A Poincare duality group of dimension $n$ has a group ring that mimics
the chains of a closed aspherical $n$-manifold. An oriented integral
$PD_n$ group $G$ carries a finite resolution of the trivial module
$\mathbb Z$ by finitely generated projective $\mathbb Z G$-modules, its
group-ring cohomology vanishes off degree $n$, and its top group is
$\mathbb Z$ with trivial right action.

Every closed aspherical orientable $n$-manifold has such a fundamental
group. Wall asked whether the algebraic condition is complete.
./pd4-citations/davis-poincare-duality-groups-survey.pdf

Without finite presentation the answer is no. Bestvina and Brady
produced finiteness counterexamples, and Davis reflected them to give
integral $PD_n$ groups that are not finitely presented for every
$n \ge 4$. With rational coefficients and without a finite classifying
space, Fowler constructed torsion-free finitely presented rational
Poincare duality groups that are not realized by closed aspherical
ANR rational homology manifolds.
./pd4-citations/fowler-2012-finiteness-rational-pd.pdf

The finitely presented integral case is Davis's Question 3.4, matching
Conjecture 7.29 of Luck's survey. Kielak records in his address at the
2026 ICM that no counterexamples were known before the construction
below.
./pd4-citations/luck-2025-farrell-jones-survey.pdf

The main theorem is the following.

Theorem. There is a finitely presented group $G$ with a finite,
oriented, four-dimensional aspherical Poincare complex $Q$, with
$$H^i(G;\mathbb Z G) = 0 \quad (i \ne 4), \qquad H^4(G;\mathbb Z G) \cong \mathbb Z$$
on trivial right action, such that no closed aspherical topological
four-manifold has fundamental group $G$. In particular $\mathbb Z$ has
a finite resolution by finitely generated free $\mathbb Z G$-modules.

The theorem refutes Wall's question in its $PD_4$ specialization. It
does not assert a counterexample in every dimension. The orientation
character causes no extra obstruction: a hypothetical aspherical
four-manifold with group $G$ would have orientation character equal to
the right action on its dualizing module, which is trivial; such a
manifold would therefore be orientable and its classifying map to $Q$
would have degree one.

The construction has two halves, joined by the reflection method of
Davis. The first half produces a compact smooth oriented four-manifold
$D$ with connected boundary $S$, free fundamental group $H$, a framed
hyperbolic basis of $\pi_2 D$ over $\mathbb Z H$, and a graph Poincare
model $P \simeq BH$ obtained by attaching one three-cell along each
basis sphere; the pair $(P,S)$ is an oriented Poincare pair of
dimension four. The chamber proposition states: the boundary $S$,
together with the marking $S \to BH$, admits no topological graph-type
filling. That is, no compact four-manifold $D'$ with $\partial D' = S$
and a homotopy equivalence $D' \to BH$ extending the boundary marking.
./pd4-citations/marked-tensor-obstruction-2026-09-24.pdf

The second half applies Davis's reflection method to the right-angled
Coxeter group $C$ on a flag triangulation of $S$. A finite-index
torsion-free subgroup $\Delta \le C$ acts freely and cocompactly,
producing a closed four-manifold $N$ and a finite aspherical Poincare
complex $Q = BG$. The complex $Q$ is obtained from $N$ by attaching
three-cells on the entire middle basis. The group $G = \pi_1 Q$ is
torsion-free and satisfies the $K$- and $L$-theoretic Farrell-Jones
conjectures with coefficients, by a support-tree construction producing
a proper cocompact CAT(0) cube complex.
./pd4-citations/davis-poincare-duality-groups-survey.pdf
./pd4-citations/bartels-luck-2012-borel-hyperbolic-CAT0.pdf
./pd4-citations/wegner-2012-K-FJ-CAT0.pdf

If $Q$ had a closed aspherical manifold model $M$, then the reflection
machinery, together with the Kasprowski-Land surgery result and Quinn's
stable product theorem, would exhibit a stable geometric realization of
all marked sphere pairs on the $N$-side. A finite chamber union,
capped along noncentral boundary factors and peeled inside a bounded
inner ball, then carries enough control for the Powell-Ray-Teichner
disc embedding machinery to construct geometric duals, and a tower
construction removes the remaining middle homology. The result is a
compact four-manifold $V$ with $\partial V = S$, $\pi_1 V = H$, and
contractible universal cover, a marked graph-type filling of $S$,
contradicting the chamber obstruction.
./pd4-citations/kasprowski-land-2020-topological-4mfds-4d-pi1.pdf
./pd4-citations/powell-ray-teichner-2025-4d-disc-embedding-dual-spheres.pdf

Throughout, orientability is in the topological category and unqualified
homology is integral. For a group $\Gamma$ and connected space $X$ with
$\pi_1 X = \Gamma$, the coefficient ring $\mathbb Z \Gamma$ denotes the
standard local system on $X$. The group-ring involution is
$\overline{\sum n_g g} = \sum n_g g^{-1}$. A hyperbolic plane over
$\mathbb Z \Gamma$ has matrix
$$\left(\begin{smallmatrix} 0 & 1 \\ 1 & 0 \end{smallmatrix}\right).$$

A framed hyperbolic system of immersed spheres carries, in addition to
the displayed intersection form and even Wall self-intersection form,
vanishing framing and Wall self-intersection obstructions. Sphere
classes and intersections below use fixed whiskers and deck
translations. All stabilizations are interior connected sums with
$S^2 \times S^2$ and leave boundary markings unchanged.

