# The first uniformly finite homology group with coefficients in
${\mathbb{Z}}$ and a characterisation of its vanishing in the transitive case

Rémi Bottinelli111Supported by the Swiss National Science Foundation project
no. PP00P2-144681/1 Tom Kaiser

###### Abstract

We study the first uniformly finite homology group of Block and Weinberger for
uniformly locally finite graphs, with coefficients in $\mathbb{Z}$ and
$\mathbb{Z}_{2}$. When the graph is a tree, or coefficients are in
$\mathbb{Z}_{2}$, a characterisation of the group is obtained. In the general
case, we describe three phenomena that entail non-vanishing of the group;
their disjunction is shown to also be necessary for non-vanishing in the case
of transitive graphs.

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries
    1. 2.1 Introduction
    2. 2.2 Definition of $H^{\mathrm{suf}}$
    3. 2.3 Definition of $H^{\mathrm{uf}}$
    4. 2.4 “Hands-on” definition of $H^{\mathrm{uf}}_{1}$
  3. 3 Trees and End-Defining Trees
    1. 3.1 Homology of Trees
    2. 3.2 Ends
  4. 4 Coefficients in $\mathbb{Z}_{2}$
    1. 4.1 $H^{\mathrm{uf}}_{1}(X,\mathbb{Z})\twoheadrightarrow H^{\mathrm{uf}}_{1}(X,\mathbb{Z}_{2})$
    2. 4.2 Decomposition of $H^{\mathrm{uf}}_{1}(X,\mathbb{Z}_{2})$
    3. 4.3 Large circuits and dimensions
  5. 5 $\mathbb{Z}$-large circuits
  6. 6 Expansion
    1. 6.1 Basics
    2. 6.2 Dimension $1$
      1. 6.2.1 $H^{\mathrm{uf}}$

##  1 Introduction

Uniformly finite homology, introduced in [1], is a coarse invariant of well-
behaved metric spaces. In [1], it is shown that vanishing of zeroth uniformly
finite homology with coefficients in ${\mathbb{Z}}$ or ${\mathbb{R}}$ is
equivalent to expansion (non-amenability) of the space at hand. Higher
uniformly finite homology groups don’t enjoy, as far as we know, such clear
cut descriptions. Partial results have been obtained, e.g. in [2, 3].

In these notes, we focus on the first homology group, with coefficients in
${\mathbb{Z}}$ of (tame enough) graphs:
$H^{\mathrm{uf}}_{1}({\cdot},{\mathbb{Z}})$. Informally, three relatively
independent phenomena are responsible for the appearance of homology classes
in dimension one:

  1. 1.

Graph theoretical ends.

  2. 2.

The existence of circuits of arbitrarily large girth. More precisely, the non-
existence of a uniform constant $R$ such that any circuit can be written as a
(possibly infinite) sum of circuits of length ${\leq}R$.

  3. 3.

A lack of $2$-simplices in the clique complex. This should be understood
quantitatively, and likened to amenability.

After recalling the necessary definitions in Section 2, we proceed in four
parts.

In Section 3, we study the first of the above phenomena. Starting with a
classification of the (first) homology of trees (Theorem 3.6) in terms of
their ends, we then show that the homology of an arbitrary graph is bounded
from below by that of an embedded tree that “realises the ends” (Proposition
3.11).

In Section 4, motivated by the observation that there is a surjection:

$H^{\mathrm{uf}}_{1}({\cdot},{\mathbb{Z}})\rightarrow
H^{\mathrm{uf}}_{1}({\cdot},{\mathbb{Z}}_{2}),$

we study $H^{\mathrm{uf}}_{1}({\cdot},{\mathbb{Z}}_{2})$ (note that the
“uniformly finite” condition becomes vacuous in ${\mathbb{Z}}_{2}$ — this may
sound like a bad omen, but turns out to clarify a lot of constructions). In
particular, we get, in Proposition 4.2, a full description of
$H^{\mathrm{uf}}_{1}({\cdot},{\mathbb{Z}}_{2})$ as a direct sum of two factors
corresponding to exactly the first and second phenomena, that is, ends and
“large circuits” respectively. Moreover, it is shown in Theorem 4.8 that the
existence of “large circuits” actually implies infinite-dimensionality of
$H^{\mathrm{uf}}_{1}({\cdot},{\mathbb{Z}}_{2})$.

In Section 5, again in $\mathbb{Z}$, we make precise the claim that “large
circuits” imply non-zero homology. Since “large circuits” is a notion which
depends on the coefficient rings, this result is _not_ a consequence of the
previous section.

In Section 6, inspired by a construction of [4], we define a notion of
“expansion” in higher dimension, which aims at capturing the third phenomenon.
It is easily verified (Lemma 6.3) that our “expansion” implies vanishing of
homology. As a (very partial) converse, we show (Theorem 6.6) that in
dimension one, and for vertex transitive graphs, vanishing implies
“expansion”.

Finally, refining the notion of expansion, we reach, for (infinite) transitive
graphs (or more generally, infinite graphs with a cocompact action by their
automorphism group), a characterisation of the non-vanishing of homology in
terms of the three phenomena of (the refined) non-expansion, ends, and large
circuits.

##  2 Preliminaries

###  2.1 Introduction

We follow the conventions of [5].

If $X$ is a simplicial complex, it is:

Uniformly Locally Finite

(ULF) if there is a uniform bound on the cardinalities of the links of its
vertices.

Uniformly Contractible

(UC) if $\forall r>0\ \exists s(r)>0$ such that any set $A\subseteq X$ of
diameter $\leq r$ is contractible in its $s(r)$-neighbourhood.

###  2.2 Definition of $H^{\mathrm{suf}}$

If $X$ is a ULF simplicial complex, and $A$ is either $\mathbb{R}$,
$\mathbb{Z}$ or $\mathbb{Z}_{2}$, define:

$C^{\mathrm{suf}}_{n}(X,A)=l^{\infty}(X_{(n)},A),$

where $X_{(n)}$ is the set of non-degenerate $n+1$-simplices of $X$. and the
boundary map:

$\displaystyle\partial:C^{\mathrm{suf}}_{n+1}(X)$ $\displaystyle\rightarrow
C^{\mathrm{suf}}_{n}(X)$ $\displaystyle f$ $\displaystyle\mapsto[\partial
f:\tau\mapsto\sum_{\tau<\sigma}i(\tau:\sigma)f(\sigma)],$

where $i(\cdot:\cdot)$ takes care of alternating signs, after choosing an
orientation for $X$.

Then, the _simplicial uniformly finite homology_ of $X$ is defined as

$H^{\mathrm{suf}}_{*}(X)=H(C^{\mathrm{suf}}_{*}(X)).$

###  2.3 Definition of $H^{\mathrm{uf}}$

If $X$ is a ULF graph and $r\in\mathbb{N}$, define the _Rips Complex_ of
radius $r$ as having $n$-simplices:

$(R_{r}(X))_{(n)}=\\{(x_{1},…x_{n})\in X_{(0)}^{n}\ :\ d(x_{i},x_{j})\leq
r\\},$

where $d$ is the graph distance (i.e. length of a shortest path). A simplex of
$R_{r}(x)$ is seen as a “virtual” simplex in $X$. Then $R_{1}(X)=X$,
$R_{r}{X}\subseteq R_{r^{\prime}}(X)$ whenever $r\leq r^{\prime}$, and
$R_{r}(X)$ is itself a ULF simplicial complex. It follows that we have a
directed system:

$\left(C^{\mathrm{suf}}_{\star}(R_{r}(X))\right)_{r\in\mathbb{N}}$

and can define:

$C^{\mathrm{uf}}_{\star}(X):=\lim_{r}C^{\mathrm{suf}}_{\star}(R_{r}(X))$

and finally, the _uniformly finite homology_ of $X$ is defined as:

$H^{\mathrm{uf}}_{\star}(X)=H(C^{\mathrm{uf}}_{\star}(X))$

As a remark, note that $H^{\mathrm{uf}}_{*}(X)$ can also be defined as the
limit of the system

$(H^{\mathrm{suf}}_{*}(R_{r}(X)))_{r}.$

In the following, we will use

$Z^{\mathrm{uf}},B^{\mathrm{uf}},Z^{\mathrm{suf}},B^{\mathrm{suf}},$

for the kernels and image of the boundary maps in the chain complexes
$C^{\mathrm{uf}}$ and $C^{\mathrm{suf}}$ respectively.

The following two important facts are proved in [5].

######  Proposition 2.1 ($H^{\mathrm{uf}}$ is QI-invariant.).

$H^{\mathrm{uf}}$ is a quasi-isometry invariant.

###### Proof.

See [5, Step 3 in the proof of Theorem 12]. ∎

######  Proposition 2.2.

If $X$ is ULF, UC, then $H^{\mathrm{suf}}(X)\cong H^{\mathrm{uf}}(X)$.

###### Proof.

See [5, Step 2 in the proof of Theorem 12]. ∎

###  2.4 “Hands-on” definition of $H^{\mathrm{uf}}_{1}$

The definition of uniformly finite homology is rather abstract; the goal in
this section is to find more concrete description of $H^{\mathrm{uf}}_{1}$.

Let $X$ be a ULF graph, and $A$ either $\mathbb{Z}$ or ${\mathbb{Z}}_{2}$.

An element of $Z^{\mathrm{suf}}_{1}(X,A)$ can be seen as a (uniformly bounded)
_flow_ on $X$; that is, each (directed) edge takes a certain value, which we
view as flowing through the edge. The condition of being a homological cycle
states that at any vertex, the sum of flows of edges directed to the vertex is
the same as the flow of edges directed out of the vertex (“in=out”). The two
building blocks for such flows are circuits and bi-infinite paths (which we
will call “bips”): Indeed, any oriented, graph theoretical circuit in $X$
defines an element of $Z^{\mathrm{suf}}_{1}(X,A)$ by taking the sum of its
(oriented) edges. Similarly, a bi-infinite path (that is, morphism of graphs
from $\mathbb{Z}$ to $X$, injective on edges for convenience) also defines an
element of $Z^{\mathrm{suf}}_{1}(X,A)$. Now, for $A$ either $\mathbb{Z}$ or
${\mathbb{Z}}_{2}$, any element of $f\in Z^{\mathrm{suf}}_{1}(X,A)$ can be
decomposed into a sum of circuits and bips by “following paths”, in such a way
that there is a uniform bound on the number of circuits/bips passing over any
given edge (Take a maximal multiset of circuits “below” f and subtract it;
Take a maximal multiset of bips “below” the result and subtract it; Nothing
remains). From now on, we will not distinguish circuits and bips as paths in
$X$ from their representatives in homology.

If a (possibly infinite) family $(c_{i})_{i\in I}$ of circuits is given, along
with coefficients $({\lambda}_{i}\in A)_{i\in I}$, then we say they define a
_thin_ sum if the family of circuits is locally finite (see [6] for a
motivation of “thinness”).

We can then define the following objects:

$\displaystyle\mathcal{C}_{r}(X,A)$ $\displaystyle:=\text{``sums of circuits
of length $\leq r$ with uniformly bounded coefficients in $A$''}$
$\displaystyle\mathcal{C}_{\infty}(X,A)$
$\displaystyle:=\cup_{r}\mathcal{C}_{r}(X,A)$ $\displaystyle\mathcal{C}(X,A)$
$\displaystyle:=\text{``thin sums of circuits with coefficients in $A$ such
that the result is uniformly bounded''}$

It is easy to see that all of those objects are actually vector subspaces of
$Z^{\mathrm{suf}}_{1}(X,A)$. Note that in the case $A=\mathbb{Z}_{2}$,
$\mathcal{C}_{r}$ can be described simply as the space of sums of circuits of
length $\leq r$, and $\mathcal{C}$ as the space of thin sums of circuits. Note
finally that, letting $\mathfrak{C}_{r}$ stand for the set of circuits of
length $\leq r$, and

$\mathfrak{e}:l^{\infty}(\mathfrak{C}_{r})\rightarrow
l^{\infty}(EX)=C^{\mathrm{suf}}_{1}(X),$

extended linearly from $c\mapsto\text{“its edges”}$, we have
$\mathcal{C}_{r}=\mathfrak{e}\left(l^{\infty}(\mathfrak{C}_{r})\right)$, and
$\mathfrak{e}$ is continuous w.r.t. the pointwise topologies.

We start with a description of $\mathcal{C}_{\infty}$ in terms of our chain
complexes.

######  Lemma 2.3.

If $X$ is a ULF graph, then, for any $r>0$:

$\displaystyle\mathcal{C}_{r}(X,A)$ $\displaystyle\subseteq
Z^{\mathrm{suf}}_{1}(X,A)\cap B^{\mathrm{suf}}_{1}(R_{r/2+1}X,A)$
$\displaystyle Z^{\mathrm{suf}}_{1}(X,A)\cap B^{\mathrm{suf}}_{1}(R_{r}X,A)$
$\displaystyle\subseteq\mathcal{C}_{3r}(X,A).$

Consequently:

$\mathcal{C}_{\infty}(X,A)=Z^{\mathrm{suf}}_{1}(X,A)\cap
B^{\mathrm{uf}}_{1}(X,A).$

###### Proof.

Drop the $A$s.

If $f\in\mathcal{C}_{r}(X)$, $f=\mathfrak{e}\phi$ for some
$\phi:\mathfrak{C}_{r}\rightarrow A$ uniformly bounded. Obviously, $f\in
Z^{\mathrm{suf}}_{1}(X)$. For any $c\in\operatorname{supp}\phi$, one can
“triangulate” $c$: that is, if $c$ is the circuit$(v_{1},…,v_{s})$ with $s\leq
r$, we can consider the sum

$\Delta c:=\sum_{i=1}^{s}(v_{1},v_{i},v_{i+1})\in
C^{\mathrm{suf}}_{2}(R_{2r+1}X),$

with addition $\mod s$. One sees that $\partial(\Delta c)=\mathfrak{e}c$, and
letting

$\Delta\phi:=\sum_{c}g(c)\cdot(\Delta c),$

we get $\Delta\phi\in C^{\mathrm{suf}}_{2}(R_{r/2+1}X)$ by ULF, and
$\partial(\Delta\phi)=\mathfrak{e}\phi=f$. Thus, $f\in
B^{\mathrm{suf}}_{2}(R_{r/2+1}X)$.

Conversely, assume $f\in Z^{\mathrm{suf}}_{1}(X)\cap
B^{\mathrm{suf}}_{1}(R_{r}X)$. Let $g\in C^{\mathrm{suf}}_{2}(R_{r}X)$ such
that $\partial g=f$.

For any two vertices $u,v$ at distance $\leq r$, fix a shortest path
$p_{u,v}=(u=u_{0},u_{1},…u_{s}=v)$ from $u$ to $v$. If $t:=(u,v,w)$ is a
triangle in $R_{r}X$, consider the circuit $Ot$ obtained by concatenating the
paths $p_{u,v},p_{v,w},p_{w,u}$. By construction, $Ot\in\mathfrak{C}_{3r}$.
Let also

$Og:=\sum_{t\in(R_{r}X)_{(3)}}g(t)\cdot(Ot)$

Then $Og\in l^{\infty}(\mathfrak{C}_{3r})$, again by ULF, and by our
assumption that $\partial g\in Z^{\mathrm{suf}}_{1}(X)$, we get:

$\mathfrak{e}(Og)=\partial g.$

It follows that $f=\mathfrak{e}(Og)\in\mathcal{C}_{3r}$.

The last equality of the lemma follows simply by taking the directed unions.

