# C*-algebras of higher-rank graphs from groups acting on buildings, and
explicit computation of their K-theory

Sam A. Mutter, Aura-Cristiana Radu, Alina Vdovina

###### Abstract

We unite elements of category theory, K-theory, and geometric group theory, by
defining a class of groups called $k$-cube groups, which act freely and
transitively on the product of $k$ trees, for arbitrary $k$. The quotient of
this action on the product of trees defines a $k$-dimensional cube complex,
which induces a higher-rank graph. We make deductions about the K-theory of
the corresponding $k$-rank graph $C^{\star}$-algebras, and give examples of
$k$-cube groups and their K-theory. These are the first explicit computations
of K-theory for an infinite family of $k$-rank graphs for $k\geq 3$, which is
not a direct consequence of the Künneth Theorem for tensor products.

##  1 Introduction

A $k$-rank graph is a combinatorial object defined by Kumjian and Pask in
[15], with motivation from Robertson and Steger [21]. To each $k$-rank graph
can be assigned a $k$-rank graph algebra, being the universal
$C^{\star}$-algebra generated by a set of partial isometries, and deductions
about the K-theory of these algebras can be made courtesy of the spectral
sequences exhibited by Evans in [6]. In general, it is difficult to calculate
K-theory for higher-rank graphs. In this article we construct an infinite
family of $k$-rank graphs, for arbitrary $k$, and remark on the K-theory when
$k=3,4,5$.

Robertson and Steger laid the foundations for these $k$-dimensional
generalisations of graph algebras, which are $C^{\star}$-algebras built with
data obtained from a directed graph. The K-theory of these algebras when $k=2$
has been detailed in [22], [12], and until now, only a small selection of
examples had been investigated in this way. Here, we further develop Robertson
and Steger’s methods, using Kumjian and Pask’s category-theoretical language,
to present an infinite family of examples arising from groups which act on the
product of $k$-many regular trees. We outline the process now.

A VH-structure on a square complex $\mathcal{M}$ is a partition of its
directed edge set $E(\mathcal{M})=E_{V}\sqcup E_{H}$ such that the link at
each vertex $x$ of $\mathcal{M}$ is the complete bipartite graph induced by
the partition. Such objects were pioneered by [27] and further studied in [3],
[4], [12]. It was shown, amongst other places, in [1], that the universal
cover of a square complex is a product of two trees if and only if the link at
each vertex is a complete bipartite graph. Let $T_{1}$, $T_{2}$ be regular
trees of constant valencies $m$, $n$, respectively, and let $\Gamma$ be a
lattice in $T_{1}\times T_{2}$, that is, a group which acts discretely and
cocompactly on $\operatorname{Aut}(T_{1})\times\operatorname{Aut}(T_{2})$,
respecting the structure of cube complexes and such that
$\Gamma\setminus(T_{1}\times T_{2})$ is a finite square complex. From [4] we
learn that $\Gamma$ corresponds uniquely to a square complex with VH-structure
of partition size $(m,n)$, up to isomorphism; indeed,
$\Gamma\setminus(T_{1}\times T_{2})$ is such a complex.

Let $\Gamma$ be a group, and let $A,B\subseteq\Gamma$ be finite subsets closed
under inverses. We say that the pair $(E_{1},E_{2})$ is a VH-structure on
$\Gamma$ if $E_{1}\cup E_{2}$ generates $\Gamma$, and the product sets
$E_{1}E_{2}$ and $E_{2}E_{1}$ are equal with size $|E_{1}|\cdot|E_{2}|$, and
without $2$-torsion. We may define a BMW-group (named for Burger, Mozes, and
Wise) as a group which admits a VH-structure, as in [3], [27]. A BMW-group
acts freely and transitively on a product of two trees, and yields a square
complex with one vertex and a VH-structure in a natural way [4]. We may
therefore freely interchange geometric and algebraic terminology.

Inspired by this, Vdovina in [25] defined a $k$-dimensional generalisation: a
partition into $k$ subsets $E_{1}\sqcup\cdots\sqcup E_{k}$ of the directed
edge set of a $k$-dimensional cube complex $\mathcal{M}$ which uniquely
corresponds to a lattice in the product of $k$ trees of constant valencies
$|E_{1}|,\ldots,|E_{k}|$. We call this an adjacency structure for
$\mathcal{M}$. Likewise, we define a $k$-cube group to be a group $\Gamma$
which admits $k$ subsets $E_{1},\ldots,E_{k}$ which satisfy a compatibility
condition, whose union generates $\Gamma$ and which, taken pairwise, have the
same properties as above. A $k$-cube group acts freely and transitively on a
product of $k$ trees—this is a $k$-rank affine building $\Delta$ which is
thick whenever $|E_{i}|>2$ for all $i$. The quotient of this action is a
$k$-dimensional cube complex with one vertex and endowed with a $k$-adjacency
structure. We identify the $k$-dimensional cells (chambers) of $\Delta$ with
the tuple of elements of $\Gamma$ which label their edges; in this way we
define the notion of a $k$-cube in the group $\Gamma$.

