# Cubic planar bipartite graphs are dispersable ††thanks: As submitted for
publication on 7/20/2020; see Postscript for additional information.

Paul C. Kainen
<EMAIL_ADDRESS>
Shannon Overbay
<EMAIL_ADDRESS>
###### Abstract

A graph is called dispersable if it has a book embedding in which each page
has maximum degree 1 and the number of pages is the maximum degree. Bernhart
and Kainen conjectured every $k$-regular bipartite graph is dispersable. Forty
years later, Alam, Bekos, Gronemann, Kaufmann, and Pupyrev have disproved this
conjecture, identifying nonplanar 3- and 4-regular bipartite graphs that are
not dispersable. They also proved all cubic planar bipartite 3-connected
graphs are dispersable and conjectured that the connectivity condition could
be relaxed. We prove that every cubic planar bipartite multigraph is
dispersable.

Key Phrases: book thickness; dispersable book embeddings; matching book
thickness; subhamiltonian vertex order; cubic bipartite planar multigraphs.

##  1 Introduction

The book thickness of a graph, [11, 15] is now well-studied and has a variety
of applications such as manipulation of stacks and queues (Even & Itai [8] and
Tarjan [17]), fault-tolerant computing [4], traffic engineering [12], RNA
folding [10], and graph drawing and visualization, e.g., [6, 7, 20]. There is
also an extensive literature on the purely mathematical aspects of this
measure of graph complexity; [19] lists over a couple of dozen and is by no
means complete.

The book thickness of a graph is the least number of colors needed for the
edges such that, for some outerplane drawing of the graph, no two edges of the
same color cross. In [3], we introduced the additional constraint that same-
color edges do not share an endpoint, so pages must be matchings.

A book embedding of a graph is an outerplane drawing (i.e., a cyclic order on
the vertices) and an edge-coloring such that edges of each color (called a
page of the embedding) induce an outerplane subgraph. A graph or vertex-order
with a 2-page book embedding is called subhamiltonian [3], [16].

The matching book thickness [13] of a graph $G$ is the least number of pages,
$mbt(G)$, needed if the pages are matchings. A graph is said to be dispersable
[3] if it has a matching book embedding in $\Delta(G)$ pages, where
$\Delta(G)$ denotes maximum degree; we refer to a dispersable book embedding.

Bernhart and Kainen [3] conjectured (in 1979):
$\mbox{\it If $G$ is regular bipartite, then $G$ has a dispersable book
embedding}.$
This BK Conjecture was recently disproved by Alam et al. [1] who showed:
There exist bipartite 3- and 4-regular graphs $G$ with $mbt(G)=4$ or $5$,
resp.
These graphs, due to Gray and Folkman, respectively, are edge-transitive but
not vertex-transitive. In fact, Tutte [18, p. 67] proved that any such graph
must be bipartite. Further, both the Gray and Folkman graphs are nonplanar.

In contrast, we know of many graphs, all vertex transitive, for which
Conjecture BK does hold, including cycles, hypercubes, complete bipartite
graphs, degree-3 circulants, and the Heawood graph [3, 16, 13]. Also, the
Cartesian product of dispersable graphs is dispersable, and if $G$ is
bipartite, then $mbt(G)$ is bounded above by the minimum sum of the page-
maximum degrees over all book embeddings and trivially by $bt(G)\Delta(G)$;
see [14].

In [1] Alam-et-al proved Conjecture BK holds if $G$ is cubic planar bipartite
3-connected and conjectured the condition, that the graph be 3-connected,
could be relaxed. We confirm their conjecture by proving that all cubic planar
bipartite multigraphs are dispersable, so Conjecture BK holds in this case.

The paper is organized as follows. Section 2 has some combinatorial
preliminaries, including two “folklore” results for regular bipartite
graphs—non-existence of a cutpoint and an “entanglement” lemma: any two cut-
edges have vertices of attachment which are in different parts of the
bipartition. Section 3 contains a proof of the main result.

##  2 Combinatorial preliminaries

Undefined graph terminology is standard [9], [5]. Let $|\cdot|$ denote
cardinality.

By a graph $G=(V,E)$ we mean a 1-dimensional simplicial complex; i.e., no
loops and at most one edge between any pair of vertices; $G$ is a multigraph
if we allow parallel edges, which are sets of more than one edge between a
pair of vertices. Thus, every graph is a multigraph. A digraph is a multigraph
whose edges have been oriented (that is, assigned a direction).

A set of edges in a multigraph $G$ is a matching if the edges are pairwise
disjoint; a matching is perfect if each vertex is in an edge of the matching.
Each page of a dispersable embedding of a regular graph is a perfect matching.

The following must be known but we haven’t found it in the literature.

######  Lemma 1.

Let $G$ be a bipartite, connected, and $k$-regular multigraph, $k\geq 3$. Let
$e^{\prime},e^{\prime\prime}$ be two disjoint edges in $G$ such that
$G\setminus\\{e^{\prime},e^{\prime\prime}\\}$ contains a connected component
$H$, and let $u^{\prime},u^{\prime\prime}$ be the endpoints of
$e^{\prime},e^{\prime\prime}$ which are in $H$. Then
$u^{\prime},u^{\prime\prime}$ are in distinct parts of the unique bipartition
of $V(G)$.

###### Proof.

The oriented degree $\deg(v)$ of a vertex in a digraph is the number of edges
in minus the number of edges out. The sum of all oriented vertex degrees for
any digraph is zero.

Let $(H$, $u^{\prime}$, $u^{\prime\prime})$ be as in the statement; give $G$
and $H$ the edge-orientation induced from the bipartition by orienting the
edges from white to black, so each vertex has oriented degree $\pm k$
according to whether it’s black or white. Put

$W:=V(H\setminus\\{u^{\prime},u^{\prime\prime}\\})=V(H)\setminus\\{u^{\prime},u^{\prime\prime}\\}.$

We have

$0\;\;=\sum_{v\in
V(H)}\deg_{H}(v)\,=\,\deg_{H}(u^{\prime})+\deg_{H}(u^{\prime\prime})+\sum_{v\in
W}\deg_{H}(v).$ (1)

For each $v\in W$, $\deg_{H}(v)=\deg_{G}(v)$, so $\sum_{v\in
W}\deg_{H}(v)\cong 0$ (mod $k$), hence, by (1),
$\deg_{H}(u^{\prime})+\deg_{H}(u^{\prime\prime})\cong 0$. As $k\geq 3$,
$2k-2\ncong 0$ (mod $k$), so the degrees of $u^{\prime}$ and
$u^{\prime\prime}$ must be of opposite sign, i.e., $u^{\prime}$ and
$u^{\prime\prime}$ are in different parts. ∎

By a similar argument, one can prove the following “folklore” result.

######  Proposition 1.

Connected regular bipartite graphs have no cutpoints.

###### Proof.

Let $G$ be a connected $k$-regular bipartite graph. If $k=0$ or $1$, then $G$
is $K_{1}$ or $K_{2}$, resp., and there are no cutpoints. For $k=2$, bipartite
or not, $G$ is a cycle and so has no cutpoint or bridge. Suppose now $k\geq
3$. Let $u$ be a cutpoint of $G$ and let $\Gamma$ be a connected component of
$G-u$.

We define $J:=\Gamma+u:=G(V(\Gamma)\cup\\{u\\})$ to be the smallest subgraph
of $G$ containing $\Gamma$ and $u$. Then again using oriented degrees,

$0\;\;=\sum_{v\in V(J)}\deg_{J}(v)\,=\,\deg_{J}(u)+\sum_{v\in
V(\Gamma)}\deg_{G}(v);$ (2)

the second summand is congruent to zero (mod $k$) but $\deg_{J}(u)\ncong 0$
(mod $k$) so no cutpoint $u$ can exist. ∎

Observe that for $k\geq 2$, there is also no bridge.

##  3 Cubic planar bipartite graphs

Alam-et-al. proved the following result [1, pp. 18–21].

######  Theorem 1 ([1]).

Let $G$ be a bipartite cubic planar $3$-connected graph. Then $G$ has a
subhamiltonian vertex ordering $\omega$ such that $mbt(G)=mbt(G,\omega)=3$.

We say that such a $G$ is subhamiltonian dispersable. It is conjectured in [1]
that $3$-connected is not needed, as we now show.

######  Theorem 2.

Bipartite cubic planar multigraphs are subhamiltonian dispersable.

Our proof of Theorem 2 uses multigraphs. The trivial cubic multigraph, which
we denote by $\Theta$, has 2 vertices joined by 3 parallel edges.

######  Lemma 2.

A cubic multigraph $\neq\Theta$ with parallel edges cannot be $3$-connected.

###### Proof.

If $G$ is cubic and $f_{1},f_{2}$ are parallel edges joining vertices $a$ and
$b$, then either (i) there is a third parallel edge $f_{3}$ and $G$ contains
the trivial cubic graph as a connected component or (ii) the remaining two
edges $f_{3}$ and $f_{4}$ at $a$ and $b$, resp., are a cutset for $G$, as they
separate $\\{a,b\\}$ from the other vertices. ∎

Figure 1 Parallel edges in a cubic graph

Hence, a nontrivial 3-connected cubic multigraph has no parallel edges and is
a cubic graph. We are now ready for the proof of Theorem 2.

###### Proof.

Let $G$ be a bipartite cubic planar multigraph. Assume without loss of
generality ([3], [16]) that $G$ is connected. We need to show that $G$ has a
subhamiltonian vertex order $\omega$ and an edge-3-coloring $c$ such that in
the outerplane drawing $(G,\omega)$, each color class of $c$ is crossing-free;
that is,

$bt(G)=bt(G,\omega)\leq 2\;\;\mbox{{\rm and}}\;\;mbt(G)=mbt(G,\omega)=3.$ (3)

Suppose that $G$ is a counterexample with the minimum number of vertices. Note
that $G$ must be a nontrivial cubic multigraph as the required $\omega$ and
$c$ do exist for $\Theta$, and $G$ cannot be $3$-connected else it is
dispersable by Theorem 1 which applies as there can be no parallel edges.

Since $G$ is bipartite and cubic, by Proposition 1 it has no cutpoint and
hence must be 2-connected. As line-connectivity and point-connectivity
coincide for cubic graphs [9, p. 55, ex. 5.6 ], there must be a cutset of two
edges. But the two edges of such a cutset cannot share a vertex – the third
edge at a common vertex would be a bridge. Hence, $G$ contains vertex-disjoint
edges
$e^{\prime}=u^{\prime}w^{\prime},\;e^{\prime\prime}=u^{\prime\prime}w^{\prime\prime}$
(neither of which is a parallel edge) such that

$G\setminus\\{e^{\prime},e^{\prime\prime}\\}=G_{1}\cup
G_{2},\;\;u^{\prime},u^{\prime\prime}\in
V(G_{1}),\;\;w^{\prime},w^{\prime\prime}\in V(G_{2}),$ (4)

where $G_{1}$ and $G_{2}$ are vertex-disjoint. Further, by Lemma 1,
$u^{\prime}$ and $u^{\prime\prime}$ are in distinct parts of the bipartition
of $G_{1}$ induced by the bipartition of $G$, and similarly for
$w^{\prime},w^{\prime\prime}$ in the corresponding bipartition of $G_{2}$.

Let $H_{1}=G_{1}+e_{1}$, where $e_{1}:=u^{\prime}u^{\prime\prime}$,
$H_{2}=G_{2}+e_{2}$, where $e_{2}:=w^{\prime}w^{\prime\prime}$. Then $H_{1}$
and $H_{2}$ are bipartite cubic planar multigraphs, either of which can
contain multiple edges even if $G$ has no parallel edges. As
$|H_{1}|,|H_{2}|<|G|$, there exist cyclic orderings $\omega_{1},\omega_{2}$ on
$V(H_{1})=V(G_{1})$ and $V(H_{2})=V(G_{2})$, resp., and edge colorings
$c_{1},c_{2}$ which constitute dispersable subhamiltonian book embeddings of
$H_{1}$ and $H_{2}$. Let $\lambda_{1}$ be one of the two possible linear
orderings obtained from $\omega_{1}$ which starts at $u^{\prime\prime}$.
Similarly, let $\lambda_{2}$ be one of the two possible linear orderings
obtained from $\omega_{2}$ which starts at $w^{\prime\prime}$.

Define $\lambda$ a linear order on $V(G)=V(H_{1})\cup V(H_{2})$ by the
equation $\lambda:=\lambda_{1}^{op}\,*\,\lambda_{2}$, where $\lambda_{1}^{op}$
is the opposite (i.e., reverse) order to $\lambda_{1}$ and $*$ denotes
concatenation. Let $\omega$ be the cyclic order determined by $\lambda$.
Observe that $u^{\prime\prime}$ and $w^{\prime\prime}$ are consecutive in the
subhamiltonian ordering of $V(G)$ given by $\omega$.