∎

We can now describe $H^{\mathrm{uf}}_{1}(X)$ in a way that only deals with
circuits and paths, and does not deal with Rips complexes:

######  Proposition 2.4.

If $X$ is a ULF graph, then, for any $r>0$, the composite homomorphism:

${\Phi}_{r}:Z^{\mathrm{suf}}_{1}(X,A)\hookrightarrow
Z^{\mathrm{suf}}_{1}(R_{r}X,A)\twoheadrightarrow
H^{\mathrm{suf}}_{1}(R_{r}X,A)$

is surjective and its kernel satisfies:

$\mathcal{C}_{2r}\leq\ker{\Phi}_{r}\leq\mathcal{C}_{3r}.$

Consequently, the composite homomorphism

${\Phi}_{\infty}:Z^{\mathrm{suf}}_{1}(X,A)\hookrightarrow
Z^{\mathrm{uf}}_{1}(X,A)\twoheadrightarrow H^{\mathrm{uf}}_{1}(X,A)$

is surjective and has kernel

$\ker{\Phi}_{\infty}=\mathcal{C}_{\infty}(X,A).$

In particular:

$H^{\mathrm{uf}}_{1}(X,A)\cong\frac{Z^{\mathrm{suf}}_{1}(X,A)}{\mathcal{C}_{\infty}(X,A)}.$

###### Proof.

Drop the $A$s.

Fix $r>0$. We first verify surjectivity of ${\Phi}_{r}$. Fix any $f\in
Z^{\mathrm{suf}}_{1}(R_{r}X)$. Choose for any two $u,v$ in $X$, a unique
shortest path $p_{u,v}=(u=u_{0},u_{1},…,u_{s}=v)$ from $u$ to $v$, and let:

$\tilde{f}:=\sum_{(u,v)\in(R_{r}X)_{(2)}}f(u,v)\cdot p_{u,v},$

that is, $\tilde{f}$ is obtained by replacing “virtual edges” (i.e. edges in
$(R_{r}X)_{(2)}$) by paths. By ULF, $\tilde{f}\in C^{\mathrm{suf}}_{1}(X)$ and
$\partial\tilde{f}=\partial f$, so that $\partial\tilde{f}=0$, hence
$\tilde{f}\in Z^{\mathrm{suf}}_{1}(X)$. We claim that $\tilde{f}-f\in
B^{\mathrm{suf}}_{1}(R_{r}X)$, which will clearly imply surjectivity of
${\Phi}_{r}$. Indeed, if for any $(u,v)$ at distance $\leq r$, one considers:

$\Delta(u,v):=\sum_{i=1}^{s-1}(u_{0},u_{i},u_{i+1})\in
C^{\mathrm{suf}}_{2}(R_{r}X)$

where $(u=u_{0},u_{1},…,u_{s}=v)$ is a path from $u$ to $v$, then
$\partial\Delta(u,v)=(u,v)-p_{u,v}$. Summing over the virtual edges $(u,v)$ in
the support of $f$, we get:

$\partial\left(\sum_{(u,v)\in(R_{r}X)_{(1)}}f(u,v)\cdot\Delta(u,v)\right)=f-\tilde{f}.$

This shows that, indeed, ${\Phi}_{r}$ is surjective, assuming the sum in the
parentheses is in $C^{\mathrm{suf}}_{2}(R_{r}X)$ which holds, as usual, by
ULF. Now, the inclusions related to $\ker{\Phi}_{r}$ remain, but these follow
at once from the first part of Lemma 2.3.

The properties of ${\Phi}_{\infty}$ follow from the second part of Lemma 2.3
and the fact that:

$H^{\mathrm{uf}}_{1}(X)=\frac{\bigcup_{r}Z^{\mathrm{suf}}_{1}(R_{r}X)}{\bigcup_{r}B^{\mathrm{suf}}_{1}(R_{r}X)}.$

∎

For later use, we note also that:

######  Corollary 2.5.

Let $X$ a ULF graph and assume that
$\mathcal{C}_{r}(X,A)=\mathcal{C}_{\infty}(X,A)$ for some $r>0$. Then, for any
$s$ large enough, the map ${\Phi}_{s}$ descends to an isomorphism:

$\tilde{\Phi}_{s}:Z^{\mathrm{suf}}_{1}(X,A)/\mathcal{C}_{r}(X,A)\rightarrow
H^{\mathrm{suf}}_{1}(R_{s}X,A),$

and so does ${\Phi}_{\infty}$:

$\tilde{\Phi}_{\infty}:Z^{\mathrm{suf}}_{1}(X,A)/\mathcal{C}_{r}(X,A)\rightarrow
H^{\mathrm{uf}}_{1}(X,A).$

Furthermore, the following diagram (of isomorphisms) commutes:

${\frac{Z^{\mathrm{suf}}_{1}(X,A)}{\mathcal{C}_{r}(X,A)}}$${H^{\mathrm{suf}}_{1}(R_{s}X,A)}$${H^{\mathrm{uf}}_{1}(X,A)}$$\scriptstyle{\tilde{\Phi}_{s}}$$\scriptstyle{\tilde{\Phi}_{\infty}}$$\scriptstyle{[\iota]}$

###### Proof.

Drop the $A$s. Consider the maps ${\Phi}_{s},{\Phi}_{\infty}$ of Proposition
2.4. Since for all $s\geq r$,
$\mathcal{C}_{\infty}(X)=\mathcal{C}_{s}(X)=\mathcal{C}_{r}(X)$, the equations
involving the respective kernels give us the desired isomorphisms.
Commutativity of the triangles follows, since all maps involved are
(appropriate quotients of) inclusions. ∎

##  3 Trees and End-Defining Trees

Expanding a result of [7], we describe the first homology of any (ULF) tree
$T$ as $l^{\infty}(B)$, for some set $B$ of bips in $T$, of cardinality the
number of ends of $t$ (minus $1$ in the finite case).

We will then show that the first homology of a sufficiently well-behaved
subtree of a graph $X$ injects into the first homology of $X$.

From now on, by a trivalent tree, we mean a tree with vertices of degree
either $2$ or $3$. We first verify that in terms of uniformly finite homology,
leaves (that is, vertices of degree $1$) can be safely forgotten.

######  Lemma 3.1.

Take $A=\mathbb{Z},\mathbb{Z}_{2}$; then
$H^{\mathrm{suf}}_{1}(T,A)=H^{\mathrm{suf}}_{1}(T^{\prime},A)$ where
$T^{\prime}$ is the maximal subtree of $T$ without leaves (i.e. vertices of
degree $1$).

###### Proof.

Since $T$ has no non-trivial circuits, we have that
$H^{\mathrm{suf}}_{1}(T,A)=Z^{\mathrm{suf}}_{1}(T,A)$ (Proposition 2.4).
Recall than an element $c\in Z^{\mathrm{suf}}_{1}(T,A)$ is a flow on $T$, and
since leaves correspond to dead-end for the flow, the edges adjacent to leaves
must have zero flow. Arguing inductively implies then that $c$ must have
support in $T^{\prime}$, and the equality is achieved. ∎

To reduce to trivalent trees, it is now enough to note that any ULF tree is
quasi-isometric to trivalent ones. Indeed suppose such a tree $T$ is given. A
trivalent “unfolding” of $T$ can be obtained by replacing each star around a
vertex as in Figure 1. The uniform bound on degrees ensures that the result,
$\tilde{T}$, is quasi-isometric to $T$. Since $H^{\mathrm{uf}}_{1}$ is QI-
invariant, it is therefore sufficient to study trivalent trees.

Figure 1: Star to Comb

###  3.1 Homology of Trees

We want to better describe the first uniformly finite homology of trees.
Recall that we essentially have two kinds of flows on a graph $X$: circuits
and bips. In case $X$ is a tree, circuits are never reduced, thus define zero
flow. The “building blocks” for a flow on $X$ are therefore bips. We will
extract from $X$ a specific set $P$ of bips which suffices to fully describe
flows on $X$. The construction of $P$ will be explicit but only some abstract
“tameness” properties of $P$ will actually be used.

In order to ease notation we consider the following construction of a
trivalent tree (with no leaves).

  * •

Start with a single vertex $v_{0}$, and call that graph $T_{0}$.

  * •

Let $T_{1}$ be the tree corresponding to $\mathbb{Z}$, containing $T_{0}$ as
$v_{0}=0$. $T_{1}$ can be viewed as two copies of $\mathbb{N}$ glued on
$v_{0}$.

And now inductively:

  * •

In $T_{n}$, choose a set $B_{n}$ of vertices that does not intersect
$T_{n-1}$; call them _branching points_ of depth $n$. To each $v\in B_{n}$,
glue a copy called $R_{v}$ of $\mathbb{N}$ at zero (a new branch). For
$i\in\mathbb{N}$, write $i_{v}$ for the vertex $i$ of $R_{v}\simeq\mathbb{N}$,
and ${\infty}_{v}$ for its end (this is not a vertex). Let $T_{n+1}$ be the
result. Formally:

$T_{n+1}=(T_{n}\sqcup\bigsqcup_{v\in B_{n}}R_{v})/\\{0_{v}\sim v\ |\ v\in
B_{n}\\}.$

Obviously, $T_{n}\leq T_{n+1}$.

Let us call each $R_{v}$ the _ray_ at $v$ and call $n$ its depth if it is in
$T_{n+1}-T_{n}$. The construction $(T_{n},B_{n})_{n\in\mathbb{N}}$ has depth
$N$ if this construction ends at step $T_{N}$, i.e. if for $k>N$ we have
$B_{k}=\emptyset$. The construction has infinite depth otherwise:
$T={\bigcup}_{n}T_{n}$.

Note that for each $n$, the set $B_{n}$ of branching points of depth $n$ can
be partitioned as:

$B_{n}=\bigsqcup_{v\in B_{n-1}}B_{n}\cap(R_{v}-\\{v\\}),$

since each new branching point has to lie on $T_{n}-T_{n-1}=\bigsqcup_{v\in
B_{n-1}}R_{v}-\\{v\\}$. Let $B_{n,v}:=B_{n}\cap R_{v}-\\{v\\}=B_{n}\cap R_{v}$
denote the branching points lying on $R_{v}$. There is an obvious ordering on
$B_{n,v}$ as a subset of $\mathbb{N}$: write $v_{v,i}$ the $i$-th element for
this ordering (where $i<\alpha$, and $\alpha\in\mathbb{N}\cup\\{\infty\\}$ is
the cardinal of $B_{n,v}$).

######  Proposition 3.2.

All (infinite) trivalent leafless trees are constructed in this fashion,
modulo one edge.

###### Proof.

Start with a vertex $v_{0}\in T$ of degree two (if such a vertex does not
exist, split an edge to obtain one). Follow any two paths starting from
$v_{0}$: this defines $T_{1}$. Let $B_{1}:=\\{v\in T_{1}:\
\mathrm{deg}(v)>2\\}$, and from each $v\in B_{1}$, follow any new ray to get
$T_{2}$. Proceed inductively like that, and let
$T_{\omega}:={\bigcup}_{n}T_{i}$. It remains to see that $T_{\omega}=T$, but
this follows from the fact that if $v\in VT$ is at distance $d$ from $v$, then
necessarily $v\in T_{d}$. ∎

We now describe how to construct the set $P$ of bips inductively, by following
the construction of $T$. Fix $T$ a trivalent tree, and consider a sequence
$(T_{n},B_{n})_{n\in\mathbb{N}}$ that matches the construction above. We
construct $P$ by induction: $P_{n+1}$ extends $P_{n}$ and is a tame set of
bips for $T_{n+1}$:

$T_{0}$:

no bip: $P_{0}=\emptyset$.

$T_{1}$:

let $P_{1}$ contain the unique bip in $P$ : $P_{1}=\\{T_{1}\\}$.

$T_{n+1}$:

Assume now $P_{i}$ has been constructed for $i\leq n$.

We will define $P_{n+1}=P_{n}\sqcup P_{n+1}^{\prime}$, with $P_{n+1}^{\prime}$
the set of _new_ bips, as follows:

First fix $u\in B_{n-1}$, and consider the ray $R_{u}$ in $T_{n}$. We recall
that the branching points of $B_{n+1}$ can be partitioned by the rays on which
they lie. Likewise, we define new bips for each ray of depth $n$:

$P_{n+1,u}^{\prime}:=\begin{cases}\\{R_{v_{u,i}}^{-1}+[v_{u,i},v_{u,i+1}]+R_{v_{u,i+1}}\
|\ i\in\mathbb{N}\\}&\text{ if $|B_{n,u}|=\infty$}\\\
\\{(R_{v_{u,i}}^{-1}+[v_{u,i},v_{u,i+1}])+R_{v_{u,i+1}}\ |\
i<k\\}\cup\\{R_{v_{u,k}}^{-1}+[v_{u,k},{\infty}_{u}]\\}&\text{ if
$|B_{n,u}|=k+1$}\end{cases}$

where $(v_{u,i})_{i<|B_{n,u}|}$ is the ordering of $B_{n,u}$. And then

$P_{n+1}^{\prime}:=\bigsqcup_{u\in B_{n-1}}P_{n+1,u}^{\prime},\qquad
P_{n+1}:=P_{n}\sqcup P_{n+1}^{\prime}.$

If $(T_{n},B_{n})_{n\in\mathbb{N}}$ has infinite depth, let
$P=\bigcup_{n}P_{n}$. We then define a (partial) order on $P$ as follows. If
$p\in P_{n}$ and $q\in P_{n+1}$ share an edge, let $p\leq q$, and take the
transitive closure. Say that $p\in P$ is the _last_ bip over an edge $e$ if
$p$ lies over $e$ and for any other $q\in P$ lying over $e$, we have $q<p$.
Write

$M_{p}:=\\{e\in ET:\ \text{$p$ is last over $e$}\\}.$

We can now describe the properties of $P$ we will use later on:

######  Proposition 3.3 ($P$ is tame).

Let $(T_{n},B_{n})_{n\in\mathbb{N}}$ be a tree construction as above, and
$(P_{n})_{n\in\mathbb{N}}$ the corresponding construction of bips. Let also
$\leq$ be the partial order on $P$ described above.

  1. 1.

Each edge is covered by at most three bips.

  2. 2.

All elements of $P_{n}^{\prime}$ lie in $T_{n+1}-T_{n-1}^{+}$ (where
$T_{k}^{+}$ is the $1$-neighbourhood of $T_{k}$).

  3. 3.

$P_{n}$ covers $T_{n}$.

  4. 4.

Each $p\in P$ has finitely many predecessors (with respect to $\leq$), and is
the last bip over at least one edge (i.e. $M_{p}\neq\emptyset$).

  5. 5.

For any edge $e$ where $p\in P_{n+1}$ is last over $e$, there exists
$e^{\prime}\in T_{n}^{+}$ such that:

$\\{q\in P\ :\ e^{\prime}\in q\\}=\\{q\in P\ :\ e\in q<p\\}=\\{q\in P_{n}\ :\
e\in q\\}=:Q_{p}.$

  6. 6.

Each set $M_{p}$ is connected.

###### Proof.

By construction. ∎

We can actually endow $P$ with a well-ordering (extending $\leq$), which will
facilitate the proof in Theorem 3.6.

######  Lemma 3.4.

If $(P,\leq)$ is a countable poset such that $\\{\leq p\\}$ is finite for all
$p$, there exist a well ordering $\preccurlyeq$ on $P$ extending $\leq$ (i.e.
$p\leq q\Rightarrow p\preccurlyeq q$) and such that $\\{\preccurlyeq p\\}$ is
still finite for any $p$.

###### Proof.

Define

$\displaystyle P_{0}$ $\displaystyle=\left\\{p\in P\ :\
\\{<p\\}=\emptyset\right\\}$ $\displaystyle P_{n+1}$
$\displaystyle=\left\\{p\in P\ :\
\\{<p\\}\subseteq\bigcup_{j=1}^{n}P_{j}\right\\}$

and note that, assuming $P$ is non-empty:

  * •

$P_{0}\neq\emptyset$ and $P=\bigcup P_{j}$, since $\\{\leq p\\}$ is finite for
all $p$;

  * •

The elements of $P_{j}$ are all pairwise incomparable (endow them all of any
well-ordering)

  * •

Each $P_{j}$ is countable (since $P$ is).