In Section 2, we construct adjacency functions on the sets of $k$-cubes of
$\Gamma$, defining two $k$-cubes to be adjacent if the cells they define in
$\Delta$ are adjacent in a certain way. This generalises the $2$-dimensional
shift system explored in [12], where two squares were considered adjacent if
they could be stacked against one another horizontally or vertically. In $k$
dimensions, there are $k$ ‘directions’ in which $k$-cubes can be stacked, and
hence we define $k$-many adjacency functions. We show in Proposition 2.17 that
the adjacency functions satisfy a Unique Common Extension Property in the
following sense: firstly, consider a $2\times 2\times 2$ arrangement of
$k$-cubes. Then, given an initial $k$-cube and three $k$-cubes adjacent to it
in three mutually-orthogonal directions, we can uniquely find four more
$k$-cubes which fill in the $2\times 2\times 2$ structure (Figure 5). In [21],
Roberston and Steger show that this $3$-dimensional commutativity of $k$-cubes
is enough to imply unique common extensions in all dimensions up to $k$. As
such, we are able to conclude in Section 3 that a $k$-cube group $\Gamma$
induces a $k$-rank graph $\mathcal{G}(\Gamma)$.

In Theorem 4.10, we show that the K-theory of our $k$-rank graph
$C^{\star}$-algebras determines the $C^{\star}$-algebras uniquely, up to
isomorphism.

We use a technique of [24] to build examples of $k$-cube groups, and uncover
enough about their K-theory to be able to distinguish their induced $k$-rank
graph $C^{\star}$-algebras. We note that a $k$-cube group $\Gamma$ is an
amalgamated product of $(k-1)$-cube groups, and the induced K-theory has no
immediately-discernible relation to that of the $(k-1)$-cube groups which
$\Gamma$ contains.

In the final section, we use Evans’ K-theory formulas and our own corollaries
to study some $k$-rank graphs arising from cube complexes in a second way,
namely as double covers of the cube complexes constructed in Section 2. This
work follows from that of [16].

##  2 $k$-cube groups

For some finite $n\geq 2$, define $T(n)$ to be the regular tree of degree $n$.
We may simply write $T$ if the degree is not important.

Let $T_{1},\ldots,T_{k}$ be regular trees, and consider the product
$T_{1}\times\cdots\times T_{k}$. This defines a $k$-dimensional cube complex
$\Delta$, which is an affine building of rank $k$.

Recall that the link at a vertex $x$ of a $k$-dimensional cell complex $G$ is
the $(k-1)$-dimensional cell complex $\operatorname{lk}_{x}(G)$ obtained as
the intersection of $G$ with a small $2$-sphere centred at $x$.

######  Proposition 2.1.

Let $\mathcal{M}$ be a $k$-dimensional cube complex. The universal cover of
$\mathcal{M}$ is a product of $k$ trees
$\tilde{\mathcal{M}}=T_{1}\times\cdots\times T_{k}$ if and only if the link at
each vertex of $\mathcal{M}$ is a clique complex of a complete $k$-partite
graph.

###### Proof.

This proposition is a generalisation of Theorem 10.2 in [2]. Observe that if
the link $\operatorname{lk}_{x}(\mathcal{M})$ at a vertex $x$ of $\mathcal{M}$
is such a clique complex, then $\operatorname{lk}_{x}(\mathcal{M})$ is a
$(k-1)$-dimensional complex such that every cycle has length at least $k$.
Hence $\operatorname{lk}_{x}(\mathcal{M})$ is $\operatorname{CAT}(1)$, and so
by the Gromov Link Condition [9, §4.2], $\mathcal{M}$ must be
$\operatorname{CAT}(0)$. The result then follows from a relatively
straightforward adaptation to Theorem 4.3 in [2]. ∎

The following definitions generalise objects from [12] and [27].

######  Definition 2.2.

Let $\mathcal{M}$ be a $k$-dimensional cube complex with vertex set $V$ and
edge set $E$. For each $x\in V$, write $E(x)$ for the set of oriented edges
originating at $x$. Suppose that we have a partition
$E=E_{1}\sqcup\cdots\sqcup E_{k}$ such that $u^{-1}\in E_{i}$ whenever $u\in
E_{i}$, and suppose that for each vertex $x\in V$, the $1$-skeleton of the
link at $x$ is the complete $k$-partite graph with vertices according to the
partition $E(x)=E(x)_{1}\sqcup\cdots\sqcup E(x)_{k}$. We say that
$E_{1},\ldots,E_{k}$ form an adjacency structure for $\mathcal{M}$.

######  Definition 2.3.

Let $\mathcal{M}$ be a $k$-dimensional cube complex with vertex set $V$, edge
set $E$, and adjacency structure $E_{1},\ldots,E_{k}$. Write
$\mathcal{S}_{2}^{\prime}=\mathcal{S}_{2}^{\prime}(\mathcal{M})$ for the set
of geometric squares of which $\mathcal{M}$ consists. We write elements of
$\mathcal{S}_{2}^{\prime}$ as ordered $4$-tuples of their oriented edge labels
$(a,b,a^{\prime},b^{\prime})$ for $a,a^{\prime}\in E_{i}$, $b,b^{\prime}\in
E_{j}$, where the map $e\mapsto e^{-1}$ reverses the orientation of the edge.
We use square brackets if we wish to emphasise that a square is labelled
according to some predetermined orientation and starting from some basepoint.
For each square $S=[a,b,a^{\prime},b^{\prime}]\in\mathcal{S}_{2}$, write:

$S_{H}:=\big{[}a^{-1},(b^{\prime})^{-1},(a^{\prime})^{-1},b^{-1}\big{]},\quad
S_{R}:=\big{[}a^{\prime},b^{\prime},a,b\big{]},\quad
S_{V}:=\big{[}(a^{\prime})^{-1},b^{-1},a^{-1},(b^{\prime})^{-1}\big{]};$

geometrically these can be interpreted as the pointed squares which lie in the
same orbit of $S$ under the actions of reflection in the $a$ direction,
rotation by $\pi$, and reflection in the $b$ direction. We define the set

$\mathcal{S}_{2}:=\\{S,S_{H},S_{R},S_{V}\mid S\in\mathcal{S}_{2}^{\prime}\\}.$

Write $F(p,q):=\\{[a,b,a^{\prime},b^{\prime}]\in\mathcal{S}_{2}\mid
a,a^{\prime}\in E_{p}\text{, and }b,b^{\prime}\in E_{q}\\}$, and identify
$F(p,q)$ with $F(q,p)$ via the map
$\varphi:[a,b,a^{\prime},b^{\prime}]\mapsto\big{[}(b^{\prime})^{-1},(a^{\prime})^{-1},b^{-1},a^{-1}\big{]}$.

Similarly, we write
$\mathcal{S}_{3}^{\prime}=\mathcal{S}_{3}^{\prime}(\mathcal{M})$ for the set
of geometric cubes which $\mathcal{M}$ comprises, and we denote elements of
$\mathcal{S}_{3}^{\prime}$ by ordered $6$-tuples of their faces
$(A,B,C,A^{\prime},B^{\prime},C^{\prime})$ for $A,A^{\prime}\in F_{ij}$,
$B,B^{\prime}\in F_{il}$, and $C,C^{\prime}\in F_{jl}$. As above, we use
square brackets to indicate that a cube is pointed and oriented, and for each
cube $S=[A,B,C,A^{\prime},B^{\prime},C^{\prime}]\in\mathcal{S}_{3}$, we write:

$\begin{array}[]{rclrcl}S_{H}&:=&\big{[}A_{H},B_{H},C_{H}^{\prime},A_{H}^{\prime},B_{H}^{\prime},C_{H}\big{]},&S_{HI}&:=&\big{[}A^{\prime},B_{R},C_{R}^{\prime},A,B_{R}^{\prime},C_{R}\big{]},\\\
S_{R}&:=&\big{[}A_{R},B^{\prime},C^{\prime},A_{R}^{\prime},B,C\big{]},&S_{RI}&:=&\big{[}A_{V}^{\prime},B_{V}^{\prime},C_{V}^{\prime},A_{V},B_{V},C_{V}\big{]},\\\
S_{V}&:=&\big{[}A_{V},B_{H}^{\prime},C_{H},A_{V}^{\prime},B_{H},C_{H}^{\prime}\big{]},&S_{VI}&:=&\big{[}A_{R}^{\prime},B_{R}^{\prime},C_{R},A_{R},B_{R},C_{R}^{\prime}\big{]},\\\
S_{I}&:=&\big{[}A_{H}^{\prime},B_{V},C_{V},A_{H},B_{V}^{\prime},C_{V}^{\prime}\big{]}.&&&\end{array}$
(1)

These are the cubes $[X_{1},\ldots,X_{6}]$ which belong to the same orbit as
$[A,B,C,A^{\prime},B^{\prime},C^{\prime}]$ under action by the symmetry group
of the cube, with the property that if $A\in F(i,j)$, then $X_{1}\in F(i,j)$
(Figure 1). Write $\mathcal{S}_{3}$ for the set which comprises each
$S\in\mathcal{S}_{3}^{\prime}$ and all of the corresponding pointed cubes
above. Write

$F(p,q,r):=\\{[A,B,C,A^{\prime},B^{\prime},C^{\prime}]\in\mathcal{S}_{3}\mid
A,A^{\prime}\in F(p,q)\\\ B,B^{\prime}\in F(p,r)\text{, and }C,C^{\prime}\in
F(q,r)\\},$

and identify $F(a,b,c)$ with $F(a,c,b)$ via the map

$(A,B,C,A^{\prime},B^{\prime},C^{\prime})\longmapsto\big{(}\varphi(A),C_{H}^{\prime},B_{H}^{\prime},\varphi(A^{\prime}),C_{H},B_{H}\big{)}.$

Likewise we are able to identify $F(a,b,c)$ with each of the sets
$F(\sigma(a,b,c))$, for each permutation $\sigma$.

For $3\leq n\leq k$, we inductively define the sets
$\mathcal{S}_{n}^{\prime}=\mathcal{S}_{n}^{\prime}(\mathcal{M})$ of geometric
$n$-cubes of which $\mathcal{M}$ consists, and we write elements of
$\mathcal{S}_{n}^{\prime}$ as ordered $(2n)$-tuples of their faces (the
incident elements of $\mathcal{S}_{n-1}$). Here, we define $\mathcal{S}_{n}$
as the set of all $(2n)$-cubes $(A_{1},\ldots,A_{2n})$ which belong to the
orbit of some
$(A_{1}^{1},\ldots,A_{1}^{n},A_{2}^{1},\ldots,A_{2}^{n})\in\mathcal{S}_{n}^{\prime}$
under the action of the group of symmetries of the $(2n)$-cube, with the
property that if $A_{1}^{1}\in F(p_{1},\ldots,p_{n-1})$, then $A_{1}\in
F(p_{1},\ldots,p_{n-1})$. Analogously to above, we write

$F(p_{1},\ldots,p_{n}):=\big{\\{}\big{[}A_{1}^{1},\ldots,A_{1}^{n},A_{2}^{1},\ldots,A_{2}^{n}\big{]}\in\mathcal{S}_{n}\bigm{|}A_{1}^{i},A_{2}^{i}\in
F\big{(}p_{1},\ldots,\hat{p}_{n-i+1},\ldots,p_{n}\big{)}\big{\\}},$

and we can identify the sets $F(\sigma(p_{1},\ldots,p_{n}))$ for each
permutation $\sigma$.