By renaming the colors, we can assume that the colorings $c_{1}$ and $c_{2}$
both use colors $\\{\alpha,\beta,\gamma\\}$ and that $e_{1}$ and $e_{2}$ are
both assigned $\gamma$. Define $c$ an edge 3-coloring of $E(G)$ to be $c_{1}$
on $E(G_{1})$, $c_{2}$ on $E(G_{2})$, and put
$c(e^{\prime})=c(e^{\prime\prime})=\gamma$. If $e\in E(G_{1})$ is adjacent to
either $e^{\prime}$ or $e^{\prime\prime}$, then $e$ was adjacent to $e_{1}$,
so $c(e)\neq c(e^{\prime})$ and similarly for $G_{2}$. Observe that
$e^{\prime\prime}$ has no crossings in $(G,\omega)$.

Figure 2 Subhamiltonian dispersable embedding of $G$

We claim $e^{\prime}$ crosses no edge colored $\gamma$. For if $e=xy$ crosses
$e^{\prime}$ in the drawing $(G,\omega)$, then $x$ and $y$ both belong to
$G_{1}$ or both belong to $G_{2}$; else $\\{e^{\prime},e^{\prime\prime}\\}$
wouldn’t be a cutset. Thus $xy$ must cross $e_{1}$ or $e_{2}$, hence
$c(xy)\neq\gamma$. So $(\omega,c)$ is a dispersable subhamiltonian book
embedding of $G$ which is impossible. ∎

##  4 Discussion

Since the examples of non-dispersable bipartite graphs given in Alam-et-al.
are nonplanar and not vertex transitive, natural questions arise. They ask [1,
question 6]: Does the BK Conjecture in [3] hold for planar bipartite graphs?

All examples we know support the following conjecture which would include a
positive answer to [1, questions 2 and 4]: Every vertex-transitive graph $G$
with $\Delta(G)=r$ has $mbt(G)=r$ or $\,r{+}1$ according to whether or not $G$
is bipartite. However, some condition on $G$ is needed as Barat et al.[2] show
that if $\Delta(G)\geq 9$, bounded degree graphs cannot have bounded book
thickness.

## References

  * [1] J. Md. Alam, M. A. Bekos, M. Gronemann, M. Kaufmann & S. Pupyrev, On dispersable book embeddings, Graph-theoretic Concepts in Computer Science (44th International Workshop, WG 2018, Cottbus, Germany), A. Brandestädt, E. Köhler, & K. Meer, Eds., LNCS 11159 (2018) 1–14, Springer, Cham, Switzerland.
  * [2] J. Barat, J. Matousek & D. R. Wood, Bounded-degree graphs have arbitrarily large geometric thickness, The Elec. J. of Comb. 13 (2006) #R13.
  * [3] F. R. Bernhart & P. C. Kainen, The book thickness of a graph, J. Comb. Theory, Series B, 27 (1979) 320–331.
  * [4] F. R. K. Chung, F. T. Leighton, & A. L. Rosenberg, Embedding graphs in books: A layout problem with applications to VLSI design, SIAM J. Alg. Discr. Meth. 8 (1) (1987) 33–58.
  * [5] R. Diestel, GraphTheory, Springer, Berlin, 2005.
  * [6] D. Eppstein, Separating geometric thickness from book thickness, 2001.
http://arXiv.org/math/0109195.

  * [7] D. Eppstein, Separating thickness from geometric thickness. In Graph Drawing, (10th International Symp. GD 2002), M. T. Goodrich & S. G. Kobourov, eds., 150–161, Springer, Berlin, 2002. Also in Towards a Theory of Geometric Graphs, Janos Pach, ed., Contemporary Math., 342 (2004) 75–86. Amer. Math. Soc., Providence, RI.
  * [8] S. Even & A. Itai, Queues, stacks, and graphs, in Theory of Machines and Computations, A. Press, Ed., pp. 71–86, 1971.
  * [9] F. Harary, Graph Theory, Addison-Wesley, Reading, MA 1969.
  * [10] C. Haslinger & P. F. Stadler, RNA structures with pseudo-knots: Graph-theoretical, combinatorial, and statistical properties, Bull. of Mathematical Biology 61 (3) (1999) 437–467.
  * [11] P. C. Kainen, Some recent results in topological graph theory, in Graphs and Combinatorics, R. A. Bari & F. Harary, Eds., Springer Lect. Notes in Math. 406, Berlin, 1974, pp. 76–108.
  * [12] P. C. Kainen, The book thickness of a graph, II, Congr. Numeratium 71 (1990), 127–132.
  * [13] P. C. Kainen, Circular layouts for crossing-free matchings, preprint 2009. http://faculty.georgetown.edu/kainen/circLayouts.pdf
  * [14] P. C. Kainen, S. Overbay, On matching book thickness, in preparation.
  * [15] L. T. Ollmann, On the book thicknesses of various graphs, Congressus Numerantium VIII (1973) 459 (Proc. 4th Southeastern Conference on Combinatorics, Graph Theory and Computing, F. Hoffman, R. B. Levow, R. S. D. Thomas, Eds.
  * [16] S. Overbay, Generalized Book Embeddings, Ph. D. Dissertation, Colorado State University, Fort Collins, CO, 1998.
  * [17] R. E. Tarjan, Sorting using networks of queues and stacks, J. Assoc. of Computing Machinery 19 (1972) 341–346.
  * [18] W. T. Tutte, Connectivity in Graphs, U. of Toronto Press, London, 1966.
  * [19] Wikipedia, Book Embedding, accessed 18 July 2020
$https://en.wikipedia.org/wiki/Book\\_embedding$

  * [20] D. R. Wood, Bounded degree book embeddings and three-dimensional orthogonal graph drawing, Graph Drawing: 9th International Symposium, P. Mutzel, M. Jünger, & S. Leipert, Eds., LNCS, 2265 (2002) 312–327 Springer, Berlin.

Paul C. Kainen, Department of Mathematics and Statistics,
Georgetown University, 37th and O Streets, N.W., Washington DC 20057

Shannon Overbay, Department of Mathematics, Gonzaga University,
502 E. Boone Ave., Spokane WA 99258

### Postscript

The previous paper [2] proves the conjecture of Alam et al. [1] made in their
2018 conference paper which we discovered in 2019. We worked during the Spring
of 2020 and on July 20, 2020, our paper was submitted for publication in the
form above. After 8 months, on March 26, 2021, we contacted the assistant to
the managing editor of the journal and were told that no referee had been
found. We then supplied names, affiliations, and emails for several (we
thought obvious) choices for possible referees, which were accepted with
thanks. On May 30, 2021, the journal rejected the paper. The referee’s report
said that the problem had been solved and included a link to the now-published
expanded version [3] of Alam et al.’s results.

Indeed, on March 3, 2020, Alam, et al. first submitted a version of the
conference paper [1] to the Elsevier journal, Theoretical Computer Science and
submitted the second version [3] on Jan. 23, 2021, and it appeared
electronically on Feb. 4, 2021. This new journal version [3], with a sixth
coauthor, Dujmović, includes a proof of the conjecture in [1].

Another proof of the Alam et al. conjecture was recently given by Shao, Liu,
and Li [4]. The paper [4] uses a different method, while [2] and the portion
of [3], which proves the conjecture, use very similar techniques.

We hope the comparison of these approaches will be helpful.

References

[1] J. Md. Alam, M. A. Bekos, M. Gronemann, M. Kaufmann & S. Pupyrev, On
dispersable book embeddings, Graph-theoretic Concepts in Computer Science
(44th International Workshop, WG 2018, Cottbus, Germany), A. Brandestädt, E.
Köhler, & K. Meer, Eds., LNCS 11159 (2018) 1–14, Springer, Cham, Switzerland.

[2] P. C. Kainen & S. Overbay, Cubic planar bipartite graphs are dispersable,
July 20, 2020 (7 pages).

[3] J. Md. Alam, M. A. Bekos, V. Dujmović, M. Gronemann, M. Kaufmann & S.
Pupyrev, On dispersable book embeddings, Theoretical Computer Science, 861
(2021), 1–22.

[4] Z. Shao, Y. Liu & Z. Li, Bipartite cubic planar graphs are dispersable,
arXiv:2107.00907v1, July 2, 2021, (10 pages).


# BNSR-Invariants of Surface Houghton Groups

Noah Torgerson and Jeremy West

(March 7, 2024)

###### Abstract

The surface Houghton groups $\mathcal{H}_{n}$ are a family of groups
generalizing Houghton groups $H_{n}$, which are constructed as asymptotically
rigid mapping class groups. We give a complete computation of the BNSR-
invariants $\Sigma^{m}(P\mathcal{H}_{n})$ of their intersection with the pure
mapping class group. To do so, we prove that the associated Stein–Farley cube
complex is CAT(0), and we adapt Zaremsky’s method for computing the BNSR-
invariants of the Houghton groups. As a consequence, we give a criterion for
when subgroups of $H_{n}$ and $P\mathcal{H}_{n}$ having the same finiteness
length as their parent group are finite index. We also discuss the failure of
some of these groups to be co-Hopfian.

##  1 Introduction

To any group $G$, one can assign a sequence of invariants111Technically,
$\Sigma^{k}$ is only defined for groups of type $F_{k}$, but one could simply
interpret this as empty when $G$ faisl to be of type $F_{k}$. $\Sigma^{k}(G)$
These invariants determine which subgroups of $G$ (containing the commutator)
share which finiteness properties of $G$. Historically, they are difficult to
compute. They were defined across several papers (see [BNS87], [BR88], and
[Ren88]), primarily by Bieri, Neumann, Strebel, and Renz, hence the names
“BNS-invariants” and “BNSR-invariants”. For the remainder of this paper, these
will be referred to as $\Sigma$-invariants for the sake of brevity. One recent
collection of groups for which these have been computed is the family of
Houghton groups, in [Zar16] and [Zar20].
Houghton defined his groups as permutation groups of infinite sets in [Hou78].
More specifically, $H_{n}$ is the group of “eventual translations” of the set
$\\{1,\dots,n\\}\times\mathbb{N}$, i.e. permutations which in each ray are
translations outside some finite set. Brown proved, via a suitable simplicial
complex, that Houghton’s group $H_{n}$ is of type $F_{n-1}$, but not of type
$F_{n}$ (in the language of “finiteness length” this is
$fl(H_{n})=n-1$)222Actually, Brown proves type $FP_{n-1}$ but not type
$FP_{n}$, and finitely presented, which together imply $F_{n}$ but not
$F_{n-1}$.. In [ABKL23], the authors define a variant of Houghton groups,
called surface Houghton groups, as asymptotically rigid mapping class groups
of surfaces with infinite genus. We denote the surface Houghton groups by
$\mathcal{H}_{n}$, and the pure (i.e. end-fixing) subgroups as
$P\mathcal{H}_{n}$. In the same paper, they also show that
$fl(\mathcal{H}_{n})=n-1$, via a cube complex analogous to the Stein–Farley
complex.
It is worth noting that this is not the only asymptotically rigid mapping
class group variant of Houghton groups. While the surface Houghton groups
replace the $\mathbb{N}$-rays with ends accumulated by genus (without
boundary), the braided Houghton groups (defined by Degenhardt in his PhD
thesis [Deg00]) can be realized as asymptotically rigid mapping class groups
as well (see [Fun07] for details). This construction replaces the
$\mathbb{N}$-rays with planar ends, accumulated by punctures; note that this
surface has non-compact boundary. These braided Houghton groups
$\operatorname{Br}H_{n}$ also have $fl(\operatorname{Br}H_{n})=n-1$ (as proven
in [GLU20]). The braided Houghton groups shall play a small role in the final
section of this paper.
In [Zar16] and [Zar20], Zaremsky computed the $\Sigma$-invariants of the
Houghton groups, using the cube complex defined implicitly in [Bro87], and
explicitly in [Lee12]. We carry out a parallel of Zaremsky’s arguments,
showing that the equivalent statement holds for pure surface Houghton groups.
Namely, we prove in Section 4 the following, where $m(\chi)$ is the number of
non-zero coefficients of $\chi$ in ascending standard form (details in Section
4):

######  Theorem 1.1.

Let $\chi$ be a non-zero character of $P\mathcal{H}_{n}$. Then
$[\chi]\in\Sigma^{m(\chi)-1}(P\mathcal{H}_{n})\setminus\Sigma^{m(\chi)}(P\mathcal{H}_{n})$.

As an application of this, we provide a partial converse (for Houghton groups
and Surface Houghton groups) of the well-known fact that if $H\leq G$ is a
finite index subgroup, then $fl(H)=fl(G)$. Specifically, we show that whenever
a subgroup of $H_{n}$ or $P\mathcal{H}_{n}$ has finiteness length $n-1$ and
intersects the corresponding commutator with finite index, it is finite index
in the full group. Further, in order to be finite index, the intersection
condition must hold, for general group theoretic reasons. (These are also true
of $\mathcal{H}_{n}$, but for trivial reasons: the commutator of
$\mathcal{H}_{n}$ is all of $\mathcal{H}_{n}$.) This is carried out in Section
5, along with some discussion of co-Hopfianness.
In order to accomplish this, we demonstrate in Section 3 that the cube complex
of [ABKL23] is CAT(0).