For definiteness, assume that $P$ is infinite countable (the finite case is
treated similarly). It remains then to construct a bijection
$\sigma:\mathbb{N}\rightarrow P$ such that $\sigma n<\sigma m\Rightarrow n<m$.

The algorithm in Algorithm 1 constructs $\sigma$. The idea is to “diagonally”
(as in the usual bijection $\mathbb{N}^{2}\leftrightarrow\mathbb{N}$) exhaust
all $P_{i}$s, while preserving the existing order.

Algorithm 1 Extending partial order to total order.

1:$i\leftarrow 0,\ \ j\leftarrow 0,\ \ k\leftarrow 0,\ \
\sigma\leftarrow\emptyset$

2:while $\exists p\in P_{j}-\operatorname{im}\sigma\text{ such that }\ \forall
q<p:q\in\operatorname{im}\sigma$ do

3: $\sigma\leftarrow\sigma\cup\\{i\mapsto p\\}$

4: $i\leftarrow i+1$

5: if $j\leq k$ then

6: $j\leftarrow j+1$

7: else

8: $j\leftarrow 0,\ \ k\leftarrow k+1$

∎

######  Corollary 3.5.

The set of bips $P$ can be assumed to be well ordered without breaking any of
the properties in Proposition 3.3 (that is, $P$ stays “tame”). In particular,
any edge has a last bip on it:

$ET=\bigsqcup_{p}M_{p}$

We prove that if $T$ is a tree (with uniformly bounded vertex degree) and $P$
is as constructed above, then we can use it to describe
$H^{\mathrm{uf}}_{1}(X)$. We invite the reader to have a quick peak at the
theorem. Before proceeding with the statement, let us first motivate the
tameness properties of $P$. The imposed order and the finiteness condition
help in making an inductive argument. The non-emptiness of the sets $M_{p}$
guarantees that the chosen bips are independent, that is, there is no bounded
sum that is zero in homology. Once we have independence of bips, the
connectedness of $M_{p}$ will guarantee generation of all homology classes.
The reader might have the feeling that we are constructing some kind of
infinite basis — this would be correct, as the theorem states. Finally the two
remaining conditions just make sure that we only have to consider uniformly
bounded sums of bips.

######  Theorem 3.6.

Given a trivalent tree $T$, a set $P$ of bips on $T$ as constructed above and
$A\in\\{\mathbb{Z},\mathbb{Z}_{2}\\}$, the following map:

$\displaystyle l^{\infty}(P)$ $\displaystyle\rightarrow
H^{\mathrm{suf}}_{1}(T,A)\cong H^{\mathrm{uf}}_{1}(T,A)$ $\displaystyle f$
$\displaystyle\mapsto c_{f}:=[e\mapsto\sum_{e\in p\in P}f(p)]$

is an isomorphism (continuous when $A=\mathbb{Z}$ and
$H^{\mathrm{suf}}_{1}(T,A)$ has the $l^{\infty}$ norm).

###### Proof.

By Corollary 3.5, we can assume that $P$ is well-ordered.

The map above is easily seen to be linear, and of norm $\leq K$ (where
$|P_{e}|\leq K$ for all $e\in E$).

Injectivity: By induction. If $f\neq 0\in l^{\infty}(P)$, there exists a bip
$p$ such that $f(p)\neq 0$; take a least such $p$. Consider then $e\in M_{p}$:

$\displaystyle c_{f}(e)=\sum_{e\in q\in P}f(q)$ $\displaystyle=\sum_{e\in
q<p}f(q)+f(p)$

Since by assumption, for each $e\in q<p$, $f(q)=0$, it follows that
$c_{f}(e)\neq 0$.

Surjectivity: Take a cycle $c\in Z^{\mathrm{suf}}_{1}(T,A)$. Then, define
inductively for $p\in P$:

$f(p):=c(e)-\sum_{e\in q<p}f(q),$

where $e\in M_{p}$. We check that this is independent of the choice of $e\in
M_{p}$. By connectedness of $M_{p}$, it is enough to verify that adjacent
edges $e_{1},e_{2}$ in $M_{p}$ yield the same value. There are only two
possible configurations locally:

$e_{1}$$e_{2}$

$e_{1}$$e_{2}$$e_{3}$

Horizontally one can see the bip $p$, for which we are defining the value
$f(p)$. The green ($G$), red ($R$) and blue ($B$) lines represent groups of
bips running over these vertices (possibly empty). Note that all these bips
are in order strictly smaller than $p$ (because $p$ monopolises
$e_{1},e_{2}$). Hence $f$ is defined for all of them. We check that $f(p)$
does not depend on whether we choose $e_{1}$ or $e_{2}$. There are two
situations possible depicted above. For the first picture the claim clearly
holds, we check the second picture:

$\displaystyle c(e_{1})-\sum_{q\in G\text{ or }q\in R}f(q)$
$\displaystyle=c(e_{2})+c(e_{3})-\sum_{q\in G\text{ or }q\in R}f(q)$
$\displaystyle=c(e_{2})+\sum_{q\in R}f(q)-\sum_{q\in B}f(q)-\sum_{q\in G\text{
or }q\in R}f(q)$ $\displaystyle=c(e_{2})-\sum_{q\in G\text{ or }q\in B}f(q)$

The second equality holds by the induction argument.

Now $f$ is uniformly bounded: from

$f(p):=c(e)-\sum_{e\in q<p}f(q),$

we see that

$|f(p)|\leq\|c\|_{\infty}+\|\sum_{e\in q<p}f(q)\|\leq 2\|c\|_{\infty},$

since the elements of $\\{e\in q<p\\}$ are assumed to share an edge (so that
the sum over $\\{e\in q<p\\}$ is exactly the value $c(e^{\prime})$ of some
edge $e^{\prime}$).

It is now easily seen that $c_{f}=c$, and since $\|f\|_{\infty}\leq
2\|c\|_{\infty}$, the desired isomorphism is reached.

∎

Note that only some of the abstract “tameness” properties of the set $P$
described in Proposition 3.3, plus well-ordering, are actually used in the
proof. Furthermore, one could work without well-ordering (well-foundedness of
$\leq$ is enough), at the cost of a less transparent proof.

As an example of necessity of the “tameness” conditions, we give Figure 2; the
set of bips chosen satisfy each condition but the fifth one — the existence of
an edge $e^{\prime}$ on which exactly the predecessors of a given bip $p$ run.
As a result the flow defined on the teeth of the comb cannot be described as a
bounded sum of the given bips.

$1$$-1$$1$$-1$$1$$-1$$1$$-1$ Figure 2:  A bi-infinite comb with bips as shown
in green and red. The cycle we consider is defined by alternating positive and
negative values on the teeth of the comb. The reader can check that the
obtained function on the bips is unbounded.

###  3.2 Ends

Now that the homology of trees is understood, we can relate it to homology of
arbitrary graphs via their ends. The core idea is that, when working with ends
of graphs, working with nicely embedded trees is enough.

We start with a definition:

######  Definition 3.7 (End-respecting (defining) trees).

Let $X$ be an arbitrary ULF graph. An end-respecting (resp. end-defining) tree
for $X$ is a leafless tree $T\leq X$ satisfying: Any end of $X$ is represented
by at most one (resp. a unique) infinite branch of $T$. Furthermore, $T$ is
said to _have separators_ if: There exists a sequence $(K_{n})_{n}$ of
increasing finite, connected subsets of $VX$ satisfying:

  * •

$T\cap K_{n}$ is connected for all $n$;

  * •

$\bigcup_{n}K_{n}=VX$;

  * •

For each infinite connected component $L$ of $X-K_{n}$, there is exactly one
edge of ${\partial}_{T}(T\cap K_{n})$ 222${\partial}_{X}Y=$ edges of $X$ with
exactly one endpoint in $Y$. incident to $L$, and vice versa. This defines a
bijection between infinite connected components of $X-K_{n}$ and
${\partial}_{T}(T\cap K_{n})$.

Fix $v_{T}$ a root for $T$, chosen in $K_{0}$.

######  Remark 3.8.

End-defining trees with separators can always be constructed. Indeed, fix $X$
and some $x\in VX$. Consider the smallest $R>0$ such that removing $B(x,R)$
from $X$ disconnects the graph in multiple infinite components. Now choose as
many edges in the boundary of the ball as there are infinite components, and a
geodesic for each of these starting in $x$. Combining these geodesics gives a
subgraph, and deleting edges as necessary yields a tree. Now one repeats this
approach inductively: Suppose we have a finite tree defined up to $B(x,R)$,
with boundary edges $l_{i}$ in correspondence to infinite connected components
$L_{i}$ (of $X-B(x,R)$). Take $\tilde{R}>R$ least such that $X-B(x,\tilde{R})$
has more components than $X-B(x,R)$. If a connected component $L_{I}$ of
$X-B(x,R)$ is further separated in components $L_{I}^{j}$ in
$X-B(x,\tilde{R})$, with $j\in\\{1\dots k\\}$ with distinguished boundary
edges $l_{I}^{j}$ for each component in the boundary of $B(x,\tilde{R})$, then
as before we take geodesics from $l_{I}$ to each $l_{I}^{j}$ and delete
possible cycles.

On the other hand, it is not clear to us whether one can always find
separators for end-respecting trees.

The following technical result is close to [1, Lemma 2.4]. Given
$\phi,\psi:EX\rightarrow\mathbb{Z}$, say that $\psi\leq\phi$ if, for any $e\in
EX$: $\phi e\geq 0$ implies $0\leq\psi e\leq\phi e$ and $\phi e\leq 0$ implies
$\phi e\leq\psi e\leq 0$.

######  Lemma 3.9.

Let $\phi\in Z^{\mathrm{suf}}_{1}(X,\mathbb{Z})$ and
$r:EX\rightarrow\mathbb{Z}$ with $r\leq\phi$, $r\notin
Z^{\mathrm{suf}}_{1}(X,\mathbb{Z})$, and $\partial r\geq 0$ Then there exists
$\hat{r}\in Z^{\mathrm{suf}}_{1}(X,\mathbb{Z})$ satisfying
$r\leq\hat{r}\leq\phi$. Furthermore, $\hat{r}$ and $r$ are different on
infinitely many edges.

Here is a picturesque interpretation: View $\phi$ as closed (in the sense that
in$=$out) flow on the graph $X$. Then $r$ is a non-closed flow that’s no
greater than $\phi$: any flow on the edges is in the same direction as $\phi$,
and not stronger. Then $r$ can be made into a closed flow by simply adding
flow to edges in the directions allowed by $\phi$ (so, no reduction of what
$r$ had and not exceeding what $\phi$ allows).

###### Proof.

We extend $r$ step by step, and see that the sequence of extensions converges:

  1. 1.

For any $v$ with $\partial r(v)>0$, we have:

$\sum_{u\sim v}\phi(u,v)=0,$

so, there must exist $e_{v}=(u,v)$ with $\phi(e_{v})<r_{v}(e)\leq 0$.

  2. 2.

Consider now $r^{1}:=r+\sum_{v\in V_{0}}e_{v}$, where $V_{0}$ is the set of
$v$s with $\partial r(v)>0$. We notice that this sum is well-defined since no
$e_{v}$ appears in two distinct $v$s, and $r\leq r^{1}\leq\phi$ by
construction. Also, $\partial r^{1}\geq 0$ since if $v\in V_{0}$, then $v$
stays (non-strictly) positive in $\partial r^{1}$, and the other $v$s can only
increase, by the direction of the added edges.

  3. 3.

Redo the argument with $r$ replaced by $r^{1}$, and construct inductively a
sequence $r=r^{0}\leq r^{1}\leq\ldots\leq\phi$.

  4. 4.

Since the sequence $(r^{i})_{i}$ is monotonous and “below” $\phi$, it must
converge to some $r\leq\hat{r}\leq\phi$, in
$Z^{\mathrm{suf}}_{1}(X,\mathbb{Z})$, by boundedness of $\phi$.

  5. 5.

The statement that $r$ and $\hat{r}$ differ on an infinite set can be deduced
from the fact that the process of extension $r^{i}\leadsto r^{i+1}$ cannot end
after finitely many enumerate. Indeed, if $v\in V_{i}$, and $(u,v)=e_{v}$ is
the edge chosen to “cancel” $v$, then $u$ becomes positive in $\partial
r^{i+1}$, so that there are always positive edges.

∎

A similar result holds for $\mathbb{Z}_{2}$. In $EX\rightarrow\mathbb{Z}_{2}$,
say $r\leq\phi$ iff $\phi(e)=0$ implies $r(e)=0$ for all $e\in EX$.

######  Lemma 3.10.

Let $X$ a (ULF) graph, and $\phi\in Z^{\mathrm{suf}}_{1}(X,\mathbb{Z}_{2})$,
$r\in C^{\mathrm{suf}}_{1}(X,\mathbb{Z}_{2})$ with $r\leq\phi$, and $\partial
r={\delta}_{v}$ for some vertex $v$. Then, there exists $r\leq\hat{r}\leq\phi$
with $\hat{r}\in Z^{\mathrm{suf}}_{1}(X,\mathbb{Z}_{2})$ and such that
$\hat{r}$ differs from $r$ on infinitely many edges.

###### Proof.

By mimicking the $\mathbb{Z}$ case. Since $\sum_{e\sim
v}\phi(e)=0\neq\sum_{e\sim v}r(e)$, there exists $e$ with $\phi(e)=1,r(e)=0$.
Thus, define $r^{1}:=r+e$ and continue by induction. ∎

We can now prove the following.

######  Proposition 3.11 (End-respecting trees inject in
$H^{\mathrm{uf}}_{1}$).

Let $X$ be ULF and $T$ an end-respecting tree with separators. Then:

$H^{\mathrm{uf}}_{1}(T,\mathbb{Z})\leq H^{\mathrm{uf}}_{1}(X,\mathbb{Z}).$

The same holds for $\mathbb{Z}_{2}$.

###### Proof.

We first consider $\mathbb{Z}$. Let $(K_{i})_{i}$ the sequence of “separators”
associated to $T$.

  1. 1.

First, recall that $H^{\mathrm{uf}}_{1}(X,\mathbb{Z})\cong
Z^{\mathrm{suf}}_{1}(X,\mathbb{Z})/\mathcal{C}_{\infty}$, so, in particular,
$H^{\mathrm{uf}}_{1}(T,\mathbb{Z})\cong Z^{\mathrm{suf}}_{1}(T,\mathbb{Z})$,
$T$ being a tree. From now on, drop the $\mathbb{Z}$. We consider the
function:

$Z^{\mathrm{suf}}_{1}(T){\overset{\subseteq}{\rightarrow}}Z^{\mathrm{suf}}_{1}(X){\overset{/}{\rightarrow}}Z^{\mathrm{suf}}_{1}(X)/\mathcal{C}_{\infty}$

so that showing injectivity amounts to showing that no non-zero function in
$Z^{\mathrm{suf}}_{1}(T)$ can be written as an element of
$\mathcal{C}_{\infty}$. We actually show more: the intersection of
$Z^{\mathrm{suf}}_{1}(T)$ and $\mathcal{C}$ is trivial.

  2. 2.

Assume, towards a contradiction, that there exists $0\neq f\in
Z^{\mathrm{suf}}_{1}(T)\cap\mathcal{C}$. In particular there exists a thin set
of circuits $C$ such that:

$f=\sum C.$

(Formally, one should add integer coefficients to the sum, but we’ll stick to
the shorthand).

  3. 3.

Since $T$ is end-respecting, there exists some $K=K_{n}$ and an edge
$e\in\operatorname{supp}f$ so that $K$ disconnects the space in multiple
infinite components, one of which – call it $L$ – is such that
$\operatorname{supp}f\cap\partial L$, consists of exactly one edge $e$.

  4. 4.

Up to taking $-f$, we may assume that $e$ is directed “away from $L$”, and
that $f$ is positive on $e$.

  5. 5.

Following $f$ backwards starting from $e$, we get a set of $k:=f(e)$ rays
$r_{1},\ldots,r_{k}$ supported in $T\cap(L\cup\\{e\\})$and all ending with
$e$.