Figure 1: For a pointed, oriented $3$-cube
$S=[A,B,C,A^{\prime},B^{\prime},C^{\prime}]\in\mathcal{S}_{3}$, the seven
corresponding cubes from (1) are defined by reflecting and rotating $S$
according to the arrows above. The transformations map the original basepoint
to a new vertex (blue), but the new cubes are given the same basepoint and
orientation as $S$ (black).

######  Definition 2.4.

Let $k\geq 2$, and let $E_{1},\ldots,E_{k}$ be finite sets of respective even
cardinalities $2m_{1},\ldots,2m_{k}$, with each $m_{i}\geq 2$. Suppose that
each set $E_{i}$ is endowed with a fixed-point-free involution, denoted
$a\mapsto a^{-1}$. For each $i,j$ with $i\neq j$, write $F(i,j):=E_{i}\times
E_{j}\times E_{i}\times E_{j}$, let $R\subseteq\bigsqcup_{i\neq j}F(i,j)$, and
define the group

$\Gamma:=\langle E_{1}\sqcup\cdots\sqcup E_{k}\mid
aba^{\prime}b^{\prime}=1\text{ whenever }(a,b,a^{\prime},b^{\prime})\in
R\rangle.$

Suppose firstly that $k=2$. We call $\Gamma$ a BMW-group if $R$ has the
following properties:

  * C1

For each element $(a,b,a^{\prime},b^{\prime})\in R$, each of
$(a^{-1},(b^{\prime})^{-1},(a^{\prime})^{-1},b^{-1})$,
$(a^{\prime},b^{\prime},a,b)$, and
$((a^{\prime})^{-1},b^{-1},a^{-1},(b^{\prime})^{-1})$ is also in $R$, and all
four $4$-tuples are distinct.

  * C2

Each of the projections of $R$ onto the subproducts of the form $E_{i}\times
E_{j}$ or $E_{j}\times E_{i}$, for all $i\neq j$, is bijective.

These groups were developed extensively in [3], [12] and [27]. In order to
generalise BMW-groups to $k\geq 3$, we require the construction of subsets
$\mathcal{S}_{n}\subseteq\prod_{\alpha=1}^{2n}R$ for each
$n\in\\{2,\ldots,k\\}$. We begin by illustrating $\mathcal{S}_{3}$.

Suppose, then, that $k\geq 3$, and fix $(a_{1},b_{1},a_{2},b_{2})\in F(p,q)$
and $(a_{1},c_{1},a_{3},c_{2})\in F(p,r)$. Also suppose that we can find some
unique elements $a_{4},b_{3},b_{4},c_{3},c_{4}\in\bigcup_{l}E_{l}$ such that
$\big{(}b_{1},c_{3}^{-1},b_{4},c_{1}^{-1}\big{)}$,
$\big{(}b_{2},c_{2}^{-1},b_{3},c_{4}\big{)}\in F(q,r)$,
$\big{(}a_{2},c_{4}^{-1},a_{4},c_{3}\big{)}\in F(p,r)$, and
$(a_{3},b_{3},a_{4},b_{4})\in F(p,q)$. Equivalently, suppose that the same is
true if we are given $(a_{1},b_{1},a_{2},b_{2})\in F(p,q)$ and
$\big{(}b_{1},c_{3}^{-1},b_{4},c_{1}^{-1}\big{)}\in F(q,r)$. Geometrically, we
can view each $4$-tuple as a square, such that each of $a_{1},\ldots,a_{4}$,
$b_{1},\ldots,b_{4}$, $c_{1},\ldots,c_{4}$ labels the edges of a cube. Write
$\mathcal{S}_{3}$ for the set of $6$-tuples of elements of $R$ which
correspond to the faces of all such cubes, pointed and oriented according to
some predetermined orientation.

We extend the notion of $\mathcal{S}_{3}$ to that of $\mathcal{S}_{n}$ as
follows. Suppose that $k\geq n$, and fix $p\in\\{1,\ldots,k\\}$. Let
$J\subseteq(\\{1,\ldots,k\\}\setminus\\{p\\})$ be some subset of cardinality
$(n-1)$, and let $L\subseteq J$ be a subset of cardinality $|L|\geq 0$, where
$|L|=0$ implies $L=\emptyset$.

Fix $(n-1)$ elements $\big{(}u,v_{j},u^{j},w_{j}\big{)}\in F(p,j)$, for each
$j\in J$. Presume that we can find unique elements $u^{L}\in E_{p}$ for each
$L$ with $1\leq|L|\leq(n-1)$, and unique $v_{i}^{L},w_{i}^{L}\in E_{i}$ for
each $L$ with $0\leq|L|\leq(n-2)$, such that:

  1. (a)

$\Big{(}u^{L},v_{j}^{L},u^{L\cup\\{j\\}},w_{j}^{L}\Big{)}\in F(p,j)$, whenever
$j\in J$ and $j\notin L$.