Where surgery arguments below are topological, the relevant manifolds
need not be smooth: the chamber that opens the construction is smooth,
and the topological extension uses Quinn's punctured smoothing.
References to Freedman-Quinn use the January 2013 reformatted,
unrevised edition throughout.
./pd4-citations/freedman-quinn-1990-topology-of-4-manifolds-reformat2013.pdf

1. A chamber with no marked graph filling

The chamber proposition is proved in four parts: an imported geometric
datum with a finite family of marked loop laws; an assembly from the
imported pieces to a smooth chamber with a specified hyperbolic middle
basis; an explicit Poincare structure on the resulting graph pair; and
a descent from an arbitrary topological filling to a forbidden cut
configuration.
./pd4-citations/marked-tensor-obstruction-2026-09-24.pdf

Proposition 1. There are a compact connected oriented smooth
four-manifold $D$, its connected boundary $S$, a finitely generated
free group $H$, and a map $c : D \to BH$ with three properties.

(1) The map $c$ is an isomorphism on fundamental groups, and $D$ has
the homotopy type of a finite graph with $2k$ two-spheres attached at
its basepoint. The module $\pi_2 D$ carries a specified hyperbolic
basis over $\mathbb Z H$, represented by framed immersed spheres with
vanishing quadratic self-intersection obstructions.

(2) Attaching a three-cell along each basis sphere gives a space $P$
with $(P,S)$ an oriented Poincare pair of formal dimension four, and
$P \simeq BH$. The boundary marking is the restriction
$c|_S : S \to BH$.

(3) There is no compact oriented topological four-manifold $D'$ with
boundary identified to $S$ and a homotopy equivalence $c' : D' \to BH$
restricting to $c|_S$ on the boundary.

Fixed whiskers, deck translations, and quadratic self-intersection are
as in the opening conventions. A hyperbolic pair has zero squares,
mutual intersection one, and vanishing framing and Wall self-
intersection obstructions.

The argument borrows three things from the marked tensor obstruction.

First, a marked plumbing body that fixes finitely many based loop
relations. Second, an exterior with algebraic dual spheres with a
group-ring intersection law before any meridians are killed. Third, a
finite tensor obstruction, one component in finitely many nilpotent
parameters, that no compatible cut configuration can satisfy. The
first two are recollected in this section; the third is recollected in
the proof of the chamber proposition itself.

Fixed integer parameters. Let $m \ge 5$ and $p \ge 9$. Start with a
four-dimensional one-handlebody on $2m+p$ handles. Attach cancelling
two-handles to the first $2m$ handles with product framings, then plumb
these two-handles in $m$ pairs, interchanging the two coordinate
directions at each plumbing. The resulting smooth oriented manifold $X$
has graph homotopy type.
./pd4-citations/marked-tensor-obstruction-2026-09-24.pdf

Let $\pi_1 X \to F_m$ be the plumbing quotient. Its associated cover is
a tree of blocks. Each block is a one-handlebody on $p$ extra
generators, and neighbouring blocks meet in a product four-ball. The
exposed boundary of a block is
$$K_v = \#^p (S^1 \times S^2) \setminus \operatorname{int} \nu(L_{2m}),$$
where $L_{2m}$ is an unlink in a ball. Its port meridians
$x_1,\dots,x_{2m}$ and extra loops $l_1,\dots,l_p$ form a free basis at
$v$. At each join the meridian on one side is the longitude on the
other.

Set $\Gamma = \{1, l_1,\dots,l_p\}$. The marked body construction
installs, for every ordered triple of distinct ports and every
independent choice
$$X_a = c_a \, x_{r_a}^{\delta_a} \, c_a^{-1}, \qquad c_a \in \Gamma, \quad \delta_a \in \{\pm 1\}, \quad a = 1,2,3,$$
a disjoint two-stage grope body. Its lower surface is a punctured
torus. One branch has a single cap; the other carries the upper
punctured torus with its two caps. All tip tracks, conjugating paths,
and product framings are specified.

The initial boundary word $R$ satisfies
$$R|_{X_a = 1} = 1 \ \ (a=1,2,3), \qquad R \equiv [X_1,[X_2,X_3]] \pmod{\Gamma_4 F(X_1,X_2,X_3)}.$$
Internal changes of the marked paths are absorbed by these identities.
The entire finite family of laws is retained; the data are not
reducible to one conjugated commutator.

Let $F_1,\dots,F_k$ be the lower punctured tori, with symplectic bases
$(a_j,b_j)$, where $a_j$ is the single-cap branch. Delete their open
product tubes, including the proper edge collars, to obtain the exterior
$E$. The marked tensor obstruction supplies truncated single caps $f_j$
and framed spheres $g_j$ in $E$ with
$$\lambda(f_i, g_j) = \delta_{ij}, \qquad \lambda(g_i, g_j) = 0, \qquad \widetilde\mu(g_j) = 0 \ \ \text{over} \ \mathbb Z[\pi_1 E].$$
The framing on $f_j$ is the surface-compression framing. The $g_j$ are
compressed normal-circle tori over the $b_j$ curves; their individual
representatives have trivial normal bundle. The displayed law is a
group-ring identity, and is asserted before any meridians are killed.