######  Theorem 1.2.

The Stein–Farley complex for $\mathcal{H}_{n}$ is CAT(0).

To do this, we use a refinement of a proposition from [GLU20], see Proposition
3.2. In Section 2, we lay out various definitions and lemmas that we need, of
the surfaces and groups we are concerned with, of the corresponding cube
complex, and of the version of discrete Morse theory we shall employ. We also
obtain various nice representatives for the vertices (Section 2) and edges
(Section 3) of the complex.
In Section 5, we finish with some applications of the $\Sigma$-invariants to
when subgroups having maximal finiteness length have finite index. This
discussion applies to $H_{n}$, $\mathcal{H}_{n}$, and $P\mathcal{H}_{n}$. To
handle the infinite index case, there is some discussion of the failure of
these groups to be co-Hopfian. In particular, we have the following theorem.

######  Theorem 1.3.

Let $H$ denote either the Houghton group, or the pure surface Houghton group,
and suppose $G<H$ has $fl(G)=fl(H)$. Then $G$ is finite index in $H$ if and
only if $G\cap H^{\prime}$ is finite index in $H^{\prime}$, where $H^{\prime}$
denotes the commutator subgroup of $H$. Furthermore, there exist subgroups $G$
with $fl(G)=fl(H)$ of both finite and infinite index.

A recent paper by Marie Abadie [Aba24] analyses the CAT(0) property for the
Stein–Farley cube-complexes of a different family of asymptotically rigid
mapping class groups, including the braided Houghton groups. Similar methods
are used, in particular a version of the same proposition as we use. Thus, one
could attempt to apply Zaremsky’s methods to compute the $\Sigma$-invariants
of $\operatorname{Br}H_{n}$ as well. It seems reasonable to guess that they
should work out the same.
The recent paper [ADL23] concerns a generalization of these surface Houghton
groups, obtained by varying the rigid structure on the same surface. As they
show that the groups they consider are finite index subgroups of our
$\mathcal{H}_{n}$, the $\Sigma$-invariants are effectively the same.
Additionally, our results in Section 5 also apply to these more general
surface Houghton groups.

## Acknowledgements

We would like to thank Jing Tao for advice and guidance on this project. We
would also like to thank Noel Brady and Justin Malestein for valuable
conversations, and Peter Patzt and Matt Zaremsky for helpful comments. The
authors were partially supported by the OU Bridge Funding Investment Program.

##  2 Definitions

###  2.1 The (Pure) Surface Houghton Group

Here we lay out the definitions necessary for the group $P\mathcal{H}_{n}$,
the pure version of the Surface Houghton Group defined in [ABKL23]. Let
$\mathcal{O}=\mathcal{O}_{n}$ be a sphere with $n$ boundary components, and
let $T$ be a torus with two boundary components, $\partial^{-}$ and
$\partial^{+}$. Fix an orientation-reversing homeomorphism
$\lambda:\partial^{-}\rightarrow\partial^{+}$, and for each $i$ an
orientation-reversing homeomorphism $\mu_{i}$ from $\partial^{-}$ to the $i$th
boundary components of $\mathcal{O}$. We construct $\Sigma_{n}$ as follows:
begin with $M^{1}=\mathcal{O}$, then glue a copy of $T$ to each boundary
component of $M^{1}$ via the $\mu_{i}$ to obtain $M^{2}$. For each $j\geq 2$,
glue a copy of $T$ to each boundary component of $M^{j}$ via $\lambda$ to
obtain $M^{j+1}$. Then the surface $\Sigma_{n}$ is the union of all the
$M^{j}$’s; $\Sigma_{n}$ is the surface with $n$ ends, all accumulated by
genus. Each of the closures of the components of $M^{j}\setminus M^{j-1}$ is
called a piece, and each piece $B$ has a canonical homeomorphism
$\iota_{B}:B\rightarrow T$. We will occasionally write $B_{k}^{j}$ to denote
the $j$th piece in the $k$th end.
Call a subsurface of $\Sigma_{n}$ suited if it is connected and the union of
$\mathcal{O}$ and finitely many pieces. Let
$\varphi:\Sigma_{n}\rightarrow\Sigma_{n}$ be a homeomorphism. Call $\varphi$
asymptotically rigid if there exists a suited subsurface $Z\subset\Sigma_{n}$
(called a defining surface for $\varphi$) such that

  * $\bullet$

$\varphi(Z)$ is also suited, and

  * $\bullet$

$\varphi$ is rigid away from $Z$, that is, for every piece
$B\subset\overline{\Sigma_{n}\setminus Z}$, we have that $\varphi(B)$ is a
piece, and $\varphi|_{B}=\iota_{\varphi(B)}^{-1}\circ\iota_{B}$.

We may sometimes say that $\varphi$ is rigid away from a piece $B$ adjacent to
$\mathcal{O}$, which we take to mean that $\varphi$ is rigid away from
$\mathcal{O}\cup B$. A special family of suited subsurfaces are those with
pieces in only one end. We denote by $L_{g}$ the suited subsurface consisting
of $\mathcal{O}$ and the first $g$-many pieces in the $n$th end.

######  Definition 2.1.

(Surface Houghton Group) The Surface Houghton Group $\mathcal{H}_{n}$ is the
subgroup of the mapping class group $\operatorname{Map}(\Sigma_{n})$ whose
elements have an asymptotically rigid representative. The Pure Surface
Houghton Group $P\mathcal{H}_{n}$ is the intersection of $\mathcal{H}_{n}$
with the pure mapping class group $\operatorname{PMap}(\Sigma_{n})$.

We define some special elements of $\mathcal{H}_{n}$. For $\sigma$ a
permutation of $\\{1,\dots,n\\}$, choose some homeomorphism of $\mathcal{O}$
permuting the boundary components in the same fashion. For the sake of
definiteness, consider $\sigma$ first as an element of
$\operatorname{Map}(S_{0,n})$, the sphere with $n$ punctures; then $\sigma$
yields a homeomorphism of $\mathcal{O}$, which we also dentoe by $\sigma$. Up
to some isotopy, we can assume that the restriction of $\sigma$ to the $i$th
boundary component of $\mathcal{O}$ is $\mu_{\sigma(i)}\circ\mu_{i}^{-1}$.
Thus, we can extend this homeomorphism to be rigid outside of $\mathcal{O}$;
call this extension again by the same name, $\sigma$. Note that such a map is
not unique: there are many choices of mapping class in the sphere with $n$
punctures which induce the same permutation.
Next, we define the handle shifts $\rho_{i}$, $\grave{a}$ $la$ [PV18]. For
$i\in\\{1,\dots,n-1\\}$, we want $\rho_{i}$ to be a homeomorphism of
$\Sigma_{n}$ which shifts the $i$th end towards the $n$th end by a single
genus. Specifically, map each piece $B_{n}^{j}$ to $B_{n}^{j+1}$, and for
$j\geq 1$ map each piece $B_{i}^{j+1}$ to $B_{i}^{j}$; choose these to be the
rigid maps. The ends other than $i$ and $n$ are unchanged, so all that remains
is to define how $\rho_{i}$ acts on $B_{i}^{1}\cup\mathcal{O}$. This is
demonstrated by figure 2.1 Note that $\rho_{i}$ is asymptotically rigid
outside of $\mathcal{O}\cup B_{i}^{1}$.

Figure 1: A handle shift being applied to a select curve.

######  Lemma 2.2.

Any map $\varphi\in\mathcal{H}_{n}$ can be written as
$\alpha\rho_{i_{1}}\cdots\rho_{i_{j}}\sigma$, where $\alpha$ is a compactly
supported mapping class, and $\sigma$ and the $\rho_{i_{k}}$’s are as above.

###### Proof.

Consider an arbitrary element $\varphi\in\mathcal{H}_{n}$: $\varphi$ clearly
induces some permutation $\sigma$ on the ends of $\Sigma_{n}$; obtain from
this $\sigma\in\mathcal{H}_{n}$. Now we have that $\varphi\circ\sigma^{-1}\in
P\mathcal{H}_{n}$, so we wish to analyse how pure mapping classes act within
ends. A rigorous examination of this is contained in Section 2.3. In the
meantime, consider how $\varphi$ acts in the complement of a defining surface:
as it does not permute the ends, and takes pieces to pieces, it must shift
each end in or out by some integer amount. From this, we obtain a sequence of
handle shifts $\rho_{i_{1}},\dots,\rho_{i_{k}}$. Finally, we have that
$\varphi\circ\sigma^{-1}\circ(\rho_{i_{1}}\cdots\rho_{i_{k}})^{-1}$ is a
compactly supported mapping class, $\alpha$. ∎

######  Remark 2.3.

The above expression obtained for $\varphi$ is far from unique: choices were
made both in the selection of $\sigma$, as well as in the selection of the
handle shifts. However, any different choices made for these will simply
change $\alpha$, and the properties of $\alpha$ are rarely (never?) relevant
for this paper.

###  2.2 The Contractible Cube Complex

The group $\mathcal{H}_{n}$ acts on a contractible cube complex $X_{n}$,
called the Stein–Farley complex. Consider ordered pairs $(Z,\varphi)$, where
$Z$ is a suited subsurface, and $\varphi\in\mathcal{H}_{n}$. Declare two such
pairs $(Z_{1},\varphi_{1})$ and $(Z_{2},\varphi_{2})$ to be equivalent if the
transition map $\varphi_{2}^{-1}\circ\varphi_{1}$ takes $Z_{1}$ to $Z_{2}$
homeomorphically, and is rigid elsewhere (i.e. in the complement of $Z_{1}$).
Denote the equivalence class of $(Z,\varphi)$ by $[Z,\varphi]$, and the set of
equivalence classes by $\SS$. The group $\mathcal{H}_{n}$ acts on $\SS$ by

$\psi\cdot[Z,\varphi]=[Z,\psi\circ\varphi].$

Define the complexity of a pair $(Z,\varphi)$ to be the genus of $Z$. This
extends to be a $\mathcal{H}_{n}$-invariant height function
$f\colon\SS\rightarrow\mathbb{Z}\subset\mathbb{R}$. Given vertices
$x_{1},x_{2}\in\SS$, we say that $x_{1}\prec x_{2}$ if there are
representatives $(Z_{i},\varphi_{i})$ of $x_{i}$ so that
$\varphi_{1}=\varphi_{2}$, $Z_{1}\subset Z_{2}$, and $\overline{Z_{2}\setminus
Z_{1}}$ is a disjoint union of pieces.
We now construct $X_{n}$ as a cube complex with vertex set $\SS$. We declare
that there is a $d$-cube whenever $x_{1}\prec x_{2}$, with
$d=f(x_{2})-f(x_{1})$. Note that as $\Sigma_{n}$ has $n$ ends, the complex
$X_{n}$ is $n$ dimensional. The height function $f$ extends affinely to a
$\mathcal{H}_{n}$-invariant height function on $X_{n}$. Also, let
$X_{n}^{f\leq k}$ denote the subcomplex spanned by vertices with height at
most $k$. We have the following properties (see [ABKL23, Theorem 4.1]):

######  Theorem 2.4.

The cube complex $X_{n}$ is contractible, and the action of $\mathcal{H}_{n}$
satisfies:

  * $\bullet$

The $\mathcal{H}_{n}$-stabilizers of cubes are finite extensions of mapping
class groups of compact surfaces.

  * $\bullet$

For $k\geq 1$, the subcomplex $X_{n}^{f\leq k}$ is
$\mathcal{H}_{n}$-cocompact.

We are actually more concerned with the action of $P\mathcal{H}_{n}$ on
$X_{n}$. Note that $P\mathcal{H}_{n}$ is a finite index subgroup of
$\mathcal{H}_{n}$, and that the finite extension in the above theorem is by
permutations of the ends. As mapping class groups of compact surfaces are of
type $F_{\infty}$, we have the following.

######  Corollary 2.5.

The action of $P\mathcal{H}_{n}$ on $X_{n}$ satisfies:

  * $\bullet$

The $P\mathcal{H}_{n}$-stabilizers of cubes are type $F_{\infty}$.

  * $\bullet$

For $k\geq 1$, the subcomplex $X_{n}^{f\leq k}$ is
$P\mathcal{H}_{n}$-cocompact.

We end this section by demonstrating that vertices and edges can be given a
nice form. First, we show that every vertex (indeed, every edge) can be
expressed with only elements of $P\mathcal{H}_{n}$, then we provide a
canonical form for the suited subsurface component of the vertex.
Specifically, we show that any edge can be represented using only elements in
the pure group:

######  Lemma 2.6.

Any edge in $X_{n}$ can be expressed as $[Z,\varphi]$—$[Z\cup B,\varphi]$,
where $\varphi\in P\mathcal{H}_{n}$.