  6. 6.

Let $C_{L}$ the set of circuits in $C$ that lie in $L$ and do not intersect
$K$.

Let $r$ be the sum of initial segments of each $r_{i}$, so that $r$ has
support not intersecting the circuits of $C$ not in $C_{L}$ (only a finite
number of them have to be taken into account since $C$ is thin and $K$
finite). Note that $r\leq f$ by construction.

Let $f_{L}$ be the restriction of $f-(\sum r_{i})$ to $L$.

  7. 7.

Remark that $f_{L}\in Z^{\mathrm{suf}}_{1}(L)$, $r\leq C_{L}-f_{L}$ and, far
enough from $K$, we have:

$r=C_{L}-f_{L}$

  8. 8.

By Lemma 3.9, there exists $\hat{r}\in Z^{\mathrm{suf}}_{1}(L)$ with
$r\leq\hat{r}\leq C_{L}-f_{L}$, $\hat{r}$ distinct from $r$ at infinitely many
edges.

  9. 9.

Then $\hat{r}$ must be distinct from $r$ away from $K$. Since
$r\leq\hat{r}\leq C_{L}-f_{L}$ and, away from $K$, $r=C_{L}-f_{L}$, we reach a
contradiction.

The case of $\mathbb{Z}_{2}$ is a simple adaptation. We follow the same
procedure up to step 5. There, we can just follow _one_ ray $r_{1}$ along $f$.
Then, at step 8, we use Lemma 3.10 instead of Lemma 3.9 to extend $r$ to some
bip $\hat{r}$, with $\hat{r}\leq\sum C_{L}-f_{L}$, and differing from $r$
arbitrarily far from $K$, from which we again deduce a contradiction.

∎

We extract the core property used in the proof for future use:

######  Corollary 3.12 (of the proof).

With the assumptions of Proposition 3.11:

$Z^{\mathrm{suf}}_{1}(T)\cap\mathcal{C}(X)=0.$

######  Corollary 3.13.

The dimension of $H^{\mathrm{uf}}_{1}(X,\mathbb{Z})$ is at least the number of
ends of $X$ minus $1$.

Recall that any finite subset in a graph induces a partition of its ends given
by the connected component in the complement of the set in which they lie.

######  Lemma 3.14.

Let $T_{1},T_{2}$ be two end-defining trees in the ULF graph $X$. For any edge
$e$ in $T_{1}$, there exists a compact $K\subset X$, such that the partition
$\mathcal{E}$ induced by $K$ on the ends of $T_{2}$ is a subpartition of the
one induced by $e$ on $T_{1}$. In particular, the shadow topology induced on
the $\partial X=\partial T_{1}=\partial T_{2}$ by $T_{1}$ is the same as the
topology induced by $T_{2}$.

###### Proof.

Write $\\{E^{+},E^{-}\\}$ for the partition of ends induced by $e$ in $T_{1}$.
Fix a vertex $x$ on $T_{1}$ and $x_{2}$ on $T_{2}$ Suppose for every ball
$B_{R}$ with radius $R$ and center $x$, the partition of ends induced by it on
$T_{2}$, indirectly given by the connected components of $T_{2}-B$, is not a
subpartition, then there are ends $\varepsilon_{i}$,
$\varepsilon_{i}^{\prime}$ such that in $T_{2}-B_{i}$ they are in the same
connected component, but $\varepsilon_{i}\in E^{-}$ and
$\varepsilon_{i}^{\prime}\in E^{+}$ Now take rays $r_{i},r_{i}^{\prime}$
starting at $x$ (resp. $\tilde{r}_{i}$, $\tilde{r}_{i}^{\prime}$ starting at
$x_{2}$) in $T_{1}$ (resp. $T_{2}$), such that their limits are the ends
$\varepsilon_{i}$ and $\varepsilon_{i}^{\prime}$ Note that since
$\tilde{r}_{i}$ and $\tilde{r}_{i}^{\prime}$ are in the same connected
component, they overlap on their first vertices inside $B_{i}$ This means that
they have the same pointwise limit On the other hand consider $r_{i}$ and
$\tilde{r}_{i}$ Take any end $\varepsilon$ of $X$, then it corresponds to a
choice of connected components of $X$ after taking away bigger and bigger
compact sets (say for example the balls $B_{R}$) However since
$r_{i}(\infty)=\tilde{r}_{i}(\infty)$, they are in the same connected
component if one goes sufficiently far However this means that any consistent
set of pointwise convergence accumulation points of $\\{r_{i}\\}_{i}$ and
$\\{\tilde{r}_{i}\\}_{i}$ have the same end After restricting to a subsequence
we can assume that both sequences converge pointwise Then, because of previous
arguments, the end of $(\lim r_{i})(\infty)=(\lim\tilde{r}_{i})(\infty)$,
which is equal to $(\lim\tilde{r}_{i}^{\prime})(\infty)$ and thus equal to
$(\lim r_{i}^{\prime})(\infty)$ However this is a contradiction, since the
ends of $r_{i}$ are in $E^{-}$ and those of $r_{i}^{\prime}$ are in $E^{+}$
Since both kinds of infinite rays are separated by the edge $e$, also the ends
of the limit rays are in these respective sets. ∎

The result in Proposition 3.11 can actually be generalised to end-respecting
trees _without_ separators.

######  Proposition 3.15.

Let $X$ be ULF and $T\leq X$ an end-respecting tree. Then:

$H^{\mathrm{uf}}_{1}(T,\mathbb{Z})\leq H^{\mathrm{uf}}_{1}(X,\mathbb{Z}).$

The same holds for $\mathbb{Z}_{2}$.

###### Proof.

Let $T\leq X$ such a tree, and $\tilde{T}$ an end-defining tree for $X$ with
separators.

  * •

We show that
$H^{\mathrm{suf}}_{1}(T,\mathbb{Z})\cap\mathcal{C}(X,\mathbb{Z})=\\{0\\}$, and
then the argument in the first step of the proof of Proposition 3.11 gives the
desired embedding.

  * •

Towards a contradiction, assume that $f\in
H^{\mathrm{suf}}_{1}(T,\mathbb{Z})\cap\mathcal{C}$ is non-zero. By the
decomposition of $H^{\mathrm{suf}}_{1}(T,\mathbb{Z})$ into $l^{\infty}(P)$ for
a set of bips $P$, one can write

$f=\sum_{b\in P}a_{b}\cdot b.$

  * •

For any bip $b\in P$, one can consider its “parallel” bip $\tilde{b}$ in the
tree $\tilde{T}$; that is, $b$ has two ends corresponding to ends of the graph
$X$, which correspond also to two ends in $\tilde{T}$ (since $\tilde{T}$ is
end-defining), and finally to a unique bip. Let $\tilde{P}:=\\{\tilde{b}\ |\
b\in P\\}$ and

$\tilde{f}=\sum_{b\in P}a_{b}\cdot\tilde{b}.$

Note that by “projecting”, one can find a collection of circuits $C_{f}$ (and
coefficients $a_{c}$) such that

$\tilde{f}=f+\sum_{c\in C_{f}}a_{c}\cdot c$

(the construction for $C_{f}$ is similar to the one in the proof of
Proposition 4.2).

  * •

We claim that $\tilde{f}$ is well-defined (but potentially unbounded) and non-
zero.

  * •

Assume for now that this is the case. Since $\tilde{f}$ is non-zero, there
exists a first separator $K_{i}$ and an edge $e$ lying on $\partial K$ such
that $\tilde{f}$ is zero on $K$ and non-zero on $e$.

  * •

Let $C_{f}^{K}$ denote those circuits in $C_{f}$ that intersect the separator
$K_{i+1}$ that follows $K_{i}$ and

$f^{\prime}:=f-\sum_{c\in C_{f}^{K}}a_{c}\cdot c.$

By construction, $f^{\prime}$ is equal to $\tilde{f}$ inside of $K_{i+1}$, so
in particular in $K_{i}$ and on $e$; hence $f^{\prime}(e)\neq 0$.

  * •

Following the argument of Proposition 3.11 starting from step 4 with function
$f^{\prime}$ shows that $f^{\prime}$ cannot be non-zero, a contradiction.

It remains to verify that $\tilde{f}$ is actually well-defined and non-zero.
Well-defined is a consequence of Lemma 3.14.

Assume that $\tilde{f}=0$, and let $e$ some edge of $T$ on which $f$ is non-
zero.

  * •

Let $E^{+}\sqcup E^{-}$ the partition of the ends of $T$ following the side of
$e$ on which they lie. By Lemma 3.14 there exists a large enough separator
$K_{i}$ in $X$ that the induced partition on ends is a subpartition of
$\\{E^{+},E^{-}\\}$

Let then $K$ such a separator and
$\\{E_{1}^{+},\ldots,E_{n}^{+}\\}\cup\\{E_{1}^{-},\ldots,E_{m}^{-}\\}$ the
subpartition induced by $K$. Let also
$e_{1}^{+},\ldots,e_{n}^{+},e_{1}^{-},e_{m}^{-}$ the edges in the boundary of
$K$, with corresponding signs. The ends in $E_{i}^{\pm}$ are then exactly
those in the component of $X-K$ corresponding to $e_{i}^{\pm}$.

For a collection of bips $B$, let $a_{B}:=\sum_{b\in B}a_{b}$. If we note
$b^{-1}$ for the bip $b$ with opposite orientation, then $a_{b^{-1}}=-a_{b}$.
For a collection of oriented bips $B$ and two disjoint sets of ends
$E_{1},E_{2}$, let $[E_{1},E_{2}]$ be $\\{b\in B\;\lvert\;b(-\infty)\in
E_{1}\text{ and }b(+\infty)\in E_{2}\\}\sqcup\\{b^{-1}\;\lvert\;b\in B\text{
and }b(-\infty)\in E_{2}\text{ and }b(+\infty)\in E_{1}\\}$. In words, the
bips having an end in $E_{1}$ and in $E_{2}$, taken with consistent
orientation.

  * •

Since $\tilde{f}=0$, we have $\tilde{f}(e_{i}^{s})=0$ for all $i$ and $s=\pm$.
This implies

$a_{[E^{+}_{i},(E^{+}_{i})^{\complement}]}=0$

  * •

We also have:

$\displaystyle f(e)$ $\displaystyle=a_{[E^{-},E^{+}]},$ and using
$[E^{-},E^{+}]=\bigsqcup_{i=1}^{m}[E_{i}^{-},E^{+}]$:
$\displaystyle=\sum_{i=1}^{m}a_{[E_{i}^{-},E^{+}]}$
$\displaystyle=\sum_{i=1}^{m}a_{[E_{i}^{-},E^{+}]}+0,$ now, using the fact
that $a_{[E_{j}^{-},E_{j}^{+}]}=-a_{[E_{j}^{+},E_{j}^{-}]}$ for any $j$:
$\displaystyle=\sum_{i=1}^{m}\left(a_{[E_{i}^{-},E^{+}]}+\sum_{j=1,j\neq
i}^{m}\overbrace{a_{[E_{i}^{-},E_{j}^{-}]}+a_{[E_{j}^{-},E_{i}^{-}]}}^{=0}\right)$
$\displaystyle=\sum_{i=1}^{m}\left(\left(a_{[E_{i}^{-},E^{+}]}+\sum_{j=1,j\neq
i}^{m}a_{[E_{i}^{-},E_{j}^{-}]}\right)+\sum_{j=1,j\neq
i}^{m}a_{[E_{j}^{-},E_{i}^{-}]}\right),$ and now by the decomposition
$[E_{i}^{-},(E_{i}^{-})^{\complement}]=[E_{i}^{-},E_{i}^{+}]\sqcup\bigsqcup_{j=1,j\neq
i}^{m}[E_{i}^{-},E_{j}^{-}]$ for any $i$:
$\displaystyle=\sum_{i=1}^{m}\left(\left(a_{[E_{i}^{-},(E_{i}^{-})^{\complement}]}\right)+\sum_{j=1,j\neq
i}^{m}a_{[E_{j}^{-},E_{i}^{-}]}\right)$ and the content of the inner
parenthesis vanishes since $\tilde{f}(e_{i}^{-})=0$:
$\displaystyle=\sum_{i=1}^{m}\sum_{j=1,j\neq i}^{m}a_{[E_{j}^{-},E_{i}^{-}]}$
$\displaystyle=\sum_{1\leq i\neq j\leq
m}a_{[E_{j}^{-},E_{i}^{-}]}+a_{[E_{i}^{-},E_{j}^{-}]}$ and finally, again
since inverting directions negates the sign: $\displaystyle=0.$

It follows therefore that $\tilde{f}=0$ implies $f(e)=0$, which is a
contradiction.

∎

##  4 Coefficients in $\mathbb{Z}_{2}$

In this section, we study the special case of
$H^{\mathrm{uf}}_{1}(\cdot,\mathbb{Z}_{2})$, or “uniformly finite homology
with coefficients in $\mathbb{Z}_{2}$”. In that case, the “uniformly finite”
condition becomes vacuous, and one may wonder if any insight can be gained
with such a restriction. This is indeed the case, as the next proposition
shows, thus motivating the study of
$H^{\mathrm{uf}}_{1}(\cdot,\mathbb{Z}_{2})$.

###  4.1 $H^{\mathrm{uf}}_{1}(X,\mathbb{Z})\twoheadrightarrow
H^{\mathrm{uf}}_{1}(X,\mathbb{Z}_{2})$

######  Proposition 4.1.

Let $X$ a (ULF) graph. Then $H^{\mathrm{uf}}_{1}(X,\mathbb{Z}_{2})$ is a
quotient of $H^{\mathrm{uf}}_{1}(X,\mathbb{Z})$.

###### Proof.

Consider the natural map:

$\Phi:Z^{\mathrm{suf}}_{1}(X,\mathbb{Z}){\overset{q_{/2}}{\rightarrow}}Z^{\mathrm{suf}}_{1}(X,\mathbb{Z}_{2}){\overset{\pi}{\twoheadrightarrow}}Z^{\mathrm{suf}}_{1}(X,\mathbb{Z}_{2})/\mathcal{C}_{\infty}(X,\mathbb{Z}_{2}).$

The first arrow is surjective. Indeed, if $f:EX\rightarrow\mathbb{Z}_{2}$ lies
in $Z^{\mathrm{suf}}_{1}(X,\mathbb{Z}_{2})$, one can “follow paths” and write

$f=\sum C+\sum B,$

for $C$ a set of circuits and $B$ a set of bips, such that the elements of
$B\cup C$ have pairwise disjoint support. Choosing signs arbitrarily, we can
lift $f$ to $\tilde{f}\in Z^{\mathrm{suf}}_{1}(X,\mathbb{Z})$ by lifting each
element of $C$ and $B$.

To get a surjection $H^{\mathrm{uf}}_{1}(X,\mathbb{Z})\twoheadrightarrow
H^{\mathrm{uf}}_{1}(X,\mathbb{Z}_{2})$, it remains to check that
$\mathcal{C}_{\infty}(X,\mathbb{Z})$ is contained in $\ker\Phi$. But if $\sum
C\in\mathcal{C}_{\infty}(X,\mathbb{Z})$ is given, then the image of $\sum C$
through $q_{/2}$ is in $\mathcal{C}_{\infty}(X,\mathbb{Z}_{2})$.