Placement refinement. Put the longitudinal tip annuli on disjoint
parallels outside sufficiently thin attaching solid tori in the initial
one-handlebody boundary, and the finitely many connectors in their
complement. The bodies are boundary joins of separated annular collars,
so this is possible by the marked handlebody model. The complement of
the attaching solid tori survives in the boundary after the two-handles
are attached, and the plumbing ports lie on the new belt-handle
portions. Thus the unpushed lower surfaces lie in the final exposed
boundary of $X$. The independent ribbon twists matching the cap
framings are made in the separated parallel tubes; they do not pass a
body across a plumbing join. Returning tip collars are lengthened back
to product parallels on the attaching regions, and then into the cores.
The construction and intersection calculation above are unchanged.

Consequently each $F_j$ is a proper inward push of a disjoint boundary
surface, with returning collars at its initial boundary curves. The
map $F_j \to X$ is null-homotopic: the three tip loops bound the
corresponding core caps in $X$, so the bottom $a_j$ and $b_j$ loops die
in $\pi_1 X$; since $X$ has graph homotopy type and $F_j$ retracts to a
graph, the whole surface map is null-homotopic.

Lemma 1. With the surface-product trivializations, the exterior and
its tube-side maps satisfy the homotopy-pushout model
$$E \simeq X \cup_{\coprod_j F_j} \coprod_j (F_j \times S^1),$$
where each gluing map to $F_j \times S^1$ is the inward section. The
displayed $F_j \times S^1$ factors represent the actual tube sides.

Proof. The constructions have disjoint collars, so one surface
suffices. Write $B = \partial X$ and $K = \partial F$, and put the
pushed surface at a horizontal collar level with vertical walls along
$K$ returning to the boundary. The inner part of the exterior has
homotopy type $X$, the outer part has homotopy type $B \setminus K$, and
their overlap has homotopy type $B \setminus F$. Product slices,
thickened by a variable positive width off the deleted surface, give
this homotopy-pushout decomposition.

In $B \setminus K$, take a two-sided neighbourhood of
$F^\circ = F \setminus K$ whose width decreases to zero near $K$.
Relative to $B \setminus F$ this neighbourhood is a bridge over
$F^\circ$, joining the two sides of the cut. Replace $F^\circ$ by a
slightly shrunken copy $F_0$; this is an objectwise homotopy
equivalence and leaves the actual deleted surface and $K$ unchanged.
In rectangular normal coordinates to the horizontal copy, split the
tube circle over $F_0$ into its outer and inward semicircles. The
outer semicircle is the bridge. The inward semicircle lies in the
inner $X$ piece and produces a specified homotopy between the bridge's
endpoint maps.

Identifying the endpoints by that homotopy replaces the bridge by
$F_0 \times S^1$, attached along its inward section. Its circle is the
outer bridge followed by the inward return, hence the actual normal
circuit. Finally $F_0 \to F$ and $F_0 \times S^1 \to F \times S^1$ are
homotopy equivalences, and the construction accounts for the proper
edge collars and side maps, not only the absolute homotopy type of $E$.

Now regard a solid torus $V_j$ as a compression body from the punctured
torus $F_j$ to a disk, compressing $a_j$. Define
$$D = E \cup_{\coprod_j (F_j \times S^1)} \coprod_j (V_j \times S^1).$$
All gluings use the specified product framings; round corners. The
remaining disk face of $\partial V_j$ contributes a solid torus to the
new boundary. Hence $S = \partial D$ is the surface-framed surgery of
$\partial X$ on the initial grope-boundary link, and is connected.

By the exterior model,
$$D \simeq X \cup_{\coprod_j F_j} \coprod_j (V_j \times S^1).$$
Each $F_j \to X$ is null-homotopic. The extra summand is the homotopy
cofiber of $F_j \to V_j \times S^1$. Retract the target to the torus on
its longitude $b_j$ and normal meridian $t_j$. The cofiber adds one
two-cell killing $b_j$ and one on the null loop $a_j$. Cancel the
$b_j$ one/two-cell pair; the torus two-cell then has null attaching map,
so the result is
$$S^1_{t_j} \vee S^2 \vee S^2.$$
Thus $H = \pi_1 D$ is the original free graph group with one new free
meridian $t_j$ per surface, and $D$ has the graph-and-spheres homotopy
type of Proposition 1.

The middle form. Close $f_j$ by a meridian disk of $V_j$ at its fixed
circle coordinate, obtaining a sphere $u_j$. The two compression
framings agree, so its second Stiefel-Whitney evaluation is zero. The
$g_j$ remain in $E$, so the exterior duality gives
$$\lambda(u_i, g_j) = \delta_{ij}, \qquad \lambda(g_i, g_j) = 0 \ \ \text{over} \ \mathbb Z H.$$
The hermitian matrix of the $u_i$ has even diagonal. As $H$ has no
element of order two, every such diagonal is $z + \bar z$: choose one
representative from each nonidentity inverse pair and halve the even
identity coefficient. Add suitable combinations of the $g_j$ to the
$u_i$, taking a triangular half off the diagonal and this half on it.
The resulting spheres $a_i$, together with $b_i = g_i$, have
hyperbolic intersection matrix.

The usual framed representative adjustments give zero Wall
self-intersections: choose normal Euler number zero using the evenness,
and then use $\lambda(a_i,a_i) = \mu(a_i) + \overline{\mu(a_i)}$
together with the absence of order-two elements. This is the ordinary
immersed-sphere calculus; no disk-embedding assertion enters it.

