $p$; this is an open subarc of the first arc
with both endpoints on $\gamma$. The closure of $\alpha$, together with
the subarc of $\gamma$ joining the two endpoints of $\alpha$, is an
embedded circle.
But $L$ has no embedded circle. So $p$ must be a cut point.
\end{proof}
\begin{proof}[Proof of Theorem~\ref{thm:main}]
Suppose $b_1^{(2)}(\widetilde Q) > 0$. By the finite-energy construction
(Sections~\ref{sec:compactification}--\ref{sec:dynamics}) and the
boundary topology (Section~\ref{sec:boundary-topology}) we have built a
nondegenerate compact metric continuum $B$ with $\dim B \leq 1$ and
$\check H^1(B;\mathbb F_2) = 0$.
If $B$ has no cut points, set $L = B$; the reduction of
Proposition~\ref{prop:block-setting} (with $P = \varnothing$) applies
directly. Otherwise, use Section~\ref{sec:pretree} and
Section~\ref{sec:blocks} to isolate a subcontinuum $L \subset B$ without
cut points, still nondegenerate, still with $\dim L \leq 1$ and
$\check H^1(L;\mathbb F_2) = 0$. In both cases,
Proposition~\ref{prop:push-local-connected} gives local connectivity of
$L$.
Then Lemma~\ref{lem:end-continuum} forces $L$ to have a cut point, so
$L$ would be a nondegenerate locally connected compact metric continuum
with a cut point — but $L$ was constructed \emph{without} cut points
(Proposition~\ref{prop:block-retraction} in the $P \ne \varnothing$
case, by definition in the $P = \varnothing$ case). Contradiction.
Therefore $b_1^{(2)}(\widetilde Q) = 0$. Lemma~\ref{lem:found-reduction}
gives $b_0^{(2)} = b_4^{(2)} = 0$, $b_3^{(2)} = b_1^{(2)} = 0$, and
vanishing in all degrees above four. Its duality argument includes
nontrivial orientation characters. Consequently
$b_p^{(2)}(\widetilde Q;\mathcal N(\Gamma)) = 0$ for every $p \ge 0$
with $p \ne 2$.
\end{proof}
\begin{proof}[Proof of Corollary~\ref{cor:singer-4d} (Singer in dimension four)]
Let $M$ be a closed connected aspherical topological four-manifold.
Lemma~\ref{lem:manifold-reduction} supplies a finite aspherical integral
Poincaré-complex homotopy model of $M$. A homotopy equivalence lifts to
an equivariant homotopy equivalence of universal covers, so it preserves
the singular-chain $L^2$-Betti numbers. Theorem~\ref{thm:main} applies
to the model.
\end{proof}
%=============================================================
\section{Euler characteristic, signature, and finite covers}
\label{sec:consequences}
%=============================================================
The main theorem determines the remaining (degree-two) $L^2$-Betti
number from a classical topological invariant. This gives constraints
on Euler characteristics and signatures of closed aspherical
four-manifolds.
\begin{corollary}[Euler characteristic]\label{cor:euler}
Let $Q$ be a finite connected aspherical integral Poincaré complex of
formal dimension four, with arbitrary orientation character. Then
\[
\chi(Q) = b_2^{(2)}(\widetilde Q) \ge 0.
\]
Moreover $\chi(Q) = 0$ if and only if every $L^2$-Betti number of
$\widetilde Q$ vanishes. The same conclusions hold for closed connected
aspherical topological four-manifolds.
\end{corollary}
\begin{proof}
The $L^2$ Euler-Poincaré formula for a finite CW complex is
\[
\chi(Q) = \sum_{p \ge 0} (-1)^p b_p^{(2)}(\widetilde Q)
\]
\cite[Theorem~1.35(2)]{Luck2002}. By Theorem~\ref{thm:main}, only
the degree-two term survives (all other $b_p^{(2)}$, $p \ne 2$, vanish,
and those $p$ are all we need consider since $b_p^{(2)} = 0$ for
$p > 4$). Its 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~\ref{lem:manifold-reduction};
the invariants are preserved.
\end{proof}
The middle term need not vanish: for closed oriented hyperbolic
surfaces $\Sigma_g, \Sigma_h$ of genera $g, h \ge 2$, the product
$\Sigma_g \times \Sigma_h$ is aspherical and has
$\chi = 4(g-1)(h-1) > 0$. On the other hand the four-torus is a flat
aspherical four-manifold with $\chi = 0$.
For an oriented closed four-manifold $M$, let $\sigma(M)$ be the
signature of the intersection form on $H^2(M;\mathbb R)$ (equivalently,
on $H_2(M;\mathbb R)$ via Poincaré duality).
\begin{corollary}[Euler-signature inequality]\label{cor:signature}
Every closed connected oriented aspherical topological four-manifold
$M$ satisfies
\[
\chi(M) \ge |\sigma(M)|.
\]
In particular, $\chi(M) = 0$ implies $\sigma(M) = 0$.
\end{corollary}
\begin{proof}
The topological $L^2$-signature theorem (which does not require
smoothness) gives
$\sigma(M) = \sigma^{(2)}(\widetilde M)$
\cite[Theorem~0.2]{LuckSchick2003}. The right side 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~\ref{cor:euler} now proves the inequality.
\end{proof}
The corollary includes non-smoothable manifolds: the cited signature
theorem is topological. Its equality between ordinary and $L^2$
signatures for a finite Poincaré complex is an additional input; we
have not assumed it here.
\begin{corollary}[Betti growth along finite covers]\label{cor:betti-growth}
Let $Q$ be as in Corollary~\ref{cor:euler}, and suppose
$\Gamma = \pi_1(Q)$ admits a descending sequence
$\Gamma = \Gamma_0 \supseteq \Gamma_1 \supseteq \cdots$ of
finite-index normal subgroups with $\bigcap_i \Gamma_i = \{1\}$. Let
$Q_i \to Q$ be the associated finite covers. For every $p \ge 0$,
\[
\lim_{i \to \infty}
\frac{b_p(Q_i;\mathbb Q)}{[\Gamma : \Gamma_i]}
= \begin{cases}
\chi(Q), & p = 2, \\
0, & p \ne 2.
\end{cases}
\]
The same statement holds for a closed connected aspherical topological
four-manifold and its corresponding covers.
\end{corollary}
\begin{proof}
Lück's approximation theorem \cite[Theorem~0.1]{Luck1994Approximation}
identifies the limit with $b_p^{(2)}(\widetilde Q)$. Apply
Theorem~\ref{thm:main} and Corollary~\ref{cor:euler}. A finite homotopy
model of a topological manifold lifts to a homotopy equivalence on each
finite cover, giving the last assertion.
\end{proof}
%=============================================================
\section{Bibliography}
%=============================================================
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\end{document}