∎

###  4.2 Decomposition of $H^{\mathrm{uf}}_{1}(X,\mathbb{Z}_{2})$

The main result of the section is the following:

######  Proposition 4.2.

Let $X$ be a (ULF) graph, and $T\leq X$ an end-defining subtree. Then:

$H^{\mathrm{uf}}_{1}(X,\mathbb{Z}_{2})\cong
H^{\mathrm{uf}}_{1}(T,\mathbb{Z}_{2})\oplus\frac{\mathcal{C}(X,\mathbb{Z}_{2})}{\mathcal{C}_{\infty}(X,\mathbb{Z}_{2})}.$

###### Proof.

Drop the $\mathbb{Z}_{2}$. Recall that
$H^{\mathrm{uf}}_{1}(X)=Z^{\mathrm{suf}}_{1}(X)/\mathcal{C}_{\infty}$ and
$H^{\mathrm{uf}}_{1}(T)\cong Z^{\mathrm{suf}}_{1}(T)$. Furthermore, both
$Z^{\mathrm{suf}}_{1}(T)$ and $\mathcal{C}$ lie in $Z^{\mathrm{suf}}_{1}(X)$.
Since $\mathcal{C}_{\infty}\leq\mathcal{C}$, it suffices therefore to show
that $Z^{\mathrm{suf}}_{1}(T)\cap\mathcal{C}=0$ and
$Z^{\mathrm{suf}}_{1}(T)+\mathcal{C}=Z^{\mathrm{suf}}_{1}(X)$. We already
know, by Corollary 3.12 that $Z^{\mathrm{suf}}_{1}(T)\cap\mathcal{C}=0$. It
remains therefore to show
$Z^{\mathrm{suf}}_{1}(T)\cap\mathcal{C}=Z^{\mathrm{suf}}_{1}(X)$.

Fix $f\in Z^{\mathrm{suf}}_{1}(X)$. We will find a thin set of circuits $D$
such that $f-\sum D$ is supported on $T$; in some sense “pushing” $f$ towards
$T$ using circuits.

By “following paths”, one can write:

$f=\sum_{\varepsilon{\neq}\varepsilon^{\prime}}\sum_{b\in
B_{\varepsilon,\varepsilon^{\prime}}}b+\sum_{b\in B_{=}}b+\sum_{c\in C}c$

with $B_{\varepsilon,\varepsilon^{\prime}}$ a set of bips going from end
$\varepsilon$ to end $\varepsilon^{\prime}$, $B_{=}$ a set of bips going to
the same ends, and $C$ a set of circuit, and so that no edge is shared between
elements of
$\bigcup_{\varepsilon{\neq}\varepsilon^{\prime}}B_{\varepsilon,\varepsilon^{\prime}}\cup
B_{=}\cup C$ – the sum is therefore (very!) thin.

If $B_{\varepsilon,\varepsilon^{\prime}}$ is finite, let, for any $b\in
B_{\varepsilon,\varepsilon^{\prime}}$, $b^{*}$ denote the unique bip in $T$
with same ends. If $B_{\varepsilon,\varepsilon^{\prime}}$ is infinite, let
$B_{\varepsilon,\varepsilon^{\prime}}^{+}\sqcup
B_{\varepsilon,\varepsilon^{\prime}}^{-}=B_{\varepsilon,\varepsilon^{\prime}}$
be a partition into two infinite parts, and choose a bijection
$B_{\varepsilon,\varepsilon^{\prime}}^{+}\rightarrow
B_{\varepsilon,\varepsilon^{\prime}}^{-}$ denoted as sending $b\in
B_{\varepsilon,\varepsilon^{\prime}}^{+}$ to $b^{*}\in
B_{\varepsilon,\varepsilon^{\prime}}^{-}$. Denote also the inverse of this
bijection with ${\cdot}^{*}$, so that $(b^{*})^{*}=b$.

Fix some $R_{0}>0$, and let
$B_{\varepsilon,\varepsilon^{\prime}}^{0},B_{=}^{0},C^{0}$ be the elements of
each set ($B_{\varepsilon,\varepsilon^{\prime}},B_{=},C$) that intersect
$B(v_{T},R_{0})$. By ULF, each of those sets is finite, and only a finite
number of pairs $\varepsilon{\neq}\varepsilon^{\prime}$ have
$B_{\varepsilon,\varepsilon^{\prime}}^{0}$ non-empty. For any $b\in
B_{\varepsilon,\varepsilon^{\prime}}^{0}$, choose two paths
$p_{b,0}^{+},p_{b,0}^{-}$ connecting the two rays of $b-B(v_{T},R_{0})$
outside of $B(v_{T},R_{0})$ to $b^{*}$. Similarly, for any $b\in B_{=}^{0}$,
find a path $p_{b,0}$ joining the two rays of $b$ outside of $B(v_{T},R_{0})$,
say $p_{b,0}$ connects $b(t_{b,0}^{+})$ to $b(t_{b,0}^{-})$. Let $R_{1}$ large
enough that $B(v_{T},R_{1})$ contains each of the path used above and each
element of $C^{0}$, let
$B_{\varepsilon,\varepsilon^{\prime}}^{1},B_{=}^{1},C^{1}$ be the elements of
each set that intersect $B(v_{T},R_{1})$ and repeat the above.

This defines circuits, say $c_{b}^{i}$ such that:

$\sum c_{b}^{i}=b-b^{*}$

if $b\in B_{\varepsilon,\varepsilon^{\prime}}$, and

$\sum c_{b}^{i}=b,$

if $b\in B_{=}$. Furthermore, at each step $i$, the circuits added do not
intersect $B(v_{T},R_{i-1})$, except for the segments corresponding to
$b^{*}$. Since those segments appear in exactly one circuit in the case of
infinite $B_{\varepsilon,\varepsilon^{\prime}}$, and in a finite number of
circuits in the case of finite $B_{\varepsilon,\varepsilon^{\prime}}$, the sum
of circuits is thin, and we have successfully pushed $f$ to the tree.

∎

######  Lemma 4.3.

If there exists $R$ such that any circuit can be written as a $\mathbb{Z}_{2}$
thin-sum of circuits of length $\leq R$, then
$\mathcal{C}(X,{\mathbb{Z}}_{2})=\mathcal{C}_{R}(X,{\mathbb{Z}}_{2})$.

###### Proof.

Using compactness. ∎

######  Corollary 4.4.

For a finitely presented group $G$:

$H^{\mathrm{uf}}_{1}(G,\mathbb{Z}_{2})\cong
H^{\mathrm{uf}}_{1}(T,\mathbb{Z}_{2}),$

which has dimension $\\#\text{ends}(G)-1$.

###### Proof.

By Proposition 4.2, the above lemma and the fact that any circuit in the
Cayley graph of $G$ can be written as a finite sums of relators. ∎

###  4.3 Large circuits and dimensions

Our goal in this section is to show that, intuitively, if a graph has circuits
of increasing length that can not be decomposed into smaller ones, then
$\mathbb{Z}_{2}$-homology cannot vanish, and will in fact be infinite
dimensional.

Recall that a collection of elements of $\mathbb{Z}_{2}^{EX}$ is said to be
_thin_ if it is locally finite. The sum of the collection (taken pointwise) is
then well-defined, and we already defined:

$\displaystyle\mathcal{C}$ $\displaystyle:=\\{\text{``thin sum of
circuits''}\\},$ $\displaystyle\mathcal{C}_{R}$
$\displaystyle:=\\{\text{``thin sum of circuits of length $\leq R$''}\\},$
$\displaystyle\mathcal{C}_{\infty}$ $\displaystyle:=\cup_{R}\mathcal{C}_{R}.$

The following are equivalent:

  1. 1.

There exists an $R>0$ such that any circuit can be written as a thin-sum sum
of circuits of length $\leq R$.

  2. 2.

$\mathcal{C}=\mathcal{C}_{R}$.

  3. 3.

For all $R^{\prime}>R$, $\mathcal{C}_{R^{\prime}}=\mathcal{C}_{R}$.

If those do not hold, we say that $X$ has _large circuits_.

For the sake of clarity, we will settle on the following notation:

  * •

A thin-sum is a well-defined sum of a potentially infinite thin collection
$A\subseteq\mathbb{Z}_{2}^{EX}$.

  * •

There is an obvious notion of being closed under thin-sums, and a vector
subspace of $\mathbb{Z}_{2}^{EX}$ is said to be a thin-sum space if it is
closed under thin-sums.

  * •

The thin-sum-span of a collection $A$ is the set:

$\\{\sum B\ |\ B\text{ a thin subset of }A\\}.$

By compactness, the thin-sum span of a collection is closed under thin-sums
(see [6]).

  * •

A collection $A$ is thin-sum-independent if no two distinct thin
subcollections of $A$ yield the same sum; equivalently, no thin subcollection
has sum $0$.

  * •

A thin-sum basis is a collection that is thin-sum-independent and a thin-sum
basis for a given thin-sum space. Note that we do not require a thin-sum basis
$A$ for a thin-sum space $\mathcal{W}$ to be thin itself, but if it is, we get
a bijection

$\mathbb{Z}_{2}^{A}\leftrightarrow\mathcal{W}.$

With those clarifications in mind, we see that each $\mathcal{C}_{R}$ is a
thin-sum space, as is $\mathcal{C}$ (as, respectively, the thin-sum span of
circuits of length $\leq R$, and all circuits), but $\mathcal{C}_{\infty}$ is
not necessarily. Finally, all the concepts above still make sense if we
replace $\mathbb{Z}_{2}^{EX}$ by $\mathbb{Z}_{2}^{W}$ for any countable set
$W$.

We will show that as soon as $X$ has large circuits, then
$H^{\mathrm{uf}}_{1}(X,{\mathbb{Z}}_{2})$ is infinite-dimensional.

To that end, we first construct a set of circuits in $X$ that is locally
finite (thin) _and_ a thin-sum basis for $\mathcal{C}$.

######  Proposition 4.5 ($\mathcal{C}$ has thin basis).

There exists a set $\mathcal{B}$ of circuits that is a thin thin-sum basis for
$\mathcal{C}$.

The proof uses an infinite variation of Gaussian elimination, which we will
use again later in a transfinite way.

Any element of $\mathcal{C}$ can be uniquely written as a function
$EX\rightarrow{\mathbb{Z}}_{2}$. Order the edges $EX$ as
$EX=(e_{i})_{i\in{\mathbb{N}}}$. For any non-zero element $f\in\mathcal{C}$,
there is then a well-defined _leading index_ $l(f)$ of $f$:

$l(f):=\min\\{i\in{\mathbb{N}}\ :\ f(i)\neq 0\\}.$

If $f=0$, we set $l(f)=\infty$ by convention, and will also set $g(\infty)=0$
for any $g\in\mathcal{C}$.

###### Proof.

Take all simple333simple $=$ no vertex appears twice in the circuit. circuits
$\mathcal{S}$ and order them: $\mathcal{S}=(f_{j})_{j\in{\mathbb{N}}}$
($\mathcal{S}$ is countable). We then apply Gaussian elimination on
$\mathcal{S}$ as follows. Define the following function inductively:

$\displaystyle F:\mathbb{N}{\times}\mathbb{N}$
$\displaystyle\rightarrow(\mathbb{N}{\rightarrow}{\mathbb{Z}}_{2})$
$\displaystyle F_{0,j}$ $\displaystyle=f_{j}$ $\displaystyle F_{t,j}$
$\displaystyle=\begin{cases}F_{j,j}&\text{if $j<t$}\\\
F_{t-1,j}+F_{t-1,j}(l(F_{t-1,t-1}))\cdot F_{t-1,t-1}&\text{if $j\geq t$}\\\
\end{cases}$

(with index notation used for readability). Note that by our convention on
leading index, the definition makes sense even for $F_{t-1,t-1}=0$. We first
remark that for any $t>j$, $F(t,j)=F(j,j)$. Let then $g_{j}:=F(j,j)$, and

$\mathcal{B}:=\\{g_{j}\ |\ j\in{\mathbb{N}},g_{j}\neq 0\\}.$

Any element of $\mathcal{B}$ is non-zero, and since for $j>i$,
$g_{j}(l(g_{i}))=0$ by construction, it follows that they are all distinct.

It remains to verify that: $\mathcal{B}$ is thin, thin-sum independent and a
thin-sum basis.

Thinness. First, note that as soon as $i=l(g_{j})$ for some $j$, then $e_{i}$
is in the support of finitely many elements of $\mathcal{B}$ by construction.

Assume now there exists an non-leading edge $e_{i}$ that is in the support of
infinitely many elements of $\mathcal{B}$, and without loss of generality that
it is actually the first such (i.e. $i^{\prime}<i$ implies $e_{i^{\prime}}$ is
in the support of finitely many circuits). Let $t$ be such that

$g_{t}(i)\neq 0\text{ and }g_{t}(i^{\prime})=0\ \forall i^{\prime}<i.$

The existence of such a $t$ is guaranteed by our assumption on $i$. Then
$g_{t}(i)\neq 0$ but $g_{t}(i^{\prime})=0$ whenever $i^{\prime}<i$, so that
$l(g_{t})=i$, which is a contradiction.

Independence. Given any non-empty set of elements $\\{g_{i}\\}_{i\in I}$ of
$\mathcal{B}$, consider $\hat{i}:=\min I$, and let $\hat{j}=l(g_{\hat{i}})$
Then $(\sum_{i\in I}g_{i})(\hat{j})=1$ since for all $I\ni i\neq\hat{i}$,
$g_{i}(\hat{j})=0$. Thus, no non-trivial sum is zero.

Basis. Fix $h:\mathbb{N}\cong EX\rightarrow\mathbb{Z}_{2}$ any element of
$\mathcal{C}$. By definition, $h$ is a thin sum of simple circuits
$\sum_{i\in\mathcal{I}}f_{i}$, where $\mathcal{I}\subset\mathbb{N}$. By
definition of $F$ we see that any $f_{i}$ can be written

$\displaystyle f_{i}$
$\displaystyle=F_{i,i}+\sum_{t=1}^{i}F_{t-1,i}(l(F_{t-1,t-1}))\cdot
F_{t-1,t-1}$ and each summand is either zero or a basis element:
$\displaystyle=g_{i}+\sum_{t=1}^{i}F_{t-1,i}(l(F_{t-1,t-1}))\cdot g_{t-1}.$

Therefore, $f_{i}$ can also be seen as a map
$\tilde{f}_{i}:\mathcal{B}\rightarrow\mathbb{Z}_{2}$, where each basis element
is mapped to its coefficient in the sum above. Define now

$\displaystyle\mathcal{I}_{s}:$ $\displaystyle=\\{i\in\mathcal{I}\ |\ i\leq
s\\}$ $\displaystyle h_{s}:$
$\displaystyle=\sum_{i\in\mathcal{I}_{s}}f_{i}\in(\mathbb{Z}_{2})^{EX}$
$\displaystyle\tilde{h}_{s}:$
$\displaystyle=\sum_{i\in\mathcal{I}_{s}}\tilde{f}_{i}\in(\mathbb{Z}_{2})^{\mathcal{B}}.$

Note that, by construction:

$h_{s}=\sum_{g}\tilde{h}_{s}(g)\cdot g.$

Note that if we fix an edge $e_{r}$, then the sequence
$\\{h_{s}(r)\\}_{s\in\mathbb{N}}$ is eventually constant (by thinness). In
other words, the sequence $\\{h_{s}\\}_{s\in\mathbb{N}}$ converges
(pointwise). Take an accumulation point and obtain the limit $\tilde{h}$ of
the sequence $\\{\tilde{h}_{s}\\}_{s}$ and a subsequence
$\\{\tilde{h}_{s_{k}}\\}_{k}$ converging to $\tilde{h}$. We now have

$h_{s_{k}}\rightarrow h,\quad\tilde{h}_{s_{k}}\rightarrow\tilde{h}.$

Define

$h^{\prime}:=\sum_{g\in\mathcal{B}}\tilde{h}(g)\cdot g.$

We see that for a fixed edge $e_{r}$,

$\displaystyle h^{\prime}(r)$
$\displaystyle=\sum_{g\in\mathcal{B}}\tilde{h}(g)\cdot g(r)$
$\displaystyle=\sum_{g\in\mathcal{B}}\left(\lim_{k}\tilde{h}_{s_{k}}(g)\right)\cdot
g(r)$
$\displaystyle=\sum_{g\in\mathcal{B}}\left(\lim_{k}\tilde{h}_{s_{k}}(g)\cdot
g(r)\right)$ and since $g(r)=0$ for all but a finitely number of
$g\in\mathcal{B}$, we can exchange limit and sum:
$\displaystyle=\lim\limits_{k}\sum_{g\in\mathcal{B}}\tilde{h}_{s_{k}}(g)\cdot
g(r)$ $\displaystyle=\lim\limits_{k}h_{s_{k}}(r)$ $\displaystyle=h(r).$

This shows that any $h\in\mathcal{C}$ can be written as a thin-sum of elements
of $\mathcal{B}$, and concludes the proof. ∎

Let $\mathcal{B}$ denote a thin thin-sum basis consisting of circuits as
constructed above. By thinness, any set of elements of $\mathcal{B}$ defines a
(legal, i.e. thin) sum, and thus, we get a bijection:

$\mathbb{Z}_{2}^{\mathcal{B}}\leftrightarrow\mathcal{C}.$

We now want to tweak $\mathcal{B}$ so as to have a sequence of increasing thin
bases $\mathcal{B}_{i}^{\prime}$ for $\mathcal{C}_{i}$ and
$\mathcal{B}^{\prime}$ for $\mathcal{C}$ such that:

$\mathcal{B}_{i}^{\prime}\subseteq\mathcal{B}_{i+1}^{\prime}\subseteq\mathcal{B}^{\prime}.$

This can be done in a somewhat abstract setting:

######  Proposition 4.6.

Assume a thin-sum space $W\subseteq\mathbb{Z}_{2}^{X}$ is given, along with a
sequence of increasing thin-sum subspaces $V_{i}$. If $\mathcal{W}$ is a
countable thin thin-sum basis for $W$, then one can construct a countable,
thin, thin-sum basis $\mathcal{B}$ for $W$, along with sequence of nested
thin-sum bases $\mathcal{B}_{i}$ for $V_{i}$.