Basis. These $2k$ classes form a basis. Indeed $\pi_2 D$ is free of
rank $2k$. The square matrix of the displayed classes has a left
inverse from their nonsingular pairing. For the right inverse, pass to
each finite quotient of $H$: the resulting matrices act on
finite-dimensional complex vector spaces, where a left inverse is a
right inverse. Residual finiteness of the free group separates the
finite support of any nonzero group-ring element, so the right-inverse
identity holds over $\mathbb Z H$. This specified basis is fixed from
this point onward.

Attach a three-cell on every $a_i$ and $b_i$ to form $P$. A map from
the graph with $2k$ two-spheres to $D$ realising this basis is a
homotopy equivalence: it induces an isomorphism on $\pi_1$ and on the
homology of simply connected covers. Thus $P \simeq BH$.

Poincare pair. Set $\Lambda = \mathbb Z H$. In the duality of $(D,S)$,
the middle sphere module maps isomorphically onto a split summand of
$H_2(D,S;\Lambda)$ by its nonsingular form. Attaching the three-cells
quotients both absolute and relative homology by these middle summands;
the attaching map is injective on them, so no new third homology
appears. On cohomology the pullbacks identify the new groups with the
annihilators of the sphere summands. If $f : (D,S) \to (P,S)$ is the
cell-attachment inclusion and $[P,S] = f_* [D,S]$, cap-product
naturality gives
$$z \frown [P,S] = f_*\bigl(f^* z \frown [D,S]\bigr).$$
Hence the old duality isomorphisms induce exactly the new ones on
these annihilators and quotients. The spaces have finite free chain
complexes over $\Lambda$, so the regular-coefficient cap-product
criterion gives Poincare duality with arbitrary local coefficients.
This proves parts (1) and (2) of Proposition 1.

Corollary. The map $\pi_1 S \to H$ is onto. If $r = \operatorname{rank} H$, then $H_1(S;\mathbb Z) \to H_1(P;\mathbb Z)$ is an
isomorphism and $H_2(S;\mathbb Z) \cong \mathbb Z^r$.

Proof. Duality and the graph type of $P$ give
$H_1(P,S;\mathbb Z H) = H_0(P,S;\mathbb Z H) = 0$. Therefore
$H_0(S;\mathbb Z H) \to H_0(P;\mathbb Z H)$ is an isomorphism. The
former group is free abelian on the cosets of the image of $\pi_1 S$,
and the latter is $\mathbb Z$; there is only one coset. With ordinary
coefficients, the pair sequence and duality give $H_2(P,S) = H_1(P,S) = 0$
and $H_3(P,S) \cong H^1(P) \cong \mathbb Z^r$.

3. Topological fillings and the cut obstruction

Part (3) of Proposition 1 remains. Suppose a compact oriented
topological four-manifold $D'$ with boundary identified to $S$, and a
homotopy equivalence $c' : D' \to BH$ extending the marking, existed.

The filling as a manifold with disk boundaries. Attach the dual
two-handles that undo the surgery on the initial link. The result is a
space $W$ with $\partial W = \partial X$. Each attaching circle
represents its prescribed new free generator $t_j$ under the marking
$D' \simeq BH$, and the corresponding one/two-cell pair cancels in the
homotopy type. Thus $W$ has the graph type of $X$, with the original
boundary-to-graph map: equality of the marked fundamental-group maps
gives equality up to homotopy because the graph is aspherical. The
two-handle cocores are disjoint proper locally flat disks bounded by
the initial link. Every original marked word is therefore represented
by a disk boundary in $W$.

Handling a topological base. Since $W$ is topological, apply Quinn's
punctured smoothing: delete an interior point away from the disks and
extend the given boundary smoothing over the complement. Approximate
the marked map smoothly over the middle intervals of the finitely many
graph edges, away from the disks and the puncture, and choose regular
values. The cuts are compact smooth three-manifolds, disjoint from the
disks. Their filled complementary pieces are topological four-manifolds
smooth away from the finite set of punctures in each finite quotient.
./pd4-citations/freedman-quinn-1990-topology-of-4-manifolds-reformat2013.pdf

Connectedness of the cuts. Form the component graph of the cuts in the
tree cover of $W$. Its fundamental group is a quotient of that of the
cover, by joining successive crossings through connected complementary
pieces; every extra boundary generator lies in one piece and therefore
maps trivially. The component graph is a tree. Its outside faces and
torus collars give an embedded principal copy of the plumbing tree.
Any remaining branch has a unique nearest principal vertex, whose
stabilizer is trivial. The quotient has finitely many components, so
these branches have finite valence and uniformly bounded depth, and
each is finite. Ignoring their closed cuts and merging them into the
principal vertices leaves connected vertex pieces and one connected
cut with each prescribed torus as its entire boundary. The disk
prescriptions are unchanged. This is the connected-tree-cuts lemma of
the marked tensor obstruction, now applied to the marked $W$ rather
than to a homeomorphism $W = X$.
./pd4-citations/marked-tensor-obstruction-2026-09-24.pdf

Parity quotient. Send every plumbing generator to the nonidentity
element of $\mathbb Z / 2$. The filled ambient space $W_2$ has the
homotopy type of the graph with two vertices and $2m$ edges, plus $p$
extra circles at each vertex. Write $A, B$ for the filled vertex
pieces and $Y_1, \dots, Y_{2m}$ for their cuts. Over any
characteristic-zero field $K$, Mayer-Vietoris gives
$$H^2(A;K) \oplus H^2(B;K) \xrightarrow{\cong} \bigoplus_{i=1}^{2m} H^2(Y_i;K),$$
because $H^2(W_2;K) = H^3(W_2;K) = 0$. In degree one the difference of
restrictions is onto. Its kernel is the image of $H^1(W_2;K)$ modulo
the $2m-1$ graph classes, and evaluation on the $p$ named extra loops
at each vertex identifies that kernel with $K^{2p}$. The restriction
map is therefore an isomorphism on the subspace where these extra
evaluations vanish.