  2. (b)

$\Big{(}\big{(}v_{i}^{L}\big{)}^{-1},v_{j}^{L},\Big{(}w_{i}^{L\cup\\{j\\}}\Big{)}^{-1},w_{j}^{L\cup\\{i\\}}\Big{)}\in
F(i,j)$, for all $i,j\in J$ with $i\neq j$ and $i,j\notin L$.

  3. (c)

$\Big{(}w_{i}^{L},\big{(}w_{j}^{L}\big{)}^{-1},v_{i}^{L\cup\\{j\\}},\Big{(}v_{j}^{L\cup\\{i\\}}\Big{)}^{-1}\Big{)}\in
F(i,j)$, for all $i,j\in J$ with $i\neq j$ and $i,j\notin L$.

We write $\square(u,v_{j},u^{j},w_{j})$ for the $2n$-tuple of $R$ comprising
the initial choices $(u,v_{j},u^{j},w_{j})$ and the elements above which they
uniquely determine, listed according to some predetermined order. We write
$\mathcal{S}_{n}=\mathcal{S}_{n}(\Gamma)$ for the set of all $2n$-tuples
$\square(u,v_{j},u^{j},w_{j})$. Elements of $\mathcal{S}_{n}$ can be regarded
as pointed, oriented $n$-cubes whose $2$-faces are labelled by elements of $R$
(Figure 2). We call elements of $\mathcal{S}_{n}$ pointed $n$-cubes.

Note that we may identify each $F(i,j)$ with $F(j,i)$ via the isomorphism
$(a,b,a^{\prime},b^{\prime})\mapsto\big{(}(b^{\prime})^{-1}),(a^{\prime})^{-1},b^{-1},a^{-1}\big{)}$,
such that we need only consider only those $i,j$ with $i<j$. We refrained from
doing so immediately in order to simplify some of the above notation.

Now, if $k\geq 3$, we call $\Gamma$ a $k$-cube group whenever, in addition to
conditions C1 and C2 above, $R$ also satisfies:

  * C3

For each set of $(n-1)$ elements $\big{(}u,v_{j},u^{j},w_{j}\big{)}\in
F(p,j)$, we can find unique elements $u^{L},v_{i}^{L},w_{i}^{L}\in E_{i}$
which satisfy (a)–(c) above, and hence are able to construct subsets
$\mathcal{S}_{n}\subseteq\prod_{\alpha=1}^{2n}R$ for each
$n\in\\{2,\ldots,k\\}$.

######  Remark 2.5.

In [11], an alternative condition was given on $R$, which is equivalent to C1
and C2:

  * C1’

The product sets $E_{i}E_{j}$ and $E_{j}E_{i}$ are equal, contain no
$2$-torsion, and have cardinality
$|E_{i}E_{j}|=|E_{i}|\cdot|E_{j}|=4m_{i}m_{j}$.

Since we are explicitly constructing $k$-cube complexes, we mainly rely on
properties C1 and C2 in this paper.

Figure 2: Let $k\geq 4$. Above is depicted a pointed $4$-cube in
$\mathcal{S}_{4}$, for some $k$-cube group with adjacency structure
$E_{1},\ldots,E_{k}$. Let $u^{L}$ be elements of $E_{1}$, and
$v_{i}^{L},w_{i}^{L}\in E_{i}$. Fix three mutually-adjacent squares: elements
of $F(1,j)$ labelled $(u,v_{j},u^{j},w_{j})$, for $j\in\\{2,3,4\\}$. Then each
of the remaining $u^{L},v_{i}^{L},w_{i}^{L}$ is uniquely-determined, such that
they label the edges of the $4$-cube above. We have condensed the notation for
the sets $L$ for clarity.

######  Lemma 2.6.

Let $\Gamma=\langle X\mid R\rangle$ be a $k$-cube group with adjacency
structure $E_{1},\ldots,E_{k}$, and let $\Gamma^{\prime}(p)=\langle
X^{\prime}(p)\mid R^{\prime}(p)\rangle\subset\Gamma$ be the subgroup obtained
by removing all of the elements of some set $E_{p}$ from the generating set
$X$. Then $\Gamma^{\prime}(p)$ is a $(k-1)$-cube group, with adjacency
structure $E_{1},\ldots,\hat{E}_{p},\ldots,E_{k}$.

By induction, we can form a $(k-m)$-cube subgroup by removing all elements of
$m$ sets $E_{p_{1}},\ldots,E_{p_{m}}$ from the generating set $X$. We denote
such a group by $\Gamma^{\prime}(p_{1},\ldots,p_{m})$.∎

######  Proposition 2.7.

Let $\Gamma$ be a $k$-cube group with adjacency structure
$E_{1},\ldots,E_{k}$, and write $\Gamma^{\prime}(p_{1},\ldots,p_{m})=\langle
X^{\prime}(p_{1},\ldots,p_{m})\mid
R^{\prime}(p_{1},\ldots,p_{m})\rangle\subset\Gamma$ to denote the $(k-m)$-cube
subgroup constructed in Lemma 2.6. Then

$\Gamma=\big{(}\big{(}\big{(}\Gamma^{\prime}(1)\ast_{\langle X^{\prime}(1)\cap
X^{\prime}(2)\rangle}\Gamma^{\prime}(2)\big{)}\ast_{\langle
X^{\prime}(3)\rangle}\Gamma^{\prime}(3)\big{)}\ast_{\langle
X^{\prime}(4)\rangle}\cdots\big{)}\ast_{\langle
X^{\prime}(k)\rangle}\Gamma^{\prime}(k),$

where $\ast_{G}$ is the amalgamated free product over the group $G$.