###### Proof.

Consider a vertex $[Z,\varphi]$, and let $B$ be a piece adjacent to $Z$, so
that $Z\cup B$ is suited. Write
$\varphi=\alpha\rho_{i_{1}}\cdots\rho_{i_{j}}\sigma$ as in Lemma 2.2. Then we
wish to show that $[Z,\varphi]=[\sigma(Z),\varphi\circ\sigma^{-1}]$, and that
$[Z\cup B,\varphi]=[\sigma(Z)\cup\sigma(B),\varphi\circ\sigma^{-1}]$. Consider
the following diagram.

$[\sigma(Z),\varphi\circ\sigma^{-1}]$$[\sigma(Z\cup
B),\varphi\circ\sigma^{-1}]$$[\sigma(Z)\cup\sigma(B),\varphi\circ\sigma^{-1}]$$[Z,\varphi]$$[Z\cup
B,\varphi]$

As $\sigma$ is rigid away from the center, we have that $\sigma(B)$ is a
piece, and of course it must be adjacent to $\sigma(Z)$, so all that remains
to be checked is that $[Z,\varphi]=[\sigma(Z),\varphi\circ\sigma^{-1}]$.
First, note that $(\varphi\circ\sigma^{-1})^{-1}\circ\varphi=\sigma$, which is
rigid away from $\mathcal{O}$, and thus takes $Z$ to $\sigma(Z)$ (a suited
subsurface), and is certainly rigid outside of $Z$. ∎

We assume from here on that any map $\varphi$ is in $P\mathcal{H}_{n}$.
Vertices have two components, a map and a surface. We now have some control
over the map, but what of the surface? For many arguments, it will be
convenient to have the following canonical form. Recall that $L_{g}$ is the
surface built by taking the center, and adding in the first $g$-many pieces in
the $n$th end. Then any edge can be represented with both surfaces an
appropriate $L_{g}$:

######  Lemma 2.7.

Any edge in $X_{n}$ can be expressed as $[L_{g},\varphi]$—$[L_{g+1},\psi]$.

###### Proof.

Suppose we are given a vertex $[Z,\varphi^{\prime}]$. Recall that the handle
shift $\rho_{i}$ is rigid away from $B_{i}^{1}$, the piece that it pushes
across the center. As $(\varphi)^{-1}\circ\varphi^{\prime}$ can be chosen to
be a composition of handle shifts taking $Z$ to $L_{g}$, the only places where
$(\varphi)^{-1}\circ\varphi^{\prime}$ can be non-rigid are pieces in $Z$. This
shows that vertices can be represented as $[L_{g},\varphi]$. To extend this to
the edge $[Z,\varphi^{\prime}]$ —$[Z\cup B,\varphi^{\prime}]$, consider the
following diagram.

$[\sigma(Z),\varphi\circ\sigma^{-1}]$$[\sigma(Z\cup
B),\varphi\circ\sigma^{-1}]$$[\sigma(Z)\cup\sigma(B),\varphi\circ\sigma^{-1}]$$[L_{g},\varphi]$$[L_{g}\cup
B_{i}^{1},\varphi]$$[L_{g+1},\varphi\circ\rho_{i}^{-1}]$

Observe that to obtain an $L_{g+1}$ form for $[Z\cup B,\varphi^{\prime}]$, we
can first push $Z$ to $L_{g}$, which sends $B$ to a piece $B_{i}^{1}$, and
then append one more $\rho_{i}$. ∎

###  2.3 Characters and $\Sigma$-Invariants

Recall that a group is of type Fm if it admits a proper cocompact action on an
$(m-1)$-connected CW-complex. Assume $Y$ is a CW complex and
$h:Y\rightarrow\mathbb{R}$ is continuous. We call the corresponding filtration
$\left(Y^{t\leq h}\right)_{t\in\mathbb{R}}$ on $Y$ essentially
($m-1$)-connected if for any $t\in\mathbb{R}$, there exists $s\leq t$ such
that inclusion $Y^{t\leq h}\hookrightarrow Y^{s\leq h}$ induces the trivial
map in $\pi_{k}$ for $k\leq m-1$.
For any group $G$, we define a character of $G$ to be a homomorphism from $G$
into the additive group $\mathbb{R}$. As characters factor through $G^{ab}$
(the abelianization of $G$), the space of characters is a vector space with
the same dimension as the rank of $G^{ab}$. Excluding the trivial character
and modding out by positive scaling, we obtain the Character Sphere
$\Sigma(G)$. The $\Sigma$-invariants (also called BNS or BNSR invariants, for
Bieri-Neumann-Strebel(-Renz)) are a filtration of the character sphere into
subspaces

$\Sigma^{0}(G)\supseteq\Sigma^{1}(G)\supseteq\Sigma^{2}(G)\supseteq\cdots,$

defined as follows (we use a slight correction of the definition in [Zar16],
as suggested in [Zar20]).

######  Definition 2.8.

Let $G$ be a group of type $F_{m}$, and let $Y$ be an $(m-1)$-connected CW
complex on which $G$ acts cocompactly. Suppose that the stabilizer of any
$k$-cell is of type $F_{m-k}$ and is contained in the kernel of every
character of $G$.333That the cell-stabilizers are contained in the kernels is
the modification. For each non-trivial
$\chi\in\operatorname{Hom}(G,\mathbb{R})$, there is a character height
function $h_{\chi}:Y\rightarrow\mathbb{R}$, a continuous map satisfying
$h_{\chi}(gy)=\chi(g)+h_{\chi}(y)$ for all $y\in Y$ and $g\in G$. Then
$\Sigma^{m}(G)$ is the set of those $[\chi]$ such that the filtration
$(Y^{t\leq h_{\chi}})_{t\in\mathbb{R}}$ is essentially $(m-1)$-connected.

Of what use are these invariants, one might ask? A standard result is the
following classification of when finiteness properties of $G$ are preserved to
subgroups containing the commutator.

######  Proposition 2.9.

Let $G$ be a group of type $F_{m}$, and let $K$ be a subgroup so that
$[G,G]\leq K\leq G$. Then $K$ is of type $F_{m}$ if and only if for every
character $\chi\in\operatorname{Hom}(G,\mathbb{R})$ such that $\chi(K)=0$, we
have $[\chi]\in\Sigma^{m}(G)$.

######  Definition 2.10.

A (Zaremsky-)Morse function on an affine cell complex $Y$ is a map
$h=(\chi,f):Y\rightarrow\mathbb{R}\times\mathbb{R}$ such that both $\chi$ and
$f$ are affine on cells. The codomain is ordered lexicographically, and we
require that $f$ take only finitely many values on $Y^{(0)}$, and that there
is some $\varepsilon>0$ such that adjacent vertices $v$ and $w$ satisfy either
$|\chi(v)-\chi(w)|\geq\varepsilon$, or $\chi(v)=\chi(w)$ and $f(v)\neq f(w)$.

We record the following general form of the Morse Lemma, as well as a more
specific version of it. Both will be used in Section 4.

######  Lemma 2.11.

Let $-\infty\leq p\leq q\leq r\leq+\infty$. If for every vertex $v\in
Y^{q<\chi\leq r}$ the descending link
$\operatorname{lk}_{Y^{p\leq\chi}}^{h\downarrow}(v)$ is $(k-1)$-connected,
then the pair $(Y^{p\leq\chi\leq r},Y^{p\leq\chi\leq q})$ is $k$-connected. If
for every vertex $v\in Y^{p\leq\chi<q}$ the ascending link
$\operatorname{lk}_{Y^{\chi\leq r}}^{h\uparrow}(v)$ is $(k-1)$-connected, then
the pair $(Y^{p\leq\chi\leq r},Y^{q\leq\chi\leq r})$ is $k$-connected.

######  Corollary 2.12.

Let $h=(\chi,f):Y\rightarrow\mathbb{R}\times\mathbb{R}$ be a Morse function.
If $Y$ is $(m-1)$-connected and for every vertex $v\in Y^{\chi<q}$ the
ascending link $\operatorname{lk}_{Y}^{h\uparrow}(v)$ is $(m-1)$-connected,
then $Y^{q\leq\chi}$ is $(m-1)$-connected.

Informally, we can think of the characters on $P\mathcal{H}_{n}$ as being
generated by counting the handle shifts $\rho_{1},\dots,\rho_{n-1}$, where
$\rho_{i}$ shifts the $i$th end into the $n$th end by one piece. Specifically,
for $1\leq i\leq n-1$, take $\chi_{i}(\varphi)$ to be the negative of the sum
of the powers of $\rho_{i}$ appearing in $\varphi$; for $i=n$, take $\chi_{n}$
to be the sum of the powers of $\rho_{1}$ through $\rho_{n-1}$. We stress that
this definition, while convenient to use, is not obviously well-defined. An
equivalent definition, which is better suited to demonstrating well-
definedness, is as follows (details can be found in Section 3 of [APV17]).
Let $\gamma$ be an oriented curve that separates one end $E$ of $\Sigma_{n}$
from the rest, oriented so that the end $E$ is on the righthand side of
$\gamma$. Then $\gamma$ defines a non-zero element of
$H_{1}^{\operatorname{sep}}(\Sigma_{n},\mathbb{Z})$. To every
$\varphi\in\operatorname{PMap}(\Sigma_{n})$ and $\gamma\in
H_{1}^{\operatorname{sep}}(\Sigma_{n},\mathbb{Z})$, associate an integer
$\theta_{[\gamma]}(\varphi)$, as a “signed genus” between $\gamma$ and
$\varphi(\gamma)$. Then the map
$\theta_{[\gamma]}:\operatorname{PMap}(\Sigma_{n})\rightarrow\mathbb{Z}$ is a
well-defined nontrivial homomorphism, depending only on the homology class of
$\gamma$. By identifying $H_{\operatorname{sep}}^{1}$ with
$\operatorname{Hom}(H_{1}^{\operatorname{sep}},\mathbb{Z})$ via the Universal
Coefficients Theorem, we obtain a map
$\Theta:\operatorname{PMap}(\Sigma_{n})\rightarrow
H_{\operatorname{sep}}^{1}(\Sigma_{n},\mathbb{Z})$, by the rule
$\Theta(\varphi)[\gamma]=\theta_{[\gamma]}(\varphi)$. Restricting to
$P\mathcal{H}_{n}$, we obtain characters, and it is not difficult to see that
they agree with the informal definition above. We emphasize: $\chi_{i}$
measures how much $\varphi$ pushes the $i$th end out.
Given a vertex, we can define the character height function
$h_{\chi_{i}}([Z,\varphi])$ by adding the number of pieces of $Z$ in the $i$th
end to $\chi_{i}(\varphi)$. This is well defined: consider two representatives
$[Z,\varphi]=[W,\psi]$. We have that $\varphi^{-1}\psi$ takes $W$ to $Z$, and
is rigid elsewhere. This composition can be further decomposed as a mapping
class of some suited subsurface composed with a sequence of handle shifts.
This compactly supported mapping class affects neither the distribution of the
pieces of $Z$ nor the value of any character on $\varphi$, so we need consider
only handle shifts. Consider
$[Z,\varphi]=[\rho_{i}(Z),\varphi\circ\rho_{i}^{-1}]$: we have moved one piece
from the $i$th end to the $n$th end, while increasing the amount of pushing
into the $i$th end and out of the $n$th end by one. These cancel out in
$h_{\chi_{i}}$ and $h_{\chi_{n}}$. All other basis characters are unchanged
under these operations, and thus we have well-defined character height
functions on $X_{n}$. For an arbitrary character $\chi$, we shall henceforth
abuse notation by writing $\chi$ when we mean $h_{\chi}$. The domain will
typically be clear.

##  3 The Stein–Farley complex $X_{n}$ is CAT(0)

We have a contractible cube complex on which $P\mathcal{H}_{n}$ acts nicely.
In this section, we show that $X$ is in fact CAT(0). To begin, we require the
following definition.

######  Definition 3.1.

A cube complex $X$ is cube-complete if whenever $X^{(1)}$ contains an embedded
copy of the 1-skeleton of a $d$-cube, $X$ contains the entire $d$-cube.

Once we have shown that $X_{n}$ is cube-complete, we seek to apply the
following proposition, which is a version of [GLU20, Proposition 4.6], where
it is taken out of the proof of [Che00, Theorem 6.1].

######  Proposition 3.2.

Let $X$ be a cube-complete cube complex, whose 1-skeleton $X^{(1)}$ is a graph
with no loops or bigons. Suppose that

  1. (a)

$X$ is simply connected,

  2. (b)

$X^{(1)}$ satisfies the $3$-square condition.

Then $X$ is CAT(0).

The $3$-square condition says that whenever the cube complex has 3 squares
intersecting in a vertex, and pairwise intersecting in edges (call such an
arrangement a 3-wheel), then they are part of a full $3$-cube.