Active edges. Half-lives-half-dies states that on $Y_i$ the image of
$H_1(\partial Y_i;K)$ in $H_1(Y_i;K)$ is a line. If both original
coordinate loops had nonzero image, their images would be nonzero
multiples of the same line. A class from either vertex annihilates the
longitude bounding in its outside face, and would therefore annihilate
both coordinates; the degree-one surjectivity above would then force
every class on $Y_i$ to do so, a contradiction. Hence one original
coordinate $\beta_i$ dies and the other $\alpha_i$ survives. The plus
endpoint is the side at which $\beta_i$ is a longitude. The parity
involution pairs the two lifts of each join and exchanges their plus
endpoints, giving exactly $m$ active edges at $A$.

Laws at the vertex pieces. The prescribed disks lie in their own
vertex pieces and give every marked law at those vertices, with all
the independently chosen conjugators and signs. The marked ambient
free group has trivial port meridians and freely assignable named
extra letters. Assign the eight nonidentity matrices among
$\exp(a \sigma e) \exp(b \sigma f)$ with $a,b \in \{0,1,2\}$ to eight
extra letters, for the standard nilpotent generators $e,f$ of
$\mathfrak{sl}_2$. Together with the identity, their adjoint operators
span all endomorphisms over $\mathbb C((\sigma))$.

Forms on a punctured vertex. Let $V$ be one of the filled compact
vertex pieces and $Z \subset V$ its finite set of punctures. In degree
zero take smooth functions on $V \setminus Z$ separately constant near
each puncture; in positive degree take smooth forms vanishing near the
punctures. This differential graded algebra computes $H^*(V;\mathbb C)$
naturally on the unpunctured faces, and supports the vertex
calculation in the Vertex Data Proposition of the marked tensor
obstruction, with the filled cohomology and the same Stokes
identities. The proof: view these forms as a sheaf on $V$ whose stalk
at a puncture is $\mathbb C$ in degree zero and zero in higher degree,
and the ordinary de Rham resolution elsewhere. Partitions of unity
constant on neighbourhoods of the finitely many punctures turn it
into a fine resolution of the constant sheaf, so the cohomology is
the filled cohomology. Linear splittings and contracting homotopies
live inside this complex. Each fixed positive-degree coefficient
vanishes near $Z$, so integration and Stokes take place on compact
subdomains; finite-coefficient operations never need a common
vanishing neighbourhood. Removing interior points in dimension four
does not change the fundamental group, and gauges can be separately
constant at distinct ends; the based holonomy and exact-gauge
arguments use the same marked group as the filled vertex. Edge
algebraicity is applied only to the unchanged compact smooth three-
manifolds $Y_i$.

Endgame. Apply the punctured-forms lemma to $A$ and $B$. The
restriction calculation and the degree-one calculations refer to their
filled cohomology, which the lemma computes. The Vertex Data Proposition
of the marked tensor obstruction now produces precisely the
characteristic-zero tensor identities whose incompatibility is the
Finite Tensor Obstruction. The obstruction contradicts their
existence. Therefore $D'$ cannot exist, which completes the proof of
Proposition 1.

4. Reflection and the duality group

The chamber pair $(D,S) \to (P,S)$ closes under Davis's reflection
method. A right-angled Coxeter group $C$ on a flag triangulation of $S$
acts on a development $\mathcal U(C,D)$, and a finite-index torsion-free
subgroup $\Delta \le C$ gives a closed four-manifold $N$ together with
a finite aspherical Poincare complex $Q = BG$. The chamber construction
ensures that all middle-dimensional classes survive the closing.

Panels. Choose a finite flag triangulation $\mathcal L$ of $S$, for
example a barycentric subdivision of a smooth triangulation. Write $V$
for its vertex set, and $S_s$ for the closed dual block of $s \in V$ in
the barycentric subdivision. These blocks are three-dimensional
panels. For nonempty $T \subseteq V$, the intersection
$S_T = \bigcap_{s \in T} S_s$ is nonempty exactly when $T$ spans a
simplex of $\mathcal L$; then it is a $(4-|T|)$-ball. Collar
neighbourhoods of these intersections have the usual orthant
structure.

Define the right-angled Coxeter group
$$C = \langle s \in V : s^2 = 1, \ st = ts \ \text{whenever} \ \{s,t\} \in \mathcal L\rangle.$$
For a collared space $B$ with these panels in its boundary, its
development is
$$\mathcal U(C,B) = (C \times B)/\sim, \quad (c,x) \sim (c',x) \iff c^{-1} c' \in C_{T(x)}.$$
Here $T(x) = \{s : x \in S_s\}$ is empty off the boundary, and $C_T$
is the subgroup generated by $T$. Each chamber embeds, since the
equivalence relation never identifies different values of $x$. For
nonempty $T(x)$, the subgroup $C_{T(x)}$ is $(\mathbb Z / 2)^{|T(x)|}$,
and its sectors fill the corresponding orthant neighbourhood.