###### Proof.

Firstly, write $G_{2}:=\Gamma^{\prime}(1)\ast_{\langle X^{\prime}(1)\cap
X^{\prime}(2)\rangle}\Gamma^{\prime}(2)$, and then

$G_{i+1}:=G_{i}\ast_{\langle X^{\prime}(i)\rangle}\Gamma^{\prime}(i),$

for all $2\leq i\leq k-1$. Then $G_{2}$ has generating set $X$, and relation
set $R^{\prime}(1)\cup R^{\prime}(2)$. At each step, we amalgamate over the
free group generated by the intersection of $X$ with $X^{\prime}(i)$, which is
$X^{\prime}(i)$. Hence each $G_{i}$ is generated by $X$, and has relation set
$R^{\prime}(1)\cup\cdots\cup R^{\prime}(i)$. But $R^{\prime}(1)\cup\cdots\cup
R^{\prime}(k)=R$, and so $G_{k}=\Gamma$. ∎

######  Remark 2.8.

It is important to note that the converse is not true—in general it is
difficult to find a family of $k$-cube groups whose amalgamated product over
the subgroups generated by their pairwise intersections forms a $(k+1)$-cube
group.

######  Proposition 2.9.

A group $\Gamma$ is a $k$-cube group if and only if it is a torsion-free
$\prod_{i=1}^{k}\tilde{A}_{1}$ group, that is, one which acts freely and
transitively on the set of vertices of the product of $k$ trees.

###### Proof.

This follows as a generalisation of Lemma 2.2 and Theorem 3.4 in [12]. ∎

######  Remark 2.10.

Vdovina, in [25], used Proposition 2.9 as the definition of a $k$-cube group.

######  Example 2.11.

The $k$-dimensional cube complex $\mathcal{M}$ with adjacency structure
$E_{1},\ldots,E_{k}$ constructed in Definition 2.3 is a $k$-cube group.
Indeed, each $k$-cube group yields such a complex with a single vertex, by a
relatively obvious process generalised from those in [4, §6.1], and [11,
§4.1]. We may henceforth regard a $k$-cube group $\Gamma$ both algebraically,
and geometrically as the corresponding cube complex with edges labelled by
elements of $\Gamma$. If a clear distinction needs to be made, we may write
$\mathcal{M}(\Gamma)$ for the geometric realisation of the cube complex.

######  Example 2.12.

Consider the group $\Gamma_{\\{3,5,7\\}}$ from Example 3.17 of [24], defined
as follows:

$\Gamma_{\\{3,5,7\\}}:=\langle
a_{1},a_{2},b_{1},b_{2},b_{3},c_{1},c_{2},c_{3},c_{4}\mid R\rangle,$

where

$\displaystyle R:=\big{\\{}$ $\displaystyle
a_{1}b_{1}a_{2}b_{2},a_{1}b_{2}a_{2}b_{1}^{-1},\,a_{1}b_{3}a_{2}^{-1}b_{1},\,a_{1}b_{3}^{-1}a_{1}b_{2}^{-1},\,a_{1}b_{1}^{-1}a_{2}^{-1}b_{3},\,a_{2}b_{3}a_{2}b_{2}^{-1},$
$\displaystyle
a_{1}c_{1}a_{2}^{-1}c_{2}^{-1},\,a_{1}c_{2}a_{1}^{-1}c_{3},\,a_{1}c_{3}a_{2}^{-1}c_{4}^{-1},\,a_{1}c_{4}a_{1}c_{1}^{-1},$
$\displaystyle
a_{1}c_{4}^{-1}a_{2}c_{2},\,a_{1}c_{3}^{-1}a_{2}c_{1},\,a_{2}c_{3}a_{2}c_{2}^{-1},\,a_{2}c_{4}a_{2}^{-1}c_{1},$
$\displaystyle
c_{1}b_{1}c_{3}b_{3}^{-1},\,c_{1}b_{2}c_{4}b_{2}^{-1},\,c_{1}b_{3}c_{4}^{-1}b_{2},\,c_{1}b_{3}^{-1}c_{4}b_{3},\,c_{1}b_{2}^{-1}c_{2}b_{1},\,c_{1}b_{1}^{-1}c_{4}b_{1}^{-1},$
$\displaystyle
c_{2}b_{2}c_{3}^{-1}b_{3}^{-1},\,c_{2}b_{3}c_{4}b_{1},\,c_{2}b_{3}^{-1}c_{3}b_{3},\,c_{2}b_{2}^{-1}c_{3}b_{2},\,c_{2}b_{1}^{-1}c_{3}b_{1}^{-1},\,c_{3}b_{1}c_{4}b_{2}\big{\\}}.$

This is a $3$-cube group with adjacency structure
$\\{a_{i},a_{i}^{-1}\\},\\{b_{i},b_{i}^{-1}\\},\\{c_{i},c_{i}^{-1}\\}$ (Figure
3).