######  Remark 3.3.

In [GLU20], there is a third condition: that the 1 skeleton not contain a copy
of the complete bipartite graph $K_{2,3}$. This third condition is redundant:
A careful examination of the proof of Theorem 6.1 in [Che00], especially the
paragraph beginning “To prove the converse”, yields that the other hypotheses
imply the non-existence of $K_{2,3}$’s. Another, smaller modification: there
was a condition that the 1-skeleton be a connected graph, which is
automatically true in a simply connected cube complex.

As we have a height function for $X_{n}$, and as adjacent vertices always
differ by exactly 1, there are two ways in which $X_{n}$ can fail to be cube-
complete.

######  Definition 3.4.

By a collapsed cube, we mean the 1-skeleton of a $d$-cube, such that the
difference between its highest and lowest heights is less than $d$. By an
empty cube, we mean the 1-skeleton of a $d$-cube that is not filled by a
$d$-cube.

Given the definition of $X_{n}$, we need to check two things: that
$X_{n}^{(1)}$ has no collapsed cubes, and that $X_{n}^{(1)}$ has no empty
cubes.

Choosing vertex representatives to have their homeomorphism components
contained in the pure group allows us a consistent definition of “direction”
for adding and removing pieces, i.e. we may consistently label the ends from
$1$ to $n$ without worrying about any potential re-indexing.
In order to show that $X_{n}$ is cube-complete, we shall proceed as follows:
first, we show that given a vertex $[Z,\varphi]$, there is only one ascending
edge per direction (see Lemma 3.5). Secondly, we show that there are no
collapsed squares. This easily gives that there are no collapsed cubes of any
dimension. Finally, we show that there are no empty squares, and from this
induct to show that there are no empty cubes.

######  Lemma 3.5.

(Uniqueness of Ascending Edges) Given a vertex $v=[Z,\varphi]$, there exists
exactly one ascending edge in each direction, i.e. for $i\in\\{1,\dots,n\\}$,
the collection $\\{[Z\cup B_{i},\varphi]\\}$, where $B_{i}$ is in the $i$th
end, exhausts the ascending link of $v$.

###### Proof.

What we wish to show is that, given two representatives
$[Z,\varphi]=[Z^{\prime},\varphi^{\prime}]$, whenever we ascend from
$[Z^{\prime},\varphi^{\prime}]$ by adding a piece in the $i$th end, we obtain
the same vertex as by adding a piece in the $i$th end to $[Z,\varphi]$.
Consider the following diagram:

$[\sigma(Z),\varphi\circ\sigma^{-1}]$$[\sigma(Z\cup
B),\varphi\circ\sigma^{-1}]$$[\sigma(Z)\cup\sigma(B),\varphi\circ\sigma^{-1}]$$[L_{g},\varphi]$$[L_{g}\cup
B_{i}^{1},\varphi]$$[L_{g+1},\varphi\circ\rho_{i}^{-1}]$$[Z,\varphi]$$[Z^{\prime},\varphi^{\prime}]$$[Z\cup
B,\varphi]$$[Z^{\prime}\cup B^{\prime},\varphi^{\prime}]$?

Our goal is to show that, assuming $B^{\prime}$ and $B$ are both in the $i$th
end, we have $[Z^{\prime}\cup B^{\prime},\varphi^{\prime}]=[Z\cup B,\varphi]$.
As $(\varphi^{\prime})^{-1}\circ\varphi$ is rigid outside $Z$, it takes $B$ to
some piece adjacent to $Z^{\prime}$. If this is $B^{\prime}$, then we have
that $(\varphi^{\prime})^{-1}\circ\varphi$ takes $Z\cup B$ to $Z^{\prime}\cup
B^{\prime}$, and is rigid elsewhere, i.e. that $[Z\cup
B,\varphi]=[Z^{\prime}\cup B^{\prime},\varphi^{\prime}]$. Thus, there can be
only one ascending edge per direction. ∎

Combining this with the $L_{g}$ representatives of Lemma 2.7, we see that the
ascending edges over $[L_{g},\varphi]$ are of the form
$[L_{g+1},\varphi\rho_{i}^{-1}]$, for $i\in\\{1,\dots,n\\}$. (Recall that
$\rho_{n}$ is just the identity.) Next, we prove that there are no collapsed
squares.

######  Lemma 3.6.

Given any square in $X_{n}^{(1)}$, the difference between the maximal and
minimal heights of vertices is 2.

###### Proof.

For the sake of contradition, suppose there is some collapsed square, which
has only two values for the heights of its vertices. Writing the lower height
vertices as $[L_{g},\varphi]$ and $[L_{g},\psi]$, we see that a collapsed
square must occur as in Figure 2. That the same representative can be used for
both edges from one of the side vertices is justified by Lemma 3.5.

$[\sigma(Z),\varphi\circ\sigma^{-1}]$$[\sigma(Z\cup
B),\varphi\circ\sigma^{-1}]$$[\sigma(Z)\cup\sigma(B),\varphi\circ\sigma^{-1}]$$[L_{g},\varphi]$$[L_{g}\cup
B_{i}^{1},\varphi]$$[L_{g+1},\varphi\circ\rho_{i}^{-1}]$$[Z,\varphi]$$[Z^{\prime},\varphi^{\prime}]$$[Z\cup
B,\varphi]$$[Z^{\prime}\cup
B^{\prime},\varphi^{\prime}]$$[L_{g},\varphi]$$[L_{g+1},\varphi\rho_{i_{1}}^{-1}]$$[L_{g+1},\psi\rho_{j_{1}}^{-1}]$$[L_{g+1},\varphi\rho_{i_{2}}^{-1}]$$[L_{g+1},\psi\rho_{j_{2}}^{-1}]$$[L_{g},\psi]$
Figure 2: A collapsed square

To show that the collapsed square in Figure 2 cannot occur, we assume that the
middle pair of vertices are distinct, and prove that the leftmost and
rightmost vertices cannot be distinct. The equality of the different forms of
the middle vertices says that the transition map
$\rho_{i_{k}}\varphi^{-1}\psi\rho_{j_{k}}^{-1}$ takes $L_{g+1}$ to itself and
is rigid elsewhere, for $i=1,2$. This implies that
$\rho_{i_{k}}\varphi^{-1}\psi\rho_{j_{k}}^{-1}$ has net-zero shifting in all
ends, and hence is compactly supported (recall that for asymptotically rigid
maps, these are the same). As the middle vertices are distinct, we know that
$i_{1}\neq i_{2}$ and $j_{1}\neq j_{2}$. Thus, we see that $i_{k}=j_{k}$ for
$k=1,2$.
We will now show that $[L_{g},\varphi]=[L_{g},\psi]$ by considering the
transition map $\varphi^{-1}\psi$ on pieces outside of $L_{g}$. The rigidity
of the $\rho_{i_{k}}\varphi^{-1}\psi\rho_{j_{k}}^{-1}$ maps outside $L_{g+1}$
immediately handles all but three pieces: $B_{i_{1}}^{1}$, $B_{i_{2}}^{1}$,
and $B_{n}^{g+1}$. The last of these is again easy: the $L_{g+1}$ transition
maps are rigid on the $g+2$nd piece, which they first push to the $g+1$st
piece, then apply $\varphi^{-1}\psi$, then push back out by one. As this must
be sent rigidly to the $g+2$nd piece, we see that $\varphi^{-1}\psi$ must be
rigid on the $g+1$st piece. For $B_{i_{1}}^{1}$, consider that
$\varphi^{-1}\psi$ and $\rho_{i_{2}}\varphi^{-1}\psi\rho_{i_{2}}^{-1}$ act the
same on it. As the latter is rigid there, so is the former. Symmetrically, we
have rigidity on $B_{i_{1}}^{1}$, and are done. Specifically, we have shown
that $\varphi^{-1}\psi$ takes $L_{g}$ to itself, and is rigid elsewhere,
implying that the collapsed square we began with was degenerate to begin with.
∎

######  Proposition 3.7.

The Stein–Farley complex $X_{n}$ is cube complete.

###### Proof.

We begin by observing that there can be no collapsed cubes: any such cube must
contain a collapsed square. Now we show that there are no empty squares. As
there are no collapsed squares, we know that the 1-skeleton of a square must
have a lowest vertex, say at height $g$, two vertices at height $g+1$, and one
vertex at height $g+2$. By the uniqueness (per end) of ascending edges (Lemma
3.5), we see that given a representative $[Z,\varphi]$ for the height $g$
vertex, the other vertices must be of the form $[-,\varphi]$, with the blank
filled by $Z\cup B_{1}$, $Z\cup B_{2}$, and $Z\cup B_{1}\cup B_{2}$, depending
on height. The only concern about the $g+2$ vertex would be that the pieces
lie in the same end, but we see that this cannot occur. Thus, there are no
empty squares.
Finally, we show that there are no empty cubes of any dimension. We do so
inductively: suppose we already have that there are no empty $(d-1)$-cubes.
Consider the $1$-skeleton of an $d$-cube $C$. Let the bottom vertex of $C$ be
$[Z,\varphi]$, at height $g$. Each of the vertices of $C$ adjacent to
$[Z,\varphi]$ is of the form $[Z\cup B_{i},\varphi]$, where $B_{i}$ is in the
$i$th end; each $i$ appears at most once by Lemma 3.5. Also, by the inductive
hypothesis, we have that $[Z,\varphi]$ connects to each of the vertices at
height $g+d-1$ via a $(d-1)$-cube, so that these vertices can be written as
$[Z\cup B_{i_{1}}\cup\cdots B_{i_{j-1}}\cup\widehat{B}_{i_{j}}\cup
B_{i_{j+1}}\cup\cdots\cup B_{i_{d}},\varphi]$, where the hat indicates
omission. Choose two such vertices, and consider their common lower vertex
(the one missing the two pieces missing in either of the chosen vertices).
Applying the doctrine of no empty squares to these 3 vertices and the apex
yields that the apex can be written as $[Z\cup B_{i_{1}}\cup\cdots\cup
B_{i_{d}},\varphi]$, and thus the cube is filled.
∎

######  Remark 3.8.

The argument above that ascending edges are unique per direction emphatically
does not hold for descending edges. In fact, for any vertex of height at least
1, there are infinitely many descending edges per direction in which it can
have a descending edge (given $[Z,\varphi]$, choose $\psi\in\mathcal{PH}_{n}$
such that $\varphi^{-1}\psi|_{Z}$ is a homeomorphism, with some nontrivial
behavior in the piece to be removed, and such that $\varphi^{-1}\psi$ is rigid
outside $Z$). It is helpful to keep in mind the following: when ascending, one
can always choose coherent representatives from a minimal vertex; when
descending, nothing is guaranteed.

######  Theorem 3.9.

Let $X_{n}$ denote the Stein–Farley complex for $\mathcal{H}_{n}$. Then
$X_{n}$ is CAT(0).

###### Proof.

By Proposition 3.7, $X_{n}$ is cube-complete. By construction, $X_{n}^{(1)}$
has no loops or bigons. As $X_{n}$ is contractible, it is simply connected, so
item $(a)$ is covered. For item $(b)$, we need to show that 3-wheels can
always be completed. There are four possibilities: the common vertex can be
lowest height, highest height, or either of the intermediate heights. In the
first case, we can have all three edges out of the common vertex use the same
representative, each adding a piece/shift in a different end, so that the cube
is completed by adding in all three at once. In the intermediate height cases,
something similar occurs: we always have the lowest height vertex of the
desired cube from which we can build upwards. It is useful to keep in mind the
following arrangement of the 1-skeleton of a 3-cube; the solid lines indicate
the 3-wheel, and the dashed lines indicate the rest of the 3-cube.

${\cdot}$${\cdot}$${\cdot}$${\cdot}$${\cdot}$${\cdot}$${\cdot}$${\cdot}$

The difficulty is in the remaining case, where the common vertex is maximal in
the cube. We begin with the diagram in Fig. 3. Our goal is to obtain a
representative $[L_{g+2},\varphi\circ\rho_{k}]$ for the height $g+2$ vertex
currently represented with both $\varphi^{\prime}$ and
$\varphi^{\prime\prime}$. This will imply the existence of a vertex
$[L_{g},\varphi\circ\rho_{i}\rho_{j}\rho_{k}]$, which realizes this 3-wheel in
a 3-cube. To begin, we have the following lemma that justifies part of the
diagram.

######  Lemma 3.10.

Whenever $[L_{g},\varphi\circ\rho_{i}]=[L_{g},\varphi^{\prime}\circ\rho_{j}]$
and $[L_{g+1},\varphi]=[L_{g+1},\varphi^{\prime}]$, we have that $i=j$.