Since $\mathcal{W}$ is thin and a basis, we have
$W\leftrightarrow\mathbb{Z}_{2}^{\mathcal{W}}$. Given any well-ordering of
$\mathcal{W}$, one can let, for any $v\in W$:

$l(v)=\min\\{i:v(i)\neq 0\text{ when $v$ is viewed as an element of
$\mathbb{Z}_{2}^{\mathcal{W}}$}\\}.$

If $v=0$, we use the same convention as above: $l(v)=\infty$, and for any $w$,
$w(\infty)=0$. In the proof, we will assume a fixed ordering on $\mathcal{W}$
and identify $W\leftrightarrow\mathbb{Z}_{2}^{\mathcal{W}}$.

###### Proof.

Well-order the set $W$ in a way that respects the subsets $V_{i}$, i.e.

$W=\\{v_{\alpha}\ |\ \alpha<\kappa\\},$

with $\kappa$ the cardinal of $W$ and such that there exist increasing
ordinals ${\alpha}_{i}$ with

$V_{i}=\\{v_{\alpha}\in W\ |\ \alpha<{\alpha}_{i+1}\\}.$

We show that we can construct $\mathcal{B}_{i}$ ($i\in\mathbb{N}$) and
$\mathcal{B}$ satisfying these requirements. We define the “Gaussian
elimination” map

$\nu:\kappa{\times}\kappa{\times}\mathbb{N}\rightarrow\mathbb{Z}_{2}$

inductively as follows:

$\displaystyle{\nu}_{0,\alpha}$ $\displaystyle=v_{\alpha},$
$\displaystyle{\nu}_{\tau,\alpha}$
$\displaystyle=\begin{cases}\lim_{\sigma<\tau}{\nu}_{\sigma,\alpha}&\text{if
$\tau$ is a limit ordinal},\\\ {\nu}_{\alpha,\alpha}&\text{otherwise and
$\tau>\alpha$},\\\
{\nu}_{\tau-1,\alpha}-{\nu}_{\tau-1,\alpha}(l({\nu}_{\tau-1,\tau-1}))\cdot{\nu}_{\tau-1,\tau-1}&\text{otherwise
and $\tau\leq\alpha$},\\\ \end{cases}$ (1)

where the limit is a pointwise limit (which means that at any coordinate $r$,
the transfinite sequence, evaluated at that coordinate, is eventually
constant).

From now on, we will write ${\nu}_{\alpha}$ for the diagonal element
${\nu}_{\alpha,\alpha}$.

The interpretation of the map $\nu$ is as follows. An element of $W$ is
uniquely determined by a map $\mathbb{N}\rightarrow\mathbb{Z}_{2}$ (a row).
Thus, the set $W$, once ordered, can be represented as a map
$\kappa{\times}\mathbb{N}\rightarrow\mathbb{Z}_{2}$ (a matrix). Gaussian
elimination, in the finite case, is an iterative process where one iterates
over the rows of a matrix, each the “pivot”, while modifying the matrix at
each step, hence a map
$\kappa{\times}\kappa{\times}\mathbb{N}\rightarrow\mathbb{Z}_{2}$ . In short,
the first coordinate represents the iteration/pivot, the second the element in
question, and the third its value at a given base coordinate.

At time zero, the matrix has not been modified. At time $\tau$, if $\tau$ is a
successor ordinal, we pivot around the element ${\nu}_{\tau-1,\tau-1}$. Note
that if the row $\alpha$ “lies before the pivot” or “is the pivot ” (i.e.
$\alpha<\tau$), it is not changed. If row $\alpha$ “lies after the pivot”
($\alpha\geq\tau$), we must subtract the pivot to what row $\alpha$ “just
was”. In case $\tau$ is a limit ordinal, then we will just re-iterate all
changes that happened before time $\tau$ and restart pivoting in its
successor.

We start by showing by induction on $\tau$ that:

  1. 1.

For any $n\in\mathbb{N}$, the number of ordinals $\sigma{\leq}\tau$ such that
${\nu}_{\sigma}(n)\neq 0$ is finite.

  2. 2.

The limit in Section 4.3 is actually well-defined.

  3. 3.

For any $\sigma<\tau,\alpha$, ${\nu}_{\tau,\alpha}(l({\nu}_{\sigma}))=0$.

  4. 4.

The value of ${\nu}_{\tau,\alpha}$ is

${\nu}_{\tau,\alpha}={\nu}_{0,\alpha}+\sum_{\sigma<\min\\{\alpha,\tau\\}}{\nu}_{\sigma,\alpha}(l({\nu}_{\sigma}))\cdot{\nu}_{\sigma}.$
(2)

Indeed, assume the above holds for any $\theta<\tau$.

If $\tau=0$: Then Items 1, 2, 3 and 4 hold trivially.

If $\tau$ is successor:

Item 1 is easy. By the induction hypothesis the number of $\sigma\leq\tau-1$
with ${\nu}_{\sigma}(n)\neq 0$ is finite. Hence it must be finite for $\tau$
too.

Item 2 does not apply successor ordinals.

For Item 3, we see that if $\tau>\alpha$, then

${\nu}_{\tau,\alpha}(l({\nu}_{\sigma}))={\nu}_{\alpha}(l({\nu}_{\sigma}))$

which is zero by induction hypothesis (i.e. Item 3 holds at $\theta=\alpha$).
If $\tau{\leq}\alpha$, then

${\nu}_{\tau,\alpha}={\nu}_{\tau-1,\alpha}-{\nu}_{\tau-1,\alpha}(l({\nu}_{\tau-1,\tau-1}))\cdot{\nu}_{\tau-1,\tau-1},$
(3)

which, evaluated in $l({\nu}_{\tau-1,\tau-1})$ is zero by construction. If
$\sigma<\tau-1$, both ${\nu}_{\tau-1,\alpha}(l({\nu}_{\sigma}))$ and
${\nu}_{\tau-1,\tau-1}(l({\nu}_{\sigma}))$ are zero, hence so is
${\nu}_{\tau,\alpha}$.

Finally, Item 4. One sees that if $\tau{\leq}\alpha$, we get
$\min\\{\tau,\alpha\\}=\tau$, and:

$\displaystyle{\nu}_{\tau,\alpha}$
$\displaystyle={\nu}_{\tau-1,\alpha}+{\nu}_{\tau-1,\alpha}(l({\nu}_{\tau-1,\tau-1}))\cdot{\nu}_{\tau-1,\tau-1}$
$\displaystyle={\nu}_{0,\alpha}+\sum_{\sigma<\tau-1}{\nu}_{\sigma,\alpha}(l({\nu}_{\sigma}))\cdot{\nu}_{\sigma}+{\nu}_{\tau-1,\alpha}(l({\nu}_{\tau-1,\tau-1}))\cdot{\nu}_{\tau-1,\tau-1},$
by definition of ${\nu}_{\tau,\alpha}$. If $\tau>\alpha$, we get
$\min\\{\tau,\alpha\\}=\alpha$, and
$\displaystyle{\nu}_{\tau,\alpha}={\nu}_{\tau-1,\alpha}$
$\displaystyle={\nu}_{0,\alpha}+\sum_{\sigma<\min\\{\alpha,\tau-1\\}}{\nu}_{\sigma,\alpha}(l({\nu}_{\sigma}))\cdot{\nu}_{\sigma}$
$\displaystyle={\nu}_{0,\alpha}+\sum_{\sigma<\min\\{\alpha,\tau\\}}{\nu}_{\sigma,\alpha}(l({\nu}_{\sigma}))\cdot{\nu}_{\sigma}$

Now moving on to the limit case.

If $\tau$ is limit: We first check Item 1. It suffices to show that for any
$n$, the number of ordinals $\sigma<\tau$ with ${\nu}_{\sigma}(n)\neq 0$ is
finite, since then the number of $\sigma{\leq}\tau$ with
${\nu}_{\sigma}(n)\neq 0$ will also be finite. If $n=l({\nu}_{\sigma})$ for
some $\sigma<\tau$, then the number of $\theta<\sigma$ for which
${\nu}_{\theta}(n)=0$ is finite by the induction hypothesis (Item 1), and for
any $\tau>\theta>\sigma$, ${\nu}_{\theta}(n)=0$, also by the induction
hypothesis (Item 3). Assume then that there exists some $n$ which is not a
leading coefficient, which is also in the support of infinitely many
${\nu}_{\sigma}$ ($\sigma<\tau$). We can take $n$ to be the least such, and
for $\sigma$ large enough, we will have:

${\nu}_{\sigma}(n)\neq 0,\quad{\nu}_{\sigma}(k)=0,\ \forall k<n,$

so that $n$ is actually the leading coefficient of ${\nu}_{\sigma}$, a
contradiction.

We now check Item 2. Let $\alpha$ arbitrary, then

${\nu}_{\tau,\alpha}=\lim_{\sigma<\tau}{\left({\nu}_{0,\alpha}+\sum_{\sigma^{\prime}<\min\\{\alpha,\sigma\\}}{\nu}_{\sigma^{\prime},\alpha}(l({\nu}_{\sigma^{\prime}}))\cdot{\nu}_{\sigma^{\prime}}\right)}$

which we need to verify to be well-defined. In other words, we need to check
that for any $n$, the sequence

$\sigma\ \mapsto\
{\nu}_{0,\alpha}(n)+\sum_{\sigma^{\prime}<\min\\{\alpha,\sigma\\}}{\nu}_{\sigma^{\prime},\alpha}(l({\nu}_{\sigma^{\prime}}))\cdot{\nu}_{\sigma^{\prime}}(n)$

stabilises starting at some ordinal. But by Item 1,
${\nu}_{\sigma^{\prime}}(n)$ is zero for $\sigma^{\prime}$ large enough, so
that the sequence indeed stabilises. Furthermore, it stabilises to:

${\nu}_{0,\alpha}(n)+\sum_{\sigma<\min\\{\alpha,\tau\\}}{\nu}_{\sigma,\alpha}(l({\nu}_{\sigma}))\cdot{\nu}_{\sigma}(n),$

which also proves Item 4.

It remains to check Item 3. But if $\sigma<\tau,\alpha$, then

${\nu}_{\tau,\alpha}(l({\nu}_{\sigma}))=\lim_{\sigma^{\prime}<\tau}{\nu}_{\sigma^{\prime},\alpha}(l({\nu}_{\sigma}))=\lim_{\sigma<\sigma^{\prime}<\tau}{\nu}_{\sigma^{\prime},\alpha}(l({\nu}_{\sigma}))=\lim
0=0.$

This closes the induction.

Let now

$\mathcal{B}:=\\{{\nu}_{\alpha}|\alpha<\kappa,{\nu}_{\alpha}{\neq}0\\},\quad\mathcal{B}_{i}=\\{{\nu}_{\alpha}|\alpha<{\alpha}_{i+1},{\nu}_{\alpha}{\neq}0\\}$

Note that by Item 1, each non-zero ${\nu}_{\alpha}$ has a well-defined leading
index $l({\nu}_{\alpha})$, and ${\nu}_{\alpha}{\neq}{\nu}_{\beta}$ implies
$l({\nu}_{\alpha})\neq l({\nu}_{\beta})$. It follows that $\mathcal{B}$ is at
most countable. We can now check the following consequences of this process:

  1. 1.

By Equation 2, $\mathcal{B}_{i}$ generates $V_{i}$ and $\mathcal{B}$ generates
$W$. Indeed, Equation 2 yields (by setting $\alpha=\tau$)

${\nu}_{\tau}={\nu}_{0,\tau}+\sum_{\sigma<\tau}{\nu}_{\sigma,\tau}(l({\nu}_{\sigma}))\cdot{\nu}_{\sigma},$

so that, by rewriting the sum, we get

${\nu}_{0,\tau}={\nu}_{\tau}-\sum_{\sigma<\tau}{\nu}_{\sigma,\tau}(l({\nu}_{\sigma}))\cdot{\nu}_{\sigma},$

so that any element of $W$ (resp. $V_{i}$) is a thin sum of elements of
$\mathcal{B}$ (resp. $\mathcal{B}_{i}$).

  2. 2.

$\mathcal{B}$ is thin, as a subset of $\mathbb{Z}_{2}^{\mathcal{W}}$, by the
same argument as the proof of Item 1 for $\tau$ limit.

  3. 3.

$\mathcal{B}$ is thin, as a subset of $\mathbb{Z}_{2}^{X}$. Every $x\in X$
appears in the support of a finite number of $v\in\mathcal{W}$, and every
$v\in\mathcal{W}$ in the support of a finite number of elements of
$\mathcal{B}$.

  4. 4.

The set $\mathcal{B}$ is thin-sum independent. Suppose not. Then, for some
$\tilde{\kappa}\subset\kappa$, we have
$\sum_{\sigma\in\tilde{\kappa}}{\nu}_{\sigma}=0$. Let
${\tau}_{0}:=\min\\{\sigma\in\tilde{\kappa}\ |\ {\nu}_{\sigma}\neq 0\\}$. Then
$\sum_{\sigma\in\tilde{\kappa}}{\nu}_{\sigma}(l({\nu}_{{\tau}_{0}}))=1$, a
contradiction.

∎

In our case, this yields:

######  Corollary 4.7.

There exists a thin thin-sum basis $\mathcal{B}$ for $\mathcal{C}$ with nested
subsets $\mathcal{B}_{i}$, each a thin-sum basis for $\mathcal{C}_{i}$.

###### Proof.

By Proposition 4.5 and Proposition 4.6. ∎

And we can now easily get:

######  Theorem 4.8.

If $X$ has large circuits, then $H^{\mathrm{uf}}_{1}(X,{\mathbb{Z}}_{2})$ is
infinite dimensional (as a vector space).

###### Proof.

Let $\mathcal{B},\mathcal{B}_{i}$ be as in Corollary 4.7.

If $X$ has large circuits, the sequence

$\mathcal{C}_{1}\subseteq\ldots\subseteq\mathcal{C}_{i}\subseteq\mathcal{C}_{i+1}\subseteq\ldots$

does not stabilise, so that there exists a sequence
$(i_{j})_{j\in{\mathbb{N}}}$ satisfying:

$\mathcal{C}_{i_{j}}\subsetneq\mathcal{C}_{i_{j+1}}.$

In particular, this implies that
$\mathcal{B}_{i_{j}}\subsetneq\mathcal{B}_{i_{j+1}}$. Choose then, for each
$j$, $b_{j}\in\mathcal{B}_{i_{j+1}}-\mathcal{B}_{i_{j}}$.