Figure 3: Geometric realisation of the pointed, oriented cube in the cube
complex corresponding to $\mathcal{S}_{3}\big{(}\Gamma_{\\{3,5,7\\}}\big{)}$
labelled
$\big{[}\big{[}a_{1},b_{1}^{-1},a_{2},b_{3}\big{]},\big{[}a_{2},c_{1}^{-1},a_{1}^{-1},c_{2}\big{]},\big{[}b_{2},c_{2},b_{1},c_{1}\big{]},\\\
\big{[}a_{2}^{-1},b_{3}^{-1},a_{2}^{-1},b_{2}\big{]},\big{[}a_{2},c_{3},a_{2},c_{2}^{-1}\big{]},\big{[}b_{3},c_{2}^{-1},b_{3}^{-1},c_{3}^{-1}\big{]}\big{]}.$
The choice of basepoints and orientations of the cubes and their faces
($2$-cells) is arbitrary, but must remain consistent across the entire
complex.

######  Definition 2.13.

Let $\Gamma$ be a $k$-cube group with adjacency structure
$E_{1},\ldots,E_{k}$, and let $\Delta$ be the $k$-rank affine building which
is the $k$-dimensional cube complex corresponding to
$T(|E_{1}|)\times\cdots\times T(|E_{k}|)$. We identify elements of $\Gamma$
with vertices of $\Delta$, such that the set $\mathcal{S}_{k}$ can be
identified with the set of pointed, oriented chambers ($k$-cubes) of $\Delta$.

Let $p\in\\{1,\ldots,k\\}$ be fixed, and let $A,B\in\mathcal{S}_{k}$ be the
pointed $k$-cubes $\square(u,v_{j},u^{j},w_{j})$,
$\square(x,y_{j},x^{j},z_{j})$, respectively, for
$j\in(\\{1,\ldots,k\\}\setminus\\{p\\})$. We define adjacency matrices
$M_{1},\ldots,M_{k}$ to be square matrices with rows and columns indexed by
$\mathcal{S}_{k}$, and with $AB$-th entries given by:

  * •

$M_{p}(A,B):=1$ if both of the following criteria are satisfied:

    1. (i)

$v_{j}^{L}=\big{(}z_{j}^{L}\big{)}^{-1}$ and
$w_{j}^{L}=\big{(}y_{j}^{L}\big{)}^{-1}$ for all $j\neq p$ with $j\notin L$,

    2. (ii)

$u^{L}\neq(x^{L})^{-1}$,

for all $L\subseteq(\\{1,\ldots,k\\}\setminus\\{p\\})$ with $|L|\geq 0$. We
define $M_{p}(A,B):=0$ otherwise.

  * •

$M_{i}(A,B):=1$, for each $i\neq p$, if all of the following criteria are
satisfied:

    1. (i)

$(x^{L})^{-1}=u^{L\cup\\{i\\}}$,

    2. (ii)

$(y_{j}^{L})^{-1}=w_{j}^{L\cup\\{i\\}}$,

    3. (iii)

$(z_{j}^{L})^{-1}=v_{j}^{L\cup\\{i\\}}$,

for all $L\subseteq(\\{1,\ldots,k\\}\setminus\\{i,p\\})$ with $|L|\geq 0$, and
all $j$ with $j\notin(L\cup\\{i,p\\})$,

    1. (iv)

$v_{i}^{L}\neq(y_{i}^{L})^{-1}$ and $w_{i}^{L}\neq(z_{i}^{L})^{-1}$, for all
$L\subseteq(\\{1,\ldots,k\\}\setminus\\{i,p\\})$ with $|L|\geq 0$.

We define $M_{i}(A,B):=0$ otherwise.

For each $i\in\\{1,\ldots,k\\}$, we say that $B$ is adjacent in the $E_{i}$
direction, or $E_{i}$-adjacent, to $A$ whenever $M_{i}(A,B)=1$ (Figure 4).

Figure 4: Let $a^{L},u^{L},x^{L}\in E_{1}$, $(\ast)_{2}^{L}\in E_{2}$, and
$(\ast)_{3}^{L}\in E_{3}$, where $E_{1},E_{2},E_{3}$ form the adjacency
structure of some $3$-cube group. Consider the pointed $3$-cubes
$A,B,C\in\mathcal{S}_{3}$. If $B$ is $E_{3}$-adjacent to $A$ (resp. $C$ is
$E_{2}$-adjacent to $A$), then the magenta (resp. blue) $2$-faces above
coincide. In general, for a $k$-cube group, $M_{i}(A,B)=1$ implies that some
$(k-1)$-faces of the geometric realisations of $A$ and $B$ in the
corresponding cube complex coincide.

######  Lemma 2.14.

Let $\Gamma$ be a $k$-cube group with adjacency structure
$E_{1},\ldots,E_{k}$. Then each of the adjacency matrices $M_{1},\ldots,M_{k}$
has entries in $\\{0,1\\}$, and has at least three non-zero entries in each
row.