Three elementary Coxeter facts are used. Special subgroups embed; a
finite special coset has a unique shortest element; the right descent
set $\operatorname{Des}(c) = \{s \in V : |cs| < |c|\}$ is a clique.
These follow from the right-angled reduced-word rule of commuting
adjacent commuting letters and cancelling equal consecutive letters.

Lemma 2 (shelling). Put $Y = \mathcal U(C,D)$ and let $D_c$ be its
chamber with label $c$. Order $C$ by nondecreasing word length,
breaking ties arbitrarily. Every finite initial segment containing the
identity is, after rounding collar corners, an iterated boundary
connected sum of its chambers. In particular
$$U_L = \bigcup_{|c| \le L} D_c$$
is a compact manifold whose boundary is the connected sum of the
corresponding copies of $S$, with fundamental group the free product of
the chamber groups. The same statements hold when chambers have been
stabilized in their interiors.

Proof. The part of a new chamber $D_c$ already present is
$$A_c = \bigcup_{s \in \operatorname{Des}(c)} S_s.$$
To handle equal-length overlaps, consider a point with panel set $T$.
Its incident chambers form the special coset $c C_T$; write $a$ for its
shortest element. Members have the form $a \prod_{s \in J} s$ with
$J \subseteq T$ and length $|a| + |J|$. If a distinct incident chamber
occurs no later than $c$, then $c$ is not shortest, so it has a
descent in $T$, and the point lies in $A_c$. Conversely, crossing each
descent panel gives a shorter adjacent chamber.

The set $A_c$ is a three-ball. A union of dual panels indexed by a
nonempty simplex is a ball: add the panels one at a time; the
intersection of a new panel with the preceding union is the
corresponding union of dual panels in its boundary link, a ball of one
lower dimension by induction on dimension. The dual-block coordinates
supply the product collars needed to glue consecutive balls.

To check that $A_c$ lies in the boundary of the preceding union, pick a
face incident to $q$ panels. Label the local orthants by subsets of
those panels, starting at the shortest incident chamber. The already
present orthants form a Boolean downset $I$: every proper subset of a
present subset has smaller word length, with any consistent tie-
breaking. When $I$ is nonempty and proper, its frontier is the PL
hypersurface $t = f(u)$ in local coordinates $x = u + t(1,\dots,1)$
with $\sum u_i = 0$, where
$$f(u) = \max_{A \text{ maximal in } I} \min_{i \notin A} (-u_i).$$
The frontier is locally a graph. Along the incoming faces, some
sectors are present and the new chamber sector is absent, so the
incoming \(3\)-ball has locally flat boundary in this three-manifold
frontier. Gluing is a boundary connected sum after rounding. Induction
proves both statements. Van Kampen gives the labelled free product,
and interior stabilizations do not affect any collar argument.

Universal development. Let $\Delta = \ker(C \to (\mathbb Z / 2)^V)$,
where each generator maps to its own coordinate vector, and put
$$N = \Delta \backslash \mathcal U(C,D), \qquad Q = \Delta \backslash \mathcal U(C,P).$$
The parity map is injective on every finite special subgroup, so
$\Delta$ has trivial intersection with every point stabilizer in
either development. The quotient has finitely many chambers. The
orthant charts show that $N$ is a closed four-manifold. Give chamber
$c$ the orientation $(-1)^{|c|}$ times the chosen chamber orientation:
adjacent chamber orientations fit across a panel, and $\Delta$ preserves
them, so $N$ is oriented. The same convention orients reflected
Poincare pairs.

Let $\widehat S \to S$ be the connected cover determined by
$\pi_1(S) \to H$, with lifted flag triangulation $\widehat{\mathcal L}$.
The lifted nerve is flag: vertices of a lifted clique project to
distinct pairwise adjacent vertices downstairs, span a simplex there,
and lifts of its edges and triangular relations place all the vertices
in one lift of that simplex. Let $\widehat C$ be the right-angled
Coxeter group on $\widehat{\mathcal L}$. The deck action of $H$
permutes its generators, and forgetting the lift gives an
$H$-invariant homomorphism $q : \widehat C \to C$. Define
$$\rho : \widehat C \rtimes H \to C, \quad \rho(\widehat c, h) = q(\widehat c), \qquad G = \rho^{-1}(\Delta).$$

Let $\widetilde P$ be the universal cover of $P$, with boundary
preimage $\widehat S$ and its lifted panels. On
$\widehat Q = \mathcal U(\widehat C,\widetilde P)$, let $(\widehat c, h)$
act by $(\widehat c, h)[\widehat d, x] = [\widehat c \, h(\widehat d), h x]$.
The map $[\widehat d, x] \mapsto [q(\widehat d), \bar x]$ descends to
the ordinary development. It is a covering map on each chamber
interior and also at each boundary point, since lifted panels map
bijectively to downstairs panels and corresponding finite special
groups map isomorphically. These charts identify $\widehat Q / G = Q$.

The action of $G$ is free: an element fixing a point has trivial
$H$-component after conjugating the chamber label, and the remaining
stabilizer is a lifted finite special subgroup of $\widehat C$ on which
$\rho$ is injective and whose image misses $\Delta$.

$\widehat Q$ is contractible. Filter its chambers by word length in
$\widehat C$. The first chamber $\widetilde P$ is contractible; every
further chamber meets the shorter ones in the nonempty ball of its
descent panels by the shelling argument, and its attachment is a
homotopy equivalence. In an infinite word-length layer, new interiors
are disjoint and their mutual intersections lie in shorter layers, so
all attachments may be performed simultaneously. The CW direct limit
is weakly contractible and hence contractible. In particular $Q$ is a
finite $K(G,1)$.