Let $\mathcal{J}$ be an infinite set of infinite pairwise disjoint subsets of
$\mathbb{N}$ and

$f_{J}:=\sum_{j\in J}b_{j},\quad J\in\mathcal{J}.$

We claim that the family $\\{[f_{J}]\in
H^{\mathrm{uf}}_{1}(X,\mathbb{Z}_{2})\\}_{J\in\mathcal{J}}$ is linearly
independent, which will imply infinite dimensionality.

We can actually show more: the family $\\{f_{J}\\}_{J\in\mathcal{J}}$ is thin,
and no thin-sum sum of the form:

$\sum_{J\in\mathcal{J}^{\prime}}f_{J},\quad\mathcal{J}^{\prime}\subseteq\mathcal{J},$

lies in $\mathcal{C}_{\infty}$. Indeed, since the $f_{J}$s have disjoint
support, any such combination can be written as an infinite sum:

$f=\sum_{j\in J^{\prime}}b_{j},$

for some $J^{\prime}\subseteq\mathbb{N}$. If $f\in\mathcal{C}_{\infty}$, then
there would exist $R$ with $f\in\mathcal{C}_{R}$, so that $f=\sum_{b\in B}b$
for $B\subseteq\mathcal{B}_{R}$, and $0=\sum_{B_{J^{\prime}}\Delta B}b$, where
$B_{j^{\prime}}=\\{b_{j}\ |\ j\in J^{\prime}\\}$. Since $J^{\prime}$ is
infinite, $B_{J^{\prime}}$ contains elements of $\mathcal{B}_{R^{\prime}}$ for
$R^{\prime}$ arbitrarily high, so that the symmetric difference above is not
empty. This is a contradiction with $\mathcal{B}$ being thin-sum independent.

∎

In $Z^{\mathrm{suf}}_{1}(X,\mathbb{Z}_{2})$, infinite sums are allowed, as
long as they are locally finite. However, in
$H^{\mathrm{uf}}_{1}(X,\mathbb{Z}_{2})$, only _finite_ sums are naturally
permitted; there is no obvious “thin-sum” space structure on quotients. Thus,
the passage to a quotient involves a major loss of information. We suggest a
very simple approach that remedies this problem in our ad hoc situation. It
makes sense to define a thin quotient of thin vector spaces. There might be a
more abstract setting about which one could say something.

######  Definition 4.9.

Fix a thin-sum space $V\leq\mathbb{Z}_{2}^{X}$, and a sub-vector space $W\leq
V$ (not necessarily closed under thin-sum). In the quotient $V/W$, we say that
a potentially infinite sum of elements $\sum a_{i}$ is _consistent_ iff there
is a collection of preimages $\tilde{a}_{i}\in a_{i}$ such that the collection
$\\{\tilde{a}_{i}\\}_{i}$ is thin (in particular, abusing notation, any $v\in
V-\\{0\\}$ appears as at most a finite number of $\tilde{a}_{i}s$), and all
such thin choices of representatives result in the same element in $V/W$:

$\sum\tilde{a}_{i}-\sum\tilde{b}_{i}\in W,$

if $\\{\tilde{a}_{i}\\}_{i},\\{\tilde{b}_{i}\\}_{i}$ are two choices of
representatives.

In order to have a cleaner theorem we also introduce the following “thin” sum.

######  Theorem 4.10.

The following are equivalent:

  1. 1.

In $H^{\mathrm{uf}}_{1}(X,\mathbb{Z}_{2})$, the infinite sum
$\\{[0]\\}_{i=0}^{\infty}$ of the zero-class is consistent.

  2. 2.

$X$ has no large circuits.

  3. 3.

Any sum of cosets that has a thin representative sum is consistent.

###### Proof.

Clearly 3 implies 1.

We show 1 implies 2. Suppose $X$ has large circuits, then there exists an
infinite thin sum of large circuits that is non-zero in homology. However
every individual circuit is finite and thus zero in homology. On the other
hand the repeated infinite sum of $0$ gives $0$. So we see that the infinite
sum of the zero class has two representing thin sums that give a different
solution. Hence it is inconsistent.

We show 2 implies 1. Consider an infinite representing thin sum
$\sum\limits_{i=1}^{\infty}\tilde{x}_{i}$ of the infinite zero class sum.
Since there are no large circuits, there is some $R>0$ such that each
$\tilde{x}_{i}$ can be written as a sum of $R$-circuits. More specifically
there is a $g_{i}$ such that $\partial g_{i}=\tilde{x}_{i}$. Now take some
accumulation point $\lambda$ of the sequence
$\left\\{\sum\limits_{i=1}^{n}g_{i}\right\\}_{n\in\mathbb{N}}$. Then
$\partial\lambda=\sum\limits_{i=1}^{\infty}\tilde{x}_{i}$. The infinite zero-
class sum is as such consistent.

We show 1 implies 3. Suppose there is some inconsistent thin sum. So two
representing sums $\sum\limits_{i=1}^{\infty}\tilde{x}_{i}$ and
$\sum\limits_{i=1}^{\infty}\tilde{y}_{i}$, such that
$[\tilde{x}_{i}]=[\tilde{y}_{i}]$. Then each
$[\tilde{x}_{i}-\tilde{y}_{i}]=[0]$. Hence by assumption
$\left[\sum\limits_{i=1}^{\infty}\tilde{x}_{i}-\tilde{y}_{i}\right]=[0]$.
Which shows that
$\left[\sum\limits_{i=1}^{\infty}\tilde{x}_{i}\right]=\left[\sum\limits_{i=1}^{\infty}\tilde{y}_{i}\right]$,
because finite sums are always well defined. ∎

######  Corollary 4.11.

If $H^{\mathrm{uf}}_{1}(X,\mathbb{Z}_{2})$ has large circuits, then every
thin-sum is inconsistent.

###### Proof.

Direct consequence of Theorem 4.10. ∎

We draw the following conclusion:

Define the _thin dimension_ $\dim_{\text{thin}}$ of
$H^{\mathrm{uf}}_{1}(X,\mathbb{Z}_{2})$ as the least cardinal $\kappa$ such
that $H^{\mathrm{uf}}_{1}(X,\mathbb{Z}_{2})$ has a subset $K$ of cardinality
$\kappa$ and such that any element of $H^{\mathrm{uf}}_{1}(X,\mathbb{Z}_{2})$
can be written as a consistent sum of elements of $K$.

######  Theorem 4.12.

  * •

$\dim_{\text{thin}}(H^{\mathrm{uf}}_{1}(X,\mathbb{Z}_{2}))$ is countable if
and only if $X$ has no large circuits. In this case it is equal to the number
of ends $-1$ (where we do not make the difference between countably many or
uncountably many ends).

  * •

$\dim_{\text{thin}}(H^{\mathrm{uf}}_{1}(X,\mathbb{Z}_{2}))$ is uncountable if
and only if $X$ has large circuits.

###### Proof.

First part. If there are no large circuits, then by Proposition 4.2 there
exists a tree $T$ inside $X$, such that
$H^{\mathrm{uf}}_{1}(T,\mathbb{Z}_{2})=H^{\mathrm{uf}}_{1}(X,\mathbb{Z}_{2})$.
The isomorphism is given by the natural inclusion map. Hence Proposition 4.2
gives us a thin thin-sum basis $\mathcal{B}$ in the tree. Note that all sums
remain well-defined when we inject $\mathcal{B}$ into $X$ by Theorem 4.10.

Second part. Suppose $X$ has large circuits, then by Corollary 4.11 all
consistent sums are finite. However we showed before that
$H^{\mathrm{uf}}_{1}(X,\mathbb{Z}_{2})$ contains an uncountable number of
elements. So a generating set must be uncountable too.

Note that since having large circuits and not having large circuits exhausts
the universe, both implications above are actually equivalences. ∎

##  5 $\mathbb{Z}$-large circuits

Recall that a ULF graph $X$ has $\mathbb{Z}$-large circuits if

$\forall r>0\
\mathcal{C}_{r}(X,\mathbb{Z})\neq\mathcal{C}_{\infty}(X,\mathbb{Z}).$

######  Proposition 5.1.

If $X$ is vertex-transitive and has $\mathbb{Z}$-large circuits, then
$H^{\mathrm{uf}}_{1}(X,\mathbb{Z})\neq 0$.

###### Proof.

Assume $X$ vertex transitive with zero homology and $\mathbb{Z}$-large
circuits. For any $r>0$, take some
$f_{r}\in\mathcal{C}_{\infty}-\mathcal{C}_{r}$.

We claim that $f_{r}$ can be assumed to have norm $1$. First, by assumption
that $f\in\mathcal{C}_{\infty}$, there exists some $s>r$ with
$f_{r}\in\mathcal{C}_{s}$, and thus $f_{r}=\mathfrak{e}g$ for some
$g:\mathfrak{C}_{s}\rightarrow\mathbb{Z}$ uniformly bounded. One can write
$g=g^{+}-g^{-}$, so just assume for now that $g\geq 0$. Now, consider the
following process: view $g$ as a multiset, and consider a maximal subset of
circuits that do not intersect pairwise, and subtract it from $g$; iterate.
This process has to end eventually (indeed, fix any circuit $c$ in the support
of $g$; by ULF, there exists some $K$ not depending on $c$ such that $c$
intersects at most $K$ other circuits in the support of $g$; if during the $K$
first steps $c$ is never removed, this means by maximality that at each step a
neighbor is removed; thus after $K+1$ steps all neighbors are removed, and
then so will $c$, by maximality), so that
$f=\sum\mathfrak{e}g^{+}_{i}-\sum\mathfrak{e}g^{-}_{i}$, with each
$f_{i}^{\pm}:=\mathfrak{e}g^{\pm}_{i}$ of norm $1$. If each $f_{i}^{\pm}$ lied
in $\mathcal{C}_{r}$, so would $f$ ($\mathcal{C}_{r}$ being a vector space).
We conclude that at least one of $f_{i}^{\pm}$ does not lie in
$\mathcal{C}_{r}$, and the claim is verified.

Fix some basepoint $v_{0}\in VX$, and for any $r$, some
$f_{r}=\mathfrak{e}g_{r}$ as above, and consider the family $(f_{r}^{i}\in
C^{\mathrm{suf}}_{1}(X,\mathbb{Z}))_{i\in\mathbb{N}}$ constructed as follows:
For any $i$, take the restriction of $g_{r}$ to circuits intersecting a ball
of radius $i$ around $v_{0}$, call this $g_{r}^{i}$ and let
$f_{r}^{i}=\mathfrak{e}g_{r}^{i}$. Note that since all circuits of $g_{r}^{i}$
are disjoint, $f_{r}^{i}$ also has norm $1$; furthermore, $f_{r}^{i}$ has
support in a ball of radius $i+s$ if $g_{r}\in l^{\infty}(\mathfrak{C}_{s})$.
Im other words, $f_{r}$ and $f_{r}^{i}$ are both elements of
$Z^{\mathrm{suf}}_{1}$ of norm ($\leq$) $1$ and agreeing on the ball of radius
$i$ around $v_{0}$. The important aspect of this family is that
$f_{r}^{i}{\overset{i}{\rightarrow}}f_{r}$ pointwise.

Since $X$ is infinite and transitive, it is now possible to move the elements
$f_{r}^{i}$ and balls $B_{r,i}$ containing their support, for $i$ _and_ $r$
varying in $X$ in such a way that the balls do not intersect pairwise. Say
${\phi}_{r,i}$ is the automorphism of $X$ used to move $f_{r}^{i}$.

Let $F$ be the well-defined sum of these (moved elements); $F$ lies in
$Z^{\mathrm{suf}}_{1}$ and has norm $1$. Since $H^{\mathrm{uf}}_{1}=0$, there
exists some $F\in\mathcal{C}_{\infty}$, and by definition,
$F\in\mathcal{C}_{s}$ for some $s$. Take therefore some
$G:\mathfrak{C}_{s}\rightarrow\mathbb{Z}$ with $\mathfrak{e}G=F$ and of
uniformly bounded norm. Fix any $r>s$ and consider the sequence
$(G_{r,i}):=(G\circ{\phi}_{r,i}^{-1})_{i}$. Note that by construction
$\mathfrak{e}G_{r,i}|B_{r,i}=f_{r,i}$ and recall that $f_{r,i}\rightarrow
f_{r}$. Up to taking a subsequence, one may assume that $G_{r,i}$ converges to
some $G_{r}$ (since the $G_{r,i}$ are uniformly bounded in norm). Therefore,
$G_{r,i}\rightarrow G_{r}$, which implies
$\mathfrak{e}G_{r,i}\rightarrow\mathfrak{e}G_{r}$, but
$\mathfrak{e}G_{r,i}\rightarrow f_{r}$ since
$\mathfrak{e}G_{r,i}|B_{r,i}=f_{r}$ and the sequence $B_{r,i}$ exhausts $X$.
Since $\mathfrak{e}G_{r}=f_{r}$, we get $f_{r}\in\mathcal{C}_{s}$, thus
contradicting $f_{r}\notin\mathcal{C}_{r}$.

∎

######  Proposition 5.2.

If $X$ is vertex-transitive and has $\mathbb{Z}$-large circuits, then
$\dim(H^{\mathrm{uf}}_{1}(X,\mathbb{Z}))=\infty$.

###### Proof.

Consider the function $f$ constructed in the proof of Proposition 5.1. In the
same way it is possible to construct instead countably many functions $f_{k}$
each defined on balls $B^{k}_{r,i}$, which are the building blocks of the
construction. Moreover these $f_{k}$ can be chosen to not have any
intersecting balls between them. By arguments of the proof before they are all
non-zero in homology and finite linear combinations between them (as well as
bounded infinite linear combinations for that matter) are also non-zero. Hence
the result follows. ∎

##  6 Expansion

In this section, we follow a different trail in our search for a description
of vanishing $H^{\mathrm{uf}}_{1}(\cdot,\mathbb{Z})$. Our motivation is
twofold: On the one hand, after playing a bit with
$H^{\mathrm{uf}}_{1}(\cdot,\mathbb{Z})$, one comes to the conclusion that,
ends and “large circuits” aside, the main ingredient in non-vanishing of
homology is a lack of triangles, compared to edges. The typical example here
is (a triangulated) $\mathbb{Z}^{2}$: Any infinite sum of parallel bips going
in one direction will define a non-zero class in homology, since there are not
enough triangles to kill all bips at the same time. On the other hand,
(classical) graph expansion can be interpreted as the ability, given any
finite subset of vertices $U$, and element $f\in Z^{\mathrm{suf}}_{0}(U)$, to
push $f$ out of $U$ in a bounded way; that is, there exists some $g\in
C^{\mathrm{suf}}_{1}(X)$ with $\partial g|U=f$ and
$\|g\|\leq{\varepsilon}^{-1}\|f\|$, where $\varepsilon$ is the Cheeger
constant for $X$. This section can then be described as an effort to try and
mimic the “homological interpretation” of expansion in higher dimension (we
call this $H^{\mathrm{suf}}_{n}$-expansion), and fine-tune it to isolate the
phenomenon of “lack of triangles” from “ends” and “large circuits” (thus
getting _pure_ $H^{\mathrm{suf}}_{n}$-expansion).

Recall that a (ULF) graph $X$ is said to be _amenable_ if it satisfies the
following condition:

$\forall\epsilon>0\ \exists U{\subseteq}_{f}VX\ :\ |\partial
U|<\varepsilon|U|.$

If $X$ is non-amenable, the least $\varepsilon>0$ satisfying $\forall
U{\subseteq}_{f}VX$, $|\partial U|\geq{\varepsilon}|U|$ is called the Cheeger
constant for $X$.