###### Proof.

First, observe that the equality
$[L_{g},\varphi\circ\rho_{i}]=[L_{g},\varphi^{\prime}\circ\rho_{j}]$ means
that $h=\rho_{i}^{-1}\varphi^{-1}\varphi^{\prime}\rho_{j}$ takes $L_{g}$ to
itself, and is rigid elsewhere; also, the equality
$[L_{g+1},\varphi]=[L_{g+1},\varphi^{\prime}]$ means that
$\varphi^{-1}\varphi^{\prime}(L_{g+1})=L_{g+1}$. Assume $j\neq n$, and
consider the piece $B^{1}_{j}$; let $\alpha$ be the boundary curve it shares
with $\mathcal{O}$. What does $h$ do to $\alpha$? First, $\rho_{j}$ deforms it
into the center and stretches it into the $n$th end, making it an essential
curve in $L_{g}$. Then $\varphi^{-1}\varphi^{\prime}$ moves it around to some
essential curve in $L_{g+1}$. Finally, $\rho_{i}^{-1}$ pushes and stretches it
towards the $i$th end. As $B_{j}^{1}$ is not in $L_{g}$, it must be sent to a
piece, and $h(\alpha)$ must be the intersection of this piece with $L_{g}$. As
all representatives here do not permute ends, it must be that
$h(\alpha)=\alpha$. If $i\neq j$, then $\rho_{i}$ fixes $\alpha$, so
$\rho_{i}(\alpha)\neq\varphi^{-1}\varphi^{\prime}\rho_{j}(\alpha)$, and
$h(\alpha)\neq\alpha$. Hence we must have that $i=j$. In the case where $j=n$,
equality is obvious. ∎

Now, we seek the following equality:
$[L_{g+2},\varphi\circ\rho_{k}]=[L_{g+2},\varphi^{\prime}\circ\rho_{k}]$. That
is, we wish to show that $\rho_{k}^{-1}\varphi^{-1}\varphi^{\prime}\rho_{k}$
takes $L_{g+2}$ to itself, and is rigid elsewhere. We already know that this
holds for $\rho_{i}^{-1}\varphi^{-1}\varphi^{\prime}\rho_{i}$. The rigidity
here occurs precisely when $\varphi^{-1}\varphi^{\prime}$ does nothing
untoward to whatever single piece gets pushed into the $n$th end, so it
doesn’t matter which index we choose! (If we have chosen the index $n$, there
is no pushing. To resolve this, simply choose a different height $g+2$ vertex
to work with. They can’t share ends, as they are in squares with lower
vertices, so we can apply Lemma 3.5.)
One small detail remains: we have a bottom vertex connecting to one of the
height $g+1$ vertices, but does it connect to the other height $g+1$ vertices?
Suppose not: then the cube between $[L_{g+3},\varphi]$ and
$[L_{g},\varphi\circ\rho_{i}\rho_{j}\rho_{k}]$ contains a square differing
from one of the original 3 squares only at its height $g+1$ vertex (that is,
at its lowest vertex). This however means that we would have two squares
agreeing on 3 vertices, disagreeing on the 4th, which would contain a
collapsed square. As this cannot occur (Lemma 3.6), our height $g$ vertex
connects to each of the height $g+1$ vertices we started with.

$[L_{g+3},\varphi]$$[L_{g+3},\varphi^{\prime}]$$[L_{g+3},\varphi^{\prime\prime}]$$[\sigma(Z),\varphi\circ\sigma^{-1}]$$[\sigma(Z\cup
B),\varphi\circ\sigma^{-1}]$$[\sigma(Z)\cup\sigma(B),\varphi\circ\sigma^{-1}]$$[L_{g},\varphi]$$[L_{g}\cup
B_{i}^{1},\varphi]$$[L_{g+1},\varphi\circ\rho_{i}^{-1}]$$[Z,\varphi]$$[Z^{\prime},\varphi^{\prime}]$$[Z\cup
B,\varphi]$$[Z^{\prime}\cup
B^{\prime},\varphi^{\prime}]$$[L_{g},\varphi]$$[L_{g+1},\varphi\rho_{i_{1}}^{-1}]$$[L_{g+1},\psi\rho_{j_{1}}^{-1}]$$[L_{g+1},\varphi\rho_{i_{2}}^{-1}]$$[L_{g+1},\psi\rho_{j_{2}}^{-1}]$$[L_{g},\psi]$$[L_{g+1},\varphi\circ\rho_{i}\rho_{j}]$$[L_{g+1},\varphi^{\prime}\circ\rho_{i}\rho_{k}]$$[L_{g+1},\varphi^{\prime\prime}\circ\rho_{j}\rho_{k}]$$[L_{g+2},\varphi\circ\rho_{i}]$$[L_{g+2},\varphi^{\prime}\circ\rho_{i}]$$[L_{g+2},\varphi\circ\rho_{j}]$$[L_{g+2},\varphi^{\prime\prime}\circ\rho_{j}]$$[L_{g+2},\varphi^{\prime}\circ\rho_{k}]$$[L_{g+2},\varphi^{\prime\prime}\circ\rho_{k}]$$[L_{g+3},\varphi]$$[L_{g+3},\varphi^{\prime}]$$[L_{g+3},\varphi^{\prime\prime}]$
Figure 3: A 3-wheel with common vertex maximal.

∎

##  4 $\Sigma$-invariants of $P\mathcal{H}_{n}$

The methods of this section are taken from [Zar16] and [Zar20]. They need only
minimal modification to work in this setting, and are presented partly for the
sake of having the entire argument in one place. We begin by noting that the
abelianization of $P\mathcal{H}_{n}$ is $\mathbb{Z}^{n-1}$, and the
abelianization map is $(\chi_{1},\dots,\chi_{n-1})$ (see Section 6
of[ABKL23]). As $\chi_{1}+\cdots+\chi_{n}=0$, any character $\chi$ can be
written (up to renumbering the ends of $\Sigma_{n}$) in ascending standard
form, i.e. $\chi=a_{1}\chi_{1}+\cdots+a_{n}\chi_{n}$, with
$a_{1}\leq\cdots\leq a_{m(\chi)}<a_{m(\chi)+1}=\cdots=a_{n}=0$. We shall
henceforth assume all characters are written in such form. Observe that
$m(\chi)$ is defined to be the maximal index so that $a_{m(\chi)}<a_{n}=0$.
Then our goal in this section is to prove the following.

######  Theorem 4.1.

Let $\chi$ be a non-zero character of $P\mathcal{H}_{n}$. Then
$[\chi]\in\Sigma^{m(\chi)-1}(P\mathcal{H}_{n})\setminus\Sigma^{m(\chi)}(P\mathcal{H}_{n})$.

###  4.1 Inclusion

We begin by showing the inclusion into the $(m(\chi)-1)$ layer of
$\Sigma(P\mathcal{H}_{n})$.

######  Theorem 4.2.

For any non-zero character
$\chi\in\operatorname{Hom}(P\mathcal{H}_{n},\mathbb{R})$, we have
$[\chi]\in\Sigma^{m(\chi)-1}(P\mathcal{H}_{n})$.

We recall here the form of the Morse lemma we shall use in this section.

######  Corollary 4.3.

Let $h=(\chi,f):Y\rightarrow\mathbb{R}\times\mathbb{R}$ be a Morse function.
If $Y$ is $(m-1)$-connected and for every vertex $v\in Y^{\chi<q}$ the
ascending link $\operatorname{lk}_{Y}^{h\uparrow}(v)$ is $(m-1)$-connected,
then $Y^{q\leq\chi}$ is $(m-1)$-connected.

From now on, we write $X$ for $X_{n}$. The role of $Y$ shall be played by
sublevel subccomplexes $X^{f\leq q}$ for sufficiently large $q$. Let
$\chi=a_{1}\chi_{1}+\cdots+a_{n}\chi_{n}$ be a non-trivial character of
$P\mathcal{H}_{n}$, written in ascending standard form. The function
$h=(\chi,f):Y\rightarrow\mathbb{R}^{2}$ is a Morse function. Observe that
between any two adjacent vertices, each basis character differs by $1$, $0$,
or $-1$. We examine $h$-ascending links in $Y$.

######  Lemma 4.4.

Let $v=[L_{g},\varphi]$ be a vertex in $Y$. An adjacent vertex $w=[L_{g\pm
1},\psi]$ is in the $h$-ascending link of $v$ if and only if one of the two
following conditions hold:

  * $\bullet$

$w=[L_{g-1},\psi]$, with $v=[L_{g},\psi\rho_{i}^{-1}]$ where $i\leq m(\chi)$

  * $\bullet$

$w=[L_{g+1},\varphi\rho_{i}^{-1}]$, where $i\geq m(\chi)+1$.

###### Proof.

If $w$ is $f$-ascending from $v$, then $w=[L_{g+1},\varphi\rho_{i}^{-1}]$ for
some $i$. If $i\geq m(\chi)+1$, then $\chi(w)=\chi(v)$, so $w$ is
$h$-ascending from $v$. If $i\leq m(\chi)$, then $w$ is $\chi$-descending from
$v$, hence $h$-descending from $v$. If $w$ is $f$-descending, then
$v=[L_{g},\psi\rho_{i}^{-1}]$ for some $i$. If $i\geq m(\chi)+1$, then we
again have $\chi(w)=\chi(v)$, so that $w$ is $h$-descending from $v$. If
$i\leq m(\chi)$, then $w$ is $\chi$-ascending from $v$, hence $h$-ascending. ∎

######  Remark 4.5.

It is worth observing a difference here from the situation in [Zar16]. For the
regular Houghton groups, there is one way to go “up” and finitely many ways to
go “down” per end (with respect to $f$); here, while there is only one way to
go up per end, there are infinitely many ways to go down. However, the
$h$-ascending link is still a join of its intersection with the $f$-ascending
and $f$-descending links, as these elements cannot share ends.

We shall need the following fact: Let $v\in X_{n}$ be a vertex. If $f(v)\geq
2n$, then the descending link of $v$ is $(n-2)$-connected.
Now set $q=3n-2$, so $Y=X^{f\leq 3n-2}$. We note that this index is one
greater than that used for the regular Houghton group. This is because the
bounds obtained for connectivity of descending links obtained in [Lee12] and
[ABKL23] differ by one (they are $2n-1$ and $2n$ respectively, see [Lee12,
Theorem 3.52] and [ABKL23, Section 5.2]). We have the following:

######  Proposition 4.6.

Let $v$ be a vertex in $Y$. Then $\operatorname{lk}_{Y}^{h\uparrow}(v)$ is
$(m(\chi)-2)$-connected.

###### Proof.

We have that $f(v)$ is between $0$ and $3n-2$. Suppose that $f(v)\leq
2n+m(\chi)-2$. Writing $v=[L_{g},\varphi]$, we have that there are $n-m(\chi)$
indices for which $v^{\prime}=[L_{g+1},\varphi\circ\rho_{i}^{-1}]$ is
$h$-ascending. As $(2n+m(\chi)-2)+(n-m(\chi))=3n-2$, the entire $f$-ascending
link of $v$ in $X$ is contained in $Y$. Thus, the $f$-ascending part of the
$h$-ascending link is an $(n-m(\chi)-1)$-simplex, which is contractible, so
that $\operatorname{lk}_{Y}^{h\uparrow}(v)$ is contractible.
Suppose instead that $2n+m(\chi)-1\leq f(v)\leq 3n-2$, and thus that $Y$ does
not contain the entire $f$-ascending part of the $h$-link of $v$. We still
have its $(3n-f(v)-3)$-skeleton, which, being a skeleton of an
$(n-m(\chi)-1)$-simplex, is $(3n-f(v)-4)$-connected. As $f(v)\geq 2n$, we have
that the entire $f$-descending part of the $h$-ascending link in is $Y$. This
is isomorphic to the $f$-descending link of a vertex of the same $f$-height as
$v$ in $X_{m(\chi)}$, which we know is $(m(\chi)-2)$-connected so long as
$f(v)\geq 2m(\chi)$. As $f(v)\geq 2n+m(\chi)-1$, we need that $2n-1\geq
m(\chi)$, which is certainly true. The join is now
$((3n-f(v)-3)+(m(\chi)-1))$-connected. As $f(v)\leq 3n-2$, we have that
$(3n-f(v)+m(\chi)-4)\geq m(\chi)-2$, and thus that
$\operatorname{lk}_{Y}^{h\uparrow}(v)$ is $(m(\chi)-2)$-connected. ∎

###  4.2 Exclusion

To demonstrate that the result of the last section is sharp, we shall need to
employ different techniques. The main result of this section is

######  Theorem 4.7.

For any non-zero character
$\chi\in\operatorname{Hom}(P\mathcal{H}_{n},\mathbb{R})$, we have
$[\chi]\notin\Sigma^{m(\chi)}(\mathcal{H}_{n})$.