Proposition 2 (reflected pair). $Q$ is a finite oriented Poincare
complex of dimension four. The natural map $f : N \to Q$ has degree
one and induces an isomorphism on fundamental groups, identifying both
with $G$. If $b$ is the number of chamber hyperbolic planes in $N$,
then $\pi_2(N) \cong (\mathbb Z G)^{2b}$ carries the specified
orthogonal hyperbolic basis, and $\widetilde H_i(\widetilde N;\mathbb Z) = 0$
for $i \ne 2$. The group $G$ is finitely presented and torsion-free,
and
$$H^i(G;\mathbb Z G) = 0 \quad (i \ne 4), \qquad H^4(G;\mathbb Z G) \cong \mathbb Z$$
with trivial right action.

Proof. Use the universal development with $\widetilde D$ in place of
$\widetilde P$. The first chamber is simply connected; attaching
subsequent simply connected chambers along balls shows the development
is the universal cover $\widetilde N$. Its reduced homology is the
direct sum of chamber reduced homologies, each concentrated in
degree two.

There are finitely many chamber orbits in $N$. For each orbit,
induction from its chamber subgroup produces
$\mathbb Z G \otimes_{\mathbb Z H} \pi_2(D) \cong (\mathbb Z G)^{2k}$.
Spheres lie in chamber interiors, so distinct chambers are orthogonal;
within a chamber the form is the given $\mathbb Z H$-form extended to
$\mathbb Z G$. Reversing a single vector in a plane absorbs any
chamber orientation sign. This exhibits the asserted basis and form.

The construction of $P$ attaches three-cells away from the boundary;
$Q$ is obtained from $N$ by attaching three-cells on this entire
middle basis. So $f$ induces the displayed identification on $\pi_1$.
The attaching three-cells split off a nonsingular middle summand in
both absolute and relative duality: on homology they quotient by the
sphere summand and on cohomology they restrict to its annihilator. With
$[Q] = f_* [N]$, naturality of the cap product identifies surviving
duality maps with those of $Q$. The finite free chain criterion
therefore gives Poincare duality for $Q$. The top fundamental class is
unchanged, so $f$ has degree one.

For the regular-coefficient calculation, the only potentially new map
in the cohomology sequence for the three-cell attachments is
$$H^2(N;\mathbb Z G) \to \operatorname{Hom}_{\mathbb Z G}(\pi_2(N),\mathbb Z G).$$
Under duality this is the adjoint of the nonsingular hyperbolic form,
hence an isomorphism. The middle cohomology disappears whereas
$H^4(Q;\mathbb Z G) = H^4(N;\mathbb Z G) = \mathbb Z$; all other degrees
vanish. The right action on the top group is trivial because the
orientation is preserved by $G$.

Finiteness and asphericity of $Q$ give a finite presentation and a
finite free resolution of $\mathbb Z$ from the chains of $\widehat Q$.
Finite integral cohomological dimension excludes torsion: restriction
to a nontrivial finite cyclic subgroup would give a finite projective
resolution for that group, contrary to nonvanishing cohomology in
arbitrarily high degrees.

CAT(0) support. When $H$ is infinite the lifted nerve has infinitely
many vertices, and its Davis complex alone need not give a cocompact
action of the semidirect product. A support-tree construction fixes
this.

Lemma 3 (support-tree CAT(0)). $\widehat C \rtimes H$, and hence its
finite-index subgroup $G$, acts properly and cocompactly by isometries
on a proper CAT(0) cube complex of dimension at most five.

Proof. Choose a locally finite tree $T$ with $H$ acting freely and
cocompactly without inversions; the trivial case is a single vertex.
Assign to each lifted-nerve vertex $s$ a point $a_s \in T^{(0)}$,
equivariantly under $H$. There are finitely many vertex and edge
orbits in the lifted nerve, so the distances $d_T(a_s,a_t)$ for
adjacent $s,t$ have a common finite bound. Choose $r$ large enough
that the closed radius-$r$ subtrees $T_s$ about $a_s$ intersect for
every adjacent pair.

For a vertex $v$ and an edge $e$ of $T$, put
$A_v = \{s : v \in T_s\}$ and $A_e = \{s : e \subset T_s\}$. Both sets
are finite of uniformly bounded size: finitely many generator orbits,
and a fixed-radius ball in the free cocompact $T$ contains uniformly
finitely many orbit points. Let $W_v$ and $W_e$ be the special
subgroups of $\widehat C$ generated by $A_v$ and $A_e$. For
$e = [v,w]$ one has $A_e = A_v \cap A_w$, so edge maps are special
subgroup inclusions.

The graph of groups over $T$ has colimit $\widehat C$: occurrences of
a generator $s$ are identified precisely over its connected support
$T_s$; every defining commutation occurs in a vertex group since the
two supports intersect at a vertex; conversely, each vertex group uses
the induced subgraph of the lifted nerve, introducing no extra
relation. All edge maps and their extensions
$W_e \subset W_v \subset \widehat C$ are injective.