The following was proved in [1]:

######  Proposition 6.1 (Part of [1, Theorem 3.1]).

A ULF graph $X$ is non-amenable iff $H^{\mathrm{uf}}_{0}(X,\mathbb{Z})=0$.

###  6.1 Basics

Let $X$ be a ULF simplicial complex. If $U\subseteq VX$ is a set of vertices,
we also use $U$ for the induced simplicial subcomplex. We then write $\partial
U$ for the simplices of $X$ adjacent to a vertex not in $U$.

######  Definition 6.2 ($H^{\mathrm{suf}}$-Expansion).

We say that $X$ has $n$-dimensional $H^{\mathrm{suf}}$-expansion if: There
exists a function $K:\mathbb{N}{\rightarrow}\mathbb{N}$ satisfying the
following. Fix any finite set of vertices $U{\subseteq}_{f}VX$, and any $f\in
C^{\mathrm{suf}}_{n}(U,\mathbb{Z})$ satisfying:

  1. 1.

$\partial f$ has support contained in $\partial U$.

  2. 2.

$f$ can be extended to some $\hat{f}\in Z^{\mathrm{suf}}_{n}(X,\mathbb{Z})$
with $\|\hat{f}\|=\|f\|$ (by extended, we mean $\hat{f}|U=f$).

Then: There exists $g\in C^{\mathrm{suf}}_{n+1}(X,\mathbb{Z})$ such that
$\partial g|_{U}=f$ and $\|g\|\leq K(\|f\|)$.

We say that a a function $K$ as above is a certificate to
$H^{\mathrm{suf}}_{n}$-expansion.

Finally, if Item 2 in the definition is replaced by

  1. 3.

$f$ can be extended to some $\hat{f}\in B^{\mathrm{suf}}_{n}(X,\mathbb{Z})$
with $\|\hat{f}\|=\|f\|$ .

and the rest is leaved as-is, then the obtained condition is called _pure_
$H^{\mathrm{suf}}_{n}$-expansion.

Intuitively, expansion can be interpreted as homological cycles locally being
homological boundaries, with control. In other words, one can always kill
“pieces of cycles” with “pieces of boundaries” in a bounded way. We will see
that this property _always_ implies that simplicial uniformly finite homology
vanishes, at the corresponding dimension. Pure expansion is slightly harder to
describe, but the core idea is that it only takes into account those $f$ that
“can actually be killed” by some boundary.

Note that

$\text{$H^{\mathrm{suf}}_{n}$-expansion}\quad\Rightarrow\quad\text{pure
$H^{\mathrm{suf}}_{n}$-expansion}.$

######  Lemma 6.3.

If $X$ has $H^{\mathrm{suf}}_{n}$-expansion, then
$H^{\mathrm{suf}}_{n}(X,\mathbb{Z})=0$.

###### Proof.

Assume that $K$ is a certificate. Fix $f\in
Z^{\mathrm{suf}}_{n}(X,\mathbb{Z})$, and $\langle U_{n}\subseteq
VX{\rangle}_{n}$ a nested sequence of finite subsets exhausting $VX$. For each
$n$, let $f_{n}$ the restriction of $f$ to $U_{n}$, and assume that
$\|f_{n}\|=\|f\|$ for each $n$ (achievable by starting with $U_{0}$ large
enough). Since $\|f\|=\|f_{n}\|$, let $g_{n}\in
C^{\mathrm{uf}}_{n+1}(X,\mathbb{Z})$ satisfying $\partial g_{n}|U_{n}=f_{n}$
and $\|g_{n}\|\leq K(\|f\|)$, as given by $H^{\mathrm{suf}}_{n}$-expansion.

Then the sequence $\langle g_{n}{\rangle}_{n}$ has an accumulation point, say
$g$, since its elements are uniformly bounded (by $K(\|f\|)$). Since $\partial
g$ is also an accumulation point of $(\partial g_{n})_{n}$, which converges to
$f$, we conclude that $\partial g=f$.

Thus, $f\in B^{\mathrm{uf}}_{n}(X,\mathbb{Z})$ and
$H^{\mathrm{suf}}_{n}(X,\mathbb{Z})=0$. ∎

######  Remark 6.4.

Consider the following condition on $X$: There exists some constant $N$ such
that for any $f\in Z^{\mathrm{suf}}_{n}(X,\mathbb{Z})$, one can write
$f=\sum_{i=1}^{N\|f\|}f_{i}$, with each $f_{i}$ of norm $1$ and in
$Z^{\mathrm{suf}}_{n}(X,\mathbb{Z})$. If this condition held, we could always
take the function $K:\mathbb{N}\rightarrow\mathbb{N}$ in
$H^{\mathrm{suf}}_{n}$-expansion to be linear. We do not know whether it
holds, even in dimension $1$ (that is: can one decompose any flow of norm $k$
into a sum of $Nk$ flows of norm $1$, for some universal $N$ ?)

In the zero-dimensional case, $H^{\mathrm{suf}}$-expansion is actually
equivalent to non-amenability. Moreover since $B_{0}^{suf}=Z_{0}^{suf}$, also
pure $H^{\mathrm{suf}}_{0}$ expansion coincides with $H^{\mathrm{suf}}_{0}$
expansion.

######  Proposition 6.5 ($H^{\mathrm{suf}}_{0}$-expansion is just expansion.).

Let $X$ be a ULF graph. Then $X$ has $H^{\mathrm{suf}}_{0}$-expansion iff $X$
is non-amenable.

###### Proof.

By Proposition 6.1 and Lemma 6.3, we get $H^{\mathrm{suf}}_{0}$-expansion
$\Rightarrow$ non-amenability.

Now for the converse: By decomposing such an $f:U\rightarrow\mathbb{Z}$ into a
sum of $0/1$-valued function, the condition reduces to:

  * •

There exists $K\in{\mathbb{N}}$ such that for all $U{\subseteq}_{f}VX$ and
$W\subseteq U$, there exists $g:EX\rightarrow\mathbb{Z}$ with $\partial
g|U={\chi}_{W}$ and $\|g\|\leq K$.

Assume now that $X$ is non-amenable, with Cheeger constant $\varepsilon$. By
adapting the proof of [4, Lemma 2.1] (more precisely, their construction of a
flow), we see that given $U{\subseteq}_{f}VX$ and $W\subseteq U$, one can
construct a $g$ satisfying $\partial g|U={\chi}_{W}$. Since in their argument,
the “flux” (=excess flow) at vertices in $W$ is $\varepsilon$ and the flow at
edges at most $1$, scaling by ${\varepsilon}^{-1}$ yields the bound of
$K:={\varepsilon}^{-1}$ for $g$.

∎

###  6.2 Dimension $1$

######  Theorem 6.6.

If $X$ is an infinite, vertex transitive ULF simplicial complex and
$H^{\mathrm{suf}}_{1}(X,\mathbb{Z})=0$, then $X$ has
$H^{\mathrm{suf}}_{1}$-expansion.

###### Proof.

Suppose $X$ does not have $H^{\mathrm{suf}}_{1}$-expansion but
$H^{\mathrm{suf}}_{1}(X,\mathbb{Z})=0$. Then, there exist sequences $\langle
U_{i}{\subseteq}_{f}VX\rangle$ and $\langle f_{i}\in
C^{\mathrm{suf}}_{1}(X,\mathbb{Z}){\rangle}_{i}$ as in the definition, with
$\|f_{i}\|\leq C$ for some $C$, but such that if $\partial g_{i}|U_{i}=f_{i}$,
then $\|g_{i}\|\geq K_{i}$, with $K_{i}\rightarrow\infty$ (the sequence
$\langle f_{i}{\rangle}_{i}$ becomes “harder” to kill with $i$ increasing).

Let $f_{i}^{\prime}\in Z^{\mathrm{suf}}_{1}(X,\mathbb{Z})$ be an extension of
$f_{i}$, as in the definition, but with finite support (their existence is
proven in Lemma 6.7). Since $X$ is infinite and transitive, one can assume
that the support of all $f_{i}^{\prime}$s are pairwise disjoint.

Let then $F=\sum_{i}f_{i}^{\prime}$. $F\in Z^{\mathrm{suf}}_{1}(X,\mathbb{Z})$
implies the existence of some $G\in C^{\mathrm{uf}}_{2}(X,\mathbb{Z})$ with
$\partial G=F$. In particular, $\partial G|U_{i}=f_{i}$, and since $\|G\|\leq
K_{i}$ for large enough $i$, we get a contradiction. ∎

Note that we do not actually need as much as transitivity, but merely being
able to move finite pieces of the graphs sufficiently far, so that, e.g.
having a cocompact action by the automorphism group is enough.

######  Lemma 6.7.

Assume $X$ is ULF and one-ended. Let $U{\subseteq}_{f}VX$ and $f\in
C^{\mathrm{suf}}_{1}(U,\mathbb{Z})$ satisfying $\partial f\subseteq\partial U$
and extensible to $\hat{f}\in Z^{\mathrm{suf}}_{1}(X,\mathbb{Z})$ with
$\|\hat{f}\|=\|f\|$. Then $\hat{f}$ can be taken with finite support.

###### Proof.

That $f$ can be extended to $\hat{f}\in Z^{\mathrm{suf}}_{1}(X,\mathbb{Z})$
implies the existence of a finite number of vertices
$v_{1},\ldots,v_{n}\in\partial U$, and integers $m_{1},\ldots,m_{n}$ such that
$\partial f=\sum m_{i}v_{i}$. Since $\partial f\subseteq\partial U$, we must
have $\sum m_{i}=0$. Note that the values at these vertices must be killed. To
do this, one can either connect two vertices of opposite sign outside of $U$,
or add an infinite ray. Because both the number of $v_{i}$ as their
coefficients are finite, there are at most finitely many rays. Consider a ball
enveloping the finite structure of $\hat{f}$, such that only the rays are
escaping. Take any ray coming from a positive vertex and any ray coming from a
negative vertex. Since $X$ is one-ended, both rays can be connected outside of
the chosen ball by a path $p$. If the path does not cross any other rays, we
can reconnect locally. If it crosses other rays we can see this as sequence of
basis vertices starting in one with a positive coefficient, ending in a
negative one. Hence there exist two consecutive rays such that their vertices
have opposite sign. This ends the proof. ∎

The above “finite extension” property for dimension $1$ is crucial in our
proof that for transitive infinite graphs, vanishing of $H^{\mathrm{suf}}_{1}$
is equivalent to $H^{\mathrm{suf}}_{1}$-expansion. Were it to also hold for
higher dimension, the proof of Theorem 6.6 would obviously generalise.

####  6.2.1 $H^{\mathrm{uf}}$

In the above, we focused on $H^{\mathrm{suf}}$, since
$H^{\mathrm{suf}}$-expansion is defined in terms of the simplicial complex at
hand. In the (coarse) setting of $H^{\mathrm{uf}}$, this is obviously
problematic. Our aim now is essentially to coarsify (pure)
$H^{\mathrm{suf}}$-expansion.

######  Definition 6.8 ($H^{\mathrm{uf}}_{1}$-expansion).

A ULF graph $X$ is said to have (pure) $H^{\mathrm{uf}}_{1}$-expansion iff its
Rips complexes eventually have (pure) $H^{\mathrm{suf}}_{1}$-expansion.

The following proposition justifies our interpretation of non-vanishing of
$H^{\mathrm{uf}}_{1}(X,\mathbb{Z})$ in terms of the three phenomena of ends,
large circuits and expansion:

######  Proposition 6.9.

If $X$ has $\mathbb{Z}$-small circuits (i.e.
$\mathcal{C}_{r}(X,\mathbb{Z})=\mathcal{C}_{\infty}(X,\mathbb{Z})$ for some
$r$) and is one-ended, then $H^{\mathrm{uf}}_{1}(X,\mathbb{Z})\neq 0$ implies
that $X$ does not have pure $H^{\mathrm{uf}}_{1}$-expansion.

In other words, if $H^{\mathrm{uf}}_{1}(X,\mathbb{Z})\neq 0$ then at least one
of the three phenomena is responsible for it. In the non-transitive case, it
is not clear that the converse holds, that is, that
$H^{\mathrm{uf}}_{1}(X,\mathbb{Z})=0$ would imply at least one of: ends, large
circuits, expansion.

###### Proof.

Assume $H^{\mathrm{uf}}_{1}(X,\mathbb{Z})\neq 0$, $X$ is one-ended and for
some $r$, $\mathcal{C}_{r}(X,\mathbb{Z})=\mathcal{C}_{\infty}(X,\mathbb{Z})$.
Then, by Corollary 2.5, for $s$ big enough,
$H^{\mathrm{suf}}_{1}(R_{s}X,\mathbb{Z})\neq 0$. By Lemma 6.3, it follows that
$R_{s}X$ does not have expansion, and it remains to check that it dos not have
_pure_ expansion either.

Fix $(U,f)$ with $f\in C^{\mathrm{suf}}_{1}(R_{s}U,\mathbb{Z})$ with $\partial
f\subseteq\partial U$ and $f$ extensible.

Since $R_{s}X$ does not have expansion, it suffices to check that there exists
some $g\in C^{\mathrm{suf}}_{2}(R_{s}X,\mathbb{Z})$ with $\partial
g|{R_{s}U}=f$ and $\|\partial g\|=\|f\|$. Let $\hat{f}$ a finite extension of
$f$, as given by Lemma 6.7. By “tracing virtual edges”, $\hat{f}$ is, up to a
finite sum of boundaries of triangles in $R_{s}X$, an element of
$Z^{\mathrm{suf}}_{1}(X,\mathbb{Z})$, of finite support. In particular, up to
a finite sum of boundaries of triangles, $\hat{f}$ lies in
$\mathcal{C}_{\infty}(X,\mathbb{Z})=\mathcal{C}_{r}(X,\mathbb{Z})$, and can be
decomposed into a sum of circuits of length $\leq r$. Now, if $s$ is large
enough, each of those circuits can itself be decomposed as a sum of boundaries
of triangles in $R_{s}X$, so that $\hat{f}$ can be written as a sum of
boundaries of triangles of $R_{s}X$. In other terms, $\hat{f}=\partial g$ for
some $g\in C^{\mathrm{suf}}_{2}(R_{s}X,\mathbb{Z})$.

∎

######  Theorem 6.10 ($H^{\mathrm{uf}}_{1}$ triad).

Let $X$ a transitive graph, and consider the following three conditions:

  1. 1.

$X$ has more than one end.

  2. 2.

$X$ has $\mathbb{Z}$-large circuits; that is,
$\mathcal{C}_{r}(X,\mathbb{Z})\neq\mathcal{C}_{\infty}(X,\mathbb{Z})$ for any
$r$.

  3. 3.

$X$ does not have pure $H^{\mathrm{uf}}_{1}$-expansion.

Then $H^{\mathrm{uf}}_{1}(X,\mathbb{Z})\neq 0$ iff either of the above holds.

###### Proof.

By the results in Section 3, know already that Item 1 implies non-zero
homology. Similarly, the results in Section 5 give that Item 2 implies non-
zero homology.

Finally, assume that Item 3 holds. If any of Items 1 and 2 hold, then
obviously $H^{\mathrm{uf}}_{1}(X,\mathbb{Z})\neq 0$ by what we already know.
So, one may assume now that for some $r$ large enough,
$\mathcal{C}_{r}(X,\mathbb{Z})=\mathcal{C}_{\infty}(X,\mathbb{Z})$. Since
$R_{r}X$ does not have (pure) expansion ($r$ large enough),
$H^{\mathrm{suf}}_{1}(R_{r}X,\mathbb{Z})\neq 0$, and by Corollary 2.5,
$H^{\mathrm{uf}}_{1}(X,\mathbb{Z})\neq 0$ either.

For the converse, it suffices to see that if
$H^{\mathrm{uf}}_{1}(X,\mathbb{Z})\neq 0$, but $X$ has one end and
$\mathbb{Z}$-small circuits (i.e. neither of Items 1 and 2 hold), then we can
apply Proposition 6.9

∎

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