We gather here the propositions and definitions from [Zar20] that we will
need, starting with the Strong Nerve Lemma:

######  Proposition 4.8.

Let $X$ be a CW-complex covered by subcomplexes $(X_{i})_{i\in I}$ and let $L$
be the nerve of the cover. Let $n\geq 1$. Suppose that any non-empty
intersection $X_{i_{1}}\cap\cdots\cap X_{i_{r}}$ is $(n-r)$-connected. Then
$H_{k}(X)\cong H_{k}(L)$ for all $k\leq n-1$, and $H_{n}(X)$ surjects onto
$H_{n}(L)$.

######  Definition 4.9.

For a set of indices $K\subseteq[n]$, consider the subcomplex $\bigcap_{i\in
K}X^{\chi_{i}\leq 0}$ of $X$. Call any connected component of such a
subcomplex a $K$-blanket. By a blanket we mean a $K$-blanket for some
unspecified $K$.

Recall that a subcomplex $Z$ of a CAT(0) cube complex $Y$ is locally
combinatorially convex if every link in $Z$ of a vertex $z$ is a full
subcomplex of the link of $z$ in $Y$, and combinatorially convex if it is
connected and locally combinatorially convex. It is known that combinatorially
convex implies CAT(0), hence contractible. In particular, this applies to
connected components of locally combinatorially convex subcomplexes.

######  Lemma 4.10.

For any $K$, $\bigcap_{i\in K}X^{\chi_{i}\leq 0}$ is locally combinatorially
convex. Thus, blankets are combinatorially convex, and hence CAT(0) and
contractible.

###### Proof.

It suffices to show that each $X^{\chi_{i}\leq 0}$ is locally combinatorially
convex. Given a pair of adjecent vertices, we can write them as
$v=[L_{g},\varphi]$ and $w=[L_{g+1},\varphi\circ\rho_{j}^{-1}]$ for some $j$.
Then $\chi_{i}(w)-\chi_{i}(v)=\delta_{i,j}$. Thus, if $C$ is a cube containing
$v$, and $w_{1},\dots,w_{k}$ are the vertices of $C$ adjacent to $v$, then the
maximum and minimum values of $\chi_{i}$ on $C$ lie in
$\\{\chi_{i}(v),\chi_{i}(w_{1}),\dots,\chi_{i}(w_{k})\\}$. Thus, whenever
$v\in X^{\chi_{i}\leq 0}$ and all these $w_{j}$’s lie in the link of $v$ in
$X^{\chi_{i}\leq 0}$, then the cube $C$ lies in $X^{\chi_{i}\leq 0}$. This
implies that the link of $v$ in $X^{\chi_{i}\leq 0}$ is a full subcomplex of
the link of $v$ in $X$. ∎

As in [Zar20], we have the immediate corollary that intersections of blankets
are blankets for the union of the sets of indices. What follows is essentially
identical to Zaremsky’s approach for the Houghton groups, reproduced here with
minor changes for completeness’ sake. Recall now the more general statement of
the Morse lemma, Lemma 2.11. For notational convenience, we write $X_{f\leq
k}$ for $X^{f\leq k}$, and $X_{f\leq k}^{t\leq\chi}$ for the intersection
$X_{f\leq k}\cap X^{t\leq\chi}$.

######  Lemma 4.11.

If $X_{f\leq 3n-2}^{0\leq\chi}$ is not $(m(\chi)-1)$-connected, then
$[\chi]\in\Sigma^{m(\chi)}(P\mathcal{H}_{n})^{c}$.

###### Proof.

If $[\chi]\in\Sigma^{m(\chi)}(P\mathcal{H}_{n})$, then the filtration
$(X_{f\leq 3n-2}^{t\leq\chi})_{t\in\mathbb{R}}$ is essentially
$(m(\chi)-1)$-connected. Every $h$-ascending link of a vertex in $X_{f\leq
3n-2}$ is $(m(\chi)-2)$-connected, so for any $s\leq t$ the inclusion
$X_{f\leq 3n-2}^{t\leq\chi}\hookrightarrow X_{f\leq 3n-2}^{s\leq\chi}$ induces
an isomorphism in $\pi_{k}$ for $k\leq m(\chi)-2$, and a surjection in
$\pi_{m(\chi)-1}$. By assumption, for any $t$ there is some $s\leq t$ such
that this inclusion induces a trivial map in $\pi_{k}$ for $k\leq m(\chi)-1$,
implying that $X_{f\leq 3n-2}^{s\leq\chi}$ is $(m(\chi)-1)$-connected.
Rescaling if necessary, we can assume that $s\in\chi(P\mathcal{H}_{n})$, and
thus we can translate to obtain $X_{f\leq 3n-2}^{s\leq\chi}\cong X_{f\leq
3n-2}^{0\leq\chi}$, so that $X_{f\leq 3n-2}^{0\leq\chi}$ is itself
$(m(\chi)-1)$-connected. ∎

In order to show that $X_{f\leq 3n-2}^{0\leq\chi}$ is not
$(m(\chi)-1)$-connected, we will apply the Strong Nerve Lemma to a covering we
now define. For $1\leq i\leq n$, let $\\{Z_{i}^{\alpha}\\}$ be the collection
of $\\{i\\}$-blankets in $X$. The $\alpha$’s are indices in some set, and this
index set is not itself important. Set also

$Y_{i}^{\alpha}=Z_{i}^{\alpha}\cap X_{f\leq 3n-2}^{0\leq\chi}.$

######  Lemma 4.12.

The $Y_{i}^{\alpha}$ with $1\leq i\leq m(\chi)$ cover $X_{f\leq
3n-2}^{0\leq\chi}$.

###### Proof.

As the $Z_{i}^{\alpha}$ are the connected components of the $X^{\chi_{i}\leq
0}$, it suffices to show that

$X^{0\leq\chi}\subseteq\bigcup_{i=1}^{m(\chi)}X^{\chi_{i}\leq 0}.$

As $\chi=a_{1}\chi_{1}+\cdots+a_{m(\chi)}\chi_{m(\chi)}$, with all
coefficients negative, any vertex $v\in X$ with $\chi(v)\geq 0$ must satisfy
$\chi_{i}(v)\leq 0$ for some $i$. This implies the inclusion on vertices.
Given a cube in $X^{0\leq\chi}$, let $v$ be its maximal vertex with respect to
$f$. Then all the vertices $w$ of the cube satisfy
$\chi_{i}(w)\leq\chi_{i}(v)$, and hence the entire cube lies in whichever
$X^{\chi_{i}\leq 0}$ contains $v$. ∎

######  Lemma 4.13.

Any non-empty intersection of subcomplexes of the form $Y_{i}^{\alpha}$ with
$1\leq i\leq m(\chi)$ is $(m(\chi)-2)$-connected.

###### Proof.

To be non-empty, such an intersection can include at most one term
$Y_{i}^{\alpha}$ for each $i$, and can thus be written
$Y=Y_{i_{1}}^{\alpha_{1}}\cap\cdots\cap Y_{i_{r}}^{\alpha_{r}}$, with the
$i_{j}$ all pairwise distinct. Let $Z=Z_{i_{1}}^{\alpha_{1}}\cap\cdots\cap
Z_{i_{r}}^{\alpha_{r}}$, so that $Y=Z\cap X_{f\leq 3n-2}^{0\leq\chi}$. We
apply Morse theoretic techinques to $Z$, this time using Lemma 2.11. As $Z$ is
an intersection of blankets, it is a blanket, and thus contractible. Given
adjacent vertices $w=[L_{g},\varphi]$ and
$v=[L_{g+1},\varphi\circ\rho_{i}^{-1}]$ with $v\in Z$, we have that $w\in Z$.
Thus, for any vertex of $Z$, the entire $f$-descending link is in $Z$. As this
is $(n-2)$-connected for $f(v)\geq 2n$, we see that $Z_{f\leq 3n-2}$ is
$(n-2)$-connected, and so is certainly $(m(\chi)-2)$-connected. Now consider
$Y$ as $Z_{f\leq 3n-2}^{0\leq\chi}$. As before, the $h$-ascending link of a
vertex $v$ is a join between its $f$-ascending and $f$-descending parts. The
latter is in $Z$ for the same reasons as above; the former is in $Z$ because
it consists of directions $i$ where $m(\chi)+1\leq i\leq n$, on which the
considered $\chi_{i_{j}}$ are constant. Thus, the $h$-ascending link of $v$ is
in $Z_{f\leq 3n-2}$. As $Z_{f\leq 3n-2}$ is $(m(\chi)-2)$-connected, the Morse
lemma tells us that $Y$ is $(m(\chi)-2)$-connected. ∎

Let $L$ be the nerve of the covering of $X_{f\leq 3n-2}^{0\leq\chi}$ by the
$Y_{i}^{\alpha}$. Since $[\chi]\in\Sigma^{m(\chi)-1}(P\mathcal{H}_{n})$, we
know that $X_{f\leq 3n-2}^{0\leq\chi}$ is $(m(\chi)-2)$-connected, so by the
Strong Nerve Lemma $L$ is $(m(\chi)-2)$-connected. The final missing piece is
to prove that $L$ is not $(m(\chi)-1)$-acyclic.

######  Lemma 4.14.

The nerve $L$ is not $(m(\chi)-1)$-acyclic.

###### Proof.

Consider vertices corresponding to $Y_{i}^{\alpha}$ and $Y_{j}^{\beta}$. These
vertices can only be adjacent if $i\neq j$, so $L$ is $(m(\chi)-1)$
dimensional. Thus, it suffices to exhibit a non-trivial $(m(\chi)-1)$-cycle.
This will come from a collection of $2m(\chi)$ vertices, 2 for each
$i\in\\{1,\dots,m(\chi)\\}$, labeled as $Y_{i}^{\epsilon_{i}}$, with
$\epsilon\in\\{1,2\\}$, with the property $Y_{1}^{\epsilon_{i}}\cap\cdots\cap
Y_{m(\chi)}^{\epsilon_{m(\chi)}}\neq\emptyset$. This will yield an embedded
$(m(\chi)-1)$-sphere in $L$, which is homologically non-trivial for
dimensional reasons.
Recall that $\mathcal{O}$ is the centerpiece of our surface. For each $i$,
take $Y_{i}^{1}$ to be the $Y_{i}^{\alpha}$ containing
$v_{0}=[\mathcal{O},\operatorname{id}]$, and take $Y_{i}^{2}$ to be the
$Y_{i}^{\alpha}$ containing $v_{i}=[\mathcal{O},\varphi_{i}]$, where
$\varphi_{i}$ is some non-trivial mapping class in the $i$th end; for the sake
of definiteness, choose some essential simple closed curve in the standard
piece, and let $\varphi_{i}$ be the Dehn twist about its image in $B_{i}^{1}$.
Any intersection $Y_{1}^{\epsilon_{1}}\cap\cdots\cap
Y_{m(\chi)}^{\epsilon_{m(\chi)}}$ includes $w=[\mathcal{O},\prod\varphi_{i}]$,
where the product (in an abitrary order) is taken over those $i$ with
$\epsilon_{i}=2$, and is therefore nonempty.
It remains to show that $Y_{i}^{1}\neq Y_{i}^{2}$ for each $i$. It suffices to
show that $Z_{i}^{1}\neq Z_{i}^{2}$. If these are equal (call them $Z_{i}$),
then we can connect $v_{0}$ to $v_{i}$ via a path in $Z_{i}$. In $X$, one
could assume that such a path is one on which $f$ first strictly increase,
then strictly decreases; since $Z_{i}$ is combinatorially convex, this
property holds for $Z_{i}$ as well. Since the path lies in $Z_{i}$, $\chi_{i}$
is non-positive on the whole path. By the uniqueness of ascending edges, this
is a path of the form

${{[Z,\operatorname{id}]}}$${{[Z^{\prime},\varphi_{i}]}}$${{[\mathcal{O},\operatorname{id}]}}$${{[\mathcal{O},\varphi_{i}]}}$

where each ascending dotted line indicates a sequence of edges which only add
pieces. Since $\chi_{i}(v_{0})=0=\chi_{i}(v_{i})$, none of the edges of the
path can be obtained by adding a piece in the $i$th end. One of $Z$ and
$Z^{\prime}$ must have at least one piece in the $i$th end, as the transition
map is $\varphi_{i}$, which has non-rigid behavior in the $i$th end. Thus, we
have a contradiction. ∎

##  5 Subgroups of maximal finiteness length

We use the notation $fl(G)=n$ to mean that $G$ is a group of type $F_{n}$ but
not of type $F_{n+1}$. We know that if $G<H$ is finite index, then
$fl(G)=fl(H)$. But what of the converse, that is, when does $fl(G)=fl(H)$
imply that $G$ is finite index in $H$? For Houghton groups, and for the (pure)
surface Houghton groups, the answer is positive for sufficiently large
subgroups, in particular for coabelian subgroups. We denote by $G^{\prime}$
the commutator subgroup of a group $G$. Recall that the commutators of the
Houghton group $H_{n}$ and the pure surface Houghton group $P\mathcal{H}_{n}$
are the finitely supported and compactly supported elements, respectively.
(Really, we could say compactly supported in both cases, using the discrete
topology for the $\mathbb{N}$-rays on which the Houghton group acts.) For
Theorem 5.1 and Proposition 5.2, let $H$ be either $H_{n}$ or
$P\mathcal{H}_{n}$.