Let $\Sigma_v$ and $\Sigma_e$ be the Davis cube complexes of these
finite-generator right-angled groups. They are CAT(0)
./pd4-citations/swiatkowski-2016-hyperbolic-coxeter-sierpinski.pdf
and $\Sigma_e \subset \Sigma_v$ is convex in the standard piecewise
Euclidean cube metric. Form the tree of spaces with vertex spaces
$\widehat C \times_{W_v} \Sigma_v$ and edge cylinders
$\widehat C \times_{W_e} (\Sigma_e \times [0,1])$ over the Bass-Serre
tree. Glue each cylinder to the vertex spaces by the special-subcomplex
maps. Convex gluing gives CAT(0) over every finite subtree, preserving
the factors as convex isometric subspaces; applying the gluing first
over finite subtrees and then passing to the union completes it, with
local finiteness below ensuring completeness.

Local finiteness. In a fixed vertex Davis complex, for one incident
support edge, translates of its special subcomplex have vertex sets
equal to the cosets of $W_e$ in $W_v$; these cosets partition the
Cayley vertices, so a finite subcomplex meets only finitely many of
them. Each support vertex has only finitely many incident edges.
Since each $\Sigma_v$ is locally finite, the whole tree of cube
complexes is locally finite; unit edge lengths and uniformly bounded
dimension make the induced length metric proper and complete.

Left multiplication gives the $\widehat C$-action, and support
equivariance extends it to $\widehat C \rtimes H$. An element
stabilizing a vertex space projects to an element of $H$ fixing its
support vertex, thus trivial; within the vertex space its point
stabilizer is a finite Coxeter stabilizer, and the same holds on edge
interiors. The cellular action on the locally finite complex is
proper. There are finitely many $H$-orbits of support vertices and
edges, and each corresponding Davis complex has compact quotient by
its special group, so the full action is cocompact. Cliques of the
lifted nerve have at most four vertices; vertex spaces have dimension
at most four and edge cylinders dimension at most five. Since $G$ has
finite index in $\widehat C \rtimes H$, its restricted action has all
the same properties.

Assembly consequences. It follows that $G$ satisfies the $K$- and
$L$-theoretic Farrell-Jones conjectures with coefficients: Bartels and
Luck apply the Borel conjecture to CAT(0) groups, and Wegner supplies
the $K$-theoretic assembly. Because $G$ is torsion-free, the
Whitehead group, reduced $K_0(\mathbb Z G)$, and negative
$K$-groups vanish, and the simple $L$-theory assembly map is an
isomorphism in every degree. No four-dimensional rigidity conclusion
is drawn from CAT(0) geometry.
./pd4-citations/bartels-luck-2012-borel-hyperbolic-CAT0.pdf
./pd4-citations/wegner-2012-K-FJ-CAT0.pdf

Gallery map. Let $\Sigma_C$ denote the ordinary Davis cube complex of
$C$. Its vertices are the chamber labels $C$, and the link of the
identity vertex is $\mathcal L$, whose realisation is $S$.

Lemma 4 (gallery map). There is a $C$-equivariant continuous map
$$\eta : Y \to \Sigma_C$$
sending the inner core of each chamber to its label, respecting
panels, with uniformly bounded image diameter on each chamber. On the
boundary collar of the central chamber its radial directions at the
identity vertex give the dual-block identification with $S$. After
interior chamber stabilizations, the same construction gives a
continuous map with the same panel, core, and diameter properties;
it is equivariant under any subgroup preserving the stabilized pattern.

Proof. The Davis chamber is the cone on the barycentric subdivision of
$\mathcal L$, with its usual dual panels on the base. Identify that
base panelwise with $S$. Map a boundary collar of $D$ to this cone by
that identification on its outer boundary and by coning to the apex on
its inner boundary; map the remainder of $D$ to the apex. The maps
agree across reflected panels and hence define $\eta$. Every chamber
image lies in the corresponding compact Davis chamber, whose diameter
is independent of its label. At the identity apex, radial projection of
the cone base to a sufficiently small metric sphere gives exactly its
link identification with $S$. Stabilization changes only the collapsed
inner core.

5. Marked stable realization

Suppose, for a contradiction, that the group $G$ is the fundamental
group of a closed aspherical topological four-manifold $M$. The
preceding construction gives a finite oriented Poincare four-complex
$Q = BG$ and a closed oriented manifold $N$, with a degree-one map
$q : N \to Q$ inducing an isomorphism on $\pi_1$. Let
$\Lambda = \mathbb Z G$. The only reduced homology of $\widetilde N$
sits in degree two, where
$$A = \pi_2(N) \cong \Lambda^{2b}, \qquad
\lambda_N \cong \bigoplus_{a=1}^b \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.$$
The displayed basis is the prescribed chamber sphere pairs. All
identifications include whiskers and orientations. The group $G$ is
torsion-free, integral $PD_4$, and satisfies the $K$- and
$L$-theoretic Farrell-Jones conjectures by the CAT(0) construction
above; these are conclusions of the construction, not additional
assumptions on a putative realization.

Both $Q$ and $M$ are $K(G,1)$ spaces. Their equivalence identifies
the orientation character with the action on the top dualizing module,
so $M$ is orientable. Choose orientations and a fundamental-group
marking so that the classifying map $c_M : M \to Q$ has degree one,
and put
$$L = M \# b(S^2 \times S^2).$$
Collapsing the simply connected summands gives a degree-one
classifying map $c_L : L \to Q$. Identify $\pi_2(L)$ with $A$ by
sending each chamber pair above to a specified standard pair in $L$.
This is an isometry of equivariant forms, and the particular isometry
is retained throughout the argument.
./pd4-citations/kasprowski-land-2020-topological-4mfds-4d-pi1.pdf