######  Theorem 5.1.

Let $G<H$ be a subgroup intersecting the commutator in a finite index
subgroup. If $fl(G)=n-1$, then $G$ is finite index in $H$.

Note that it suffices to show for groups containing the commutator. We begin
by showing that when $G$ contains the commutator, the image of $G$ in the
abelianization is a maximal rank sublattice, which will then imply finite
index.

######  Proposition 5.2.

Let $G<H$ be as above, i.e. $fl(G)=n-1$, and $G$ contains the commutator
$G^{\prime}$. Write $F$ for the abelianization map to $\mathbb{Z}^{n-1}$. Then
$F(G)$ is finite index in $\mathbb{Z}^{n-1}$.

###### Proof.

We start by observing that $F=(\chi_{1},\dots,\chi_{n-1})$. As
$\Sigma^{n-1}(H)$ is empty, the only way for $G$ to be of type $F_{n-1}$ is
that it cannot be killed by any non-zero character of $H$. We shall construct
a sequence of maps
$F_{i}=(F_{i}^{1},\dots,F_{i}^{n-1}):H\rightarrow\mathbb{Z}^{n-1}$ with
$F=F_{1}$, and a sequence of elements $g_{i}\in G$ such that
$F_{n}^{j}(g_{i})$ is positive when $i=j$, and zero when $i<j$. Then
$F_{n}(G)$ will be a maximal rank sublattice of $\mathbb{Z}^{n-1}$, obtainable
from $F_{1}(G)$ by integer matrix transformations.
As $\chi_{1}(G)\neq 0$, we have some smallest positive integer
$c_{1}\in\chi_{1}(G)$. Let $g_{1}\in\chi_{1}^{-1}(c_{1})\cap G$, and write
$F_{1}(g_{1})=(a_{1}^{1},\dots,a_{n-1}^{1})$ (note that $a_{1}^{1}=c_{1}$).
Set
$F_{2}=(\chi_{1},a_{1}^{1}\chi_{2}-a_{2}^{1}\chi_{1},\dots,a_{1}^{1}\chi_{n-1}-a_{n-1}^{1}\chi_{1})$.
Then $F_{2}(g_{1})=(c_{1},0,\dots,0)$. By the same argument, we can choose
$c_{2}>0$ minimal in the image of the character
$a_{1}^{1}\chi_{2}-a_{2}^{1}\chi_{1}$, and then $g_{2}\in G$ in its preimage.
Carrying on, we obtain $F_{i}$ from $F_{i-1}$ by modifying only the components
from $i$ onwards, and build a collection $g_{1},\dots,g_{n-1}\in G$ such that
the matrix obtained by applying $F_{n}$ to this collection is lower triangular
with integer entries, with positive values on the main diagonal. ∎

We now prove Theorem 5.1, using the characterization of the commutator as the
elements of compact support. Note that for the Houghton group, the ends
correspond to the rays in the obvious way.

###### Proof.

Suppose first that $G$ contains the commutator. Consider the map of coset
spaces $\Psi:H/G\rightarrow\mathbb{Z}^{n-1}/(\textbf{c}\mathbb{Z}^{n-1})$
given by $hG\mapsto F(h)\textbf{c}\mathbb{Z}^{n-1}$. To see that this is well-
defined, suppose $h_{1}G=h_{2}G$: then $h_{1}=h_{2}g$ for some $g\in G$, and
$F(h_{1})=F(h_{2})+F(g)$. As $F(g)\in\textbf{c}\mathbb{Z}^{n-1}$, we see that
$\Psi(h_{1}G)=\Psi(h_{2}G)$. We now wish to show that $\Psi$ is injective.
Suppose that $\Psi(h_{1}G)=\Psi(h_{2}G)$. Then
$F(h_{1}h_{2}^{-1})=F(h_{1})-F(h_{2})=F(g)$ for some $g\in G$. So our question
becomes: does $F(h_{1}h_{2}^{-1})\in F(G)$ imply $h_{1}h_{2}^{-1}\in G$? The
element $h_{1}h_{2}^{-1}g^{-1}$ will have no net translation in any end, and
hence is compactly supported, and therefore is in $G$.
∎

As there are clearly subgroups of finite index in both cases (take the pre-
images of a finite index subgroup of $\mathbb{Z}^{n-1}$, there is one loose
end remaining: the infinite index case. Specifically, do there exist subgroups
$G<H_{n}$ (or $G<P\mathcal{H}_{n}$) whose intersection with the commutator
have infinite index, and such that $fl(G)=n-1$. First, we see that $G$ is
necessarily infinite index in $H_{n}$ (or $P\mathcal{H}_{n}$), so all that
remains is to check whether this case actually occurs. We thank Noel Brady for
the proofs of the following lemmas.

######  Lemma 5.3.

Suppose $G,K\leq H$ are subgroups. If $G$ is finite index in $H$, then $G\cap
K$ is finite index in $K$.

###### Proof.

Suppose $[H:G]=m<\infty$. Write $H=Gh_{1}\cup\cdots\cup Gh_{m}$, as a disjoint
union. For each index $i$ such that $K\cap Gh_{i}\neq\emptyset$, there is some
$g_{i}\in G$ and $k_{i}\in K$ such that $k_{i}=h_{i}g_{i}$. Then we can write

$Gh_{i}=Gg_{i}^{-1}k_{i}=Gk_{i}.$

Thus,

$K\cap Gh_{i}=K\cap Gk_{i}=Kk_{i}\cap Gk_{i}=(K\cap G)k_{i}.$

Intersecting this with our original disjoint decomposition for $H$, We have

$K=(K\cap G)k_{1}\cup\cdots\cup(K\cap G)k_{s},$

where $s\leq m$ is the number of indices for which $K\cap
Gh_{i}\neq\emptyset$. ∎

######  Lemma 5.4.

Suppose that a group $H$ has an exhaustion by subgroups, i.e. there exists a
nested chain of subgroups $K_{1}\leq K_{2}\leq\cdots\leq H$ such that $H$ is
the union of all the $K_{i}$’s. Let $G\leq H$. Then $[H:G]$ is finite if and
only if the sequence $([K_{i}:G\cap K_{i}])$ is eventually constant, in which
case it is the limiting value.

###### Proof.

By Lemma 5.3, we see that the sequence $([K_{i}:G\cap K_{i}])$ is bounded by
$[H:G]=m$. It is not hard to see that it is non-decreasing, so we need only
show that it can’t stabilize below $m$. Choosing a decomposition of $H$ into
$G$-cosets, say $H=Gh_{1}\cup\cdots\cup Gh_{m}$, we see that there must be
some index $j$ such that $K_{j}$ contains all of $\\{h_{1},\dots,h_{m}\\}$.
Therefore, the sequence of indices stabilizes at $m$.
For the converse, consider a decomposition $H=Gh_{1}\cup Gh_{2}\cup\cdots$.
Intersecting this with $K_{i}$, we see as before that the index $[K_{i}:G\cap
K_{i}]$ is the numebr of indices $j$ such that $Gh_{j}\cap K_{i}$ is non-
empty. As every $h_{j}$ must be in some $K_{i}$, we see that if $[H:G]$ is
infinite, then the sequence $([K_{i}:G\cap K_{i}])$ goes to infinity. ∎

We return to our question on the existence of infinite index subgroups of
Houghton groups with finiteness length $n-1$, We must now split off the
ordinary Houghton group from the surface Houghton group. For $H_{n}$, the
answer to our existence question is easily yes: there are even copies of
$H_{n}$ itself as infinite index subgroups of $H_{n}$! Consider the stabilizer
of a single point in $[n]\times\mathbb{N}$: ignoring the fixed point, and
sliding its ray back by one to fill in, we have a natural bijection between
$[n]\times\mathbb{N}$, on which $H_{n}$ acts, and
$([n]\times\mathbb{N})\setminus\\{(i,j)\\}$, on which
$\operatorname{Stab}(i,j)$ acts, and we see that these actions are the same.
(This fact was known to Houghton, see [Hou78].)
By the same argument, the subgroup fixing pointwise any finite set will be a
copy of $H_{n}$, and the subgroup fixing (as a set, not pointwise) any finite
set will be a finite extension of $H_{n}$.
In [Cor16], Yves Cornulier defined a stronger failure of co-Hopfianness, which
he called “dis-cohopfian”. Call a group $G$ dis-co-Hopfian if there is some
injective homomorphism $\eta:G\rightarrow G$ such that the intersection of all
iterated images of this homomorphism is trivial, i.e.

$\bigcap_{n=1}^{\infty}\eta^{n}(G)=\\{e\\}.$

By taking the finite set being stabilized to be the first point in each
$\mathbb{N}$-ray, we obtain the following:

######  Proposition 5.5.

The Houghton groups $H_{n}$ are not co-Hopfian, and are in fact dis-co-
Hopfian.

To extend these results to Surface Houghton Groups, we shall work out more
carefully the embedding $\operatorname{Br}H_{n}\hookrightarrow\mathcal{H}_{n}$
suggested by [ABKL23]. For the braided Houghton group, we take the
asymptotically rigid mapping class group of the following surface (see [Fun07]
for details). Begin with a $2n$-gon as the center piece, and take a punctured
square for the attached pieces (see Figure 4). In [GLU21], it was shown that
the braided Houghton group $\operatorname{Br}H_{n}$ is type $F_{n-1}$ but not
type $FP_{n}$. This makes it a good candidate for an infinite index subgroup
of $\mathcal{H}_{n}$ with $fl(\operatorname{Br}H_{n})=fl(\mathcal{H}_{n})$.

Figure 4: Left: the defining surface for $\operatorname{Br}H_{n}$; Right: the
surface $\Sigma^{\prime}$

To obtain a homomorphism $\operatorname{Br}H_{n}\rightarrow\mathcal{H}_{n}$,
we replace each puncture with a boundary component, and then pass to the
double. Write $\Sigma^{\prime}$ for the surface obtained by replacing
punctures with boundary components. As for why this yields an injective
homomorphism, consider the following diagram, where the upper row is exact:

${1}$${K}$${\operatorname{Map}(\Sigma^{\prime})}$${\operatorname{Br}H_{n}}$${1}$${\mathcal{H}_{n}}$$\scriptstyle{a}$$\scriptstyle{b}$

By the construction of Ivanov-McCarthy, the map $b$ is an injective
homomorphism. The kernel of $a$ is generated by half-twists about the compact
boundary components, which are trivial under $b$. Therefore, we obtain an
injective homomorphism $\operatorname{Br}H_{n}\rightarrow\mathcal{H}_{n}$. As
$\operatorname{Br}H_{n}$ acts trivially on the maximal (i.e. non-puncture)
ends, this image lies in $P\mathcal{H}_{n}$.
The image of this homomorphism is a subgroup which fixes (up to isotopy) the
set of curves which correspond to the following curve via the canonical maps
$\iota_{B}$ of Section 2.1 [Add picture of curve circumscribing the genus of a
torus with two boundary components]. All that remains is to see that this
subgroup has infinite index in $P\mathcal{H}_{n}$. Consider its intersection
with the mapping class group of any suited subsurface: here, it must fix the
multicurve consisting of the above curves, and is therefore of infinite index.
As infinite index in a subgroup implies infinite index in the full group, we
have our result.
It is easy to see that $\operatorname{Br}H_{n}$ is not co-Hopfian: there is an
inclusion map from the defining surface of $\operatorname{Br}H_{n}$ to itself,
sliding the first puncture in some end out, and with all elements moving
things only on one side of the skipped puncture (see Figure 5). As with the
Houghton groups, doing such a move in all ends simultaneously yields a
homomorphism whose iterated images act trivially on arbitrarily large compact
subsurfaces.

Figure 5: Pushing one puncture “off to the side”

These approaches are not immediately available for the surface Houghton group,
as there is no inclusion of surfaces which skips over a single genus. In fact,
in light of the results of [ALM23], if the pure surface Houghton group were to
fail to be co-Hopfian, then it must fail either by a non-twist-preserving
homomorphism, or by a homomorphism which is not the restriction of a
homomorphism on the level of pure mapping class groups.

All told, we have the following for co-Hopfianness of the asymptotically rigid
mapping class group versions of Houghton groups.

######  Theorem 5.6.

The braided Houghton group $\operatorname{Br}H_{n}$ is not co-Hopfian, and is
in fact dis-co-Hopfian.

######  Question 5.7.

Is the pure surface Houghton group $P\mathcal{H}_{n}$ co-Hopfian? If not, is
it dis-co-Hopfian?

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Noah Torgerson
Department of Mathematics
University of Oklahoma
<EMAIL_ADDRESS>|  Jeremy West
Department of Mathematics
University of Oklahoma
<EMAIL_ADDRESS>
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