###### Proof.

Consider the pointed $k$-cube
$A:=\square(u,v_{j},u^{j},w_{j})\in\mathcal{S}_{k}(\Gamma)$, as constructed in
Definition 2.4. Since $|E_{i}|\geq 4$ for all $i$, and by property C1, we can
find some $k$-cube
$B:=\square\big{(}x^{-1},(z_{j})^{-1},(x^{j})^{-1},(v_{j})^{-1}\big{)}$, where
$x^{-1}\neq u$. Then C2 implies that $(x^{j})^{-1}\neq u^{j}$ for all $j$. It
follows that $M_{p}(A,B)=1$, and a similar argument can be used to find a
$k$-cube $C$ with $M_{i}(A,C)=1$, for $i\neq p$.

Hence, in each row in each $M_{1},\ldots,M_{k}$, there are at least three non-
zero entries, and by definition these are $\\{0,1\\}$-matrices. ∎

Figure 5:

  1. (a)

The Unique Common Extension Property says that, given four adjacent $k$-cubes
$A,B_{1},B_{2},B_{3}$ arranged as above (grey), we can uniquely find four more
$k$-cubes $C_{12},C_{13},C_{23},D$ (dashed) to complete a $2\times 2\times 2$
arrangement.

  2. (b)

In the second figure, we assign each cube from 5(a) a vertex, and draw a
directed arrow of colour $i$ between two vertices $X,Y$ whenever
$M_{i}(X,Y)=1$, that is, if $Y$ is $E_{i}$-adjacent to $X$. Then, given three
arrows originating at $A$, we can find unique arrows (dashed) to complete the
commuting cube diagram.

######  Definition 2.15.

Let $k\geq 3$, and let $\Gamma$ be a $k$-cube group with adjacency structure
$E_{1},\ldots E_{k}$ and adjacency matrices $M_{1},\ldots,M_{k}$. Let
$A,B_{p},B_{q},B_{r}$ be pointed $k$-cubes in $\mathcal{S}_{k}(\Gamma)$ such
that $M_{p}(A,B_{p})=M_{q}(A,B_{q})=M_{r}(A,B_{r})=1$ for some $p,q,r$. We say
that the matrices $M_{i}$ satisfy the Unique Common Extension Property if we
can find unique $k$-cubes $C_{pq},C_{pr},C_{qr},D\in\mathcal{S}_{k}$ such that
each of

$M_{p}(B_{q},C_{pq}),\,M_{p}(B_{r},C_{pr}),\,M_{q}(B_{p},C_{pq}),\,M_{q}(B_{r},C_{pr}),\,M_{r}(B_{p},C_{pr}),\,M_{r}(B_{q},C_{qr}),$

and each of

$M_{p}(C_{qr},D),\,M_{q}(C_{pr},D),\,M_{r}(C_{pq},D)$

is equal to $1$ (Figure 5(a)). In the case where $k=2$, let
$A,B,C\in\mathcal{S}_{2}$ be such that $M_{1}(A,B)=M_{2}(A,C)=1$. Then
$M_{1},M_{2}$ satisfy the Unique Common Extension Property if there exists a
unique $D\in\mathcal{S}_{2}$ such that $M_{2}(B,D)=M_{1}(C,D)=1$.

Referring to the example in Figure 4, the Unique Common Extension Property
would suggest the existence of a unique cube $D$ which is simultaneously
$E_{2}$-adjacent to $B$, and $E_{3}$-adjacent to $C$.

######  Remark 2.16.

  1. (i)

One might notice that the definition for the Unique Common Extension Property
could be extended to deal with $B_{1},\ldots,B_{k}\in\mathcal{S}_{k}$, such
that each $B_{i}$ is $E_{i}$-adjacent to $A$. It turns out by Lemma 1.4 in
[21], however, that having unique common extensions given three $k$-cubes
$B_{p},B_{q},B_{r}$ initially adjacent to $A$ as above, is enough to imply
unique common extensions for any number of initial $B_{i}$.

  2. (ii)

The Unique Common Extension Property is formulated slightly differently to the
factorisation property of $k$-rank graphs (c.f. Definition 3.1, and [15,
1.1]). By property C2, any two adjacent sides of a square in the complex
$\mathcal{M}(\Gamma)$ uniquely define the square. Then, any three adjacent and
mutually perpendicular edges uniquely determine a cube.

The link at each vertex of $\mathcal{M}(\Gamma)$ is a clique complex of a
complete $k$-partite graph, so given $k$-cubes $A$ and $D$ arranged as in
Figure 5, then the remaining cubes $B_{i}$, $C_{ij}$ are determined by $A$ and
$D$ (Figure 6).

Figure 6: Given cubes $A$ and $D$ arranged as above, then $B_{2}$ is uniquely-
determined by edges $x,y,z$. In turn, each of the remaining cubes in the
diagram are determined. This is equivalent to the associativity property of
$k$-rank graphs.

######  Proposition 2.17.

Let $\mathcal{M}$ be the $k$-dimensional cube complex with adjacency structure
$E_{1},\ldots,E_{k}$, as constructed in Definition 2.3. If we regard
$\mathcal{M}$ as a $k$-cube group, then its adjacency matrices
$M_{1},\ldots,M_{k}$ commute, and satisfy the Unique Common Extension
Property.

###### Proof.

If $k=2$, this is Lemma 4.1 from [12].

Suppose, then, that $k\geq 3$. Fix $p\in\\{1,\ldots,k\\}$ and let
$A:=\square(u,v_{j},u^{j},w_{j})\in\mathcal{S}_{k}$, for
$j\in(\\{1,\ldots,k\\}\setminus\\{p\\})$. Define three more $k$-cubes
$B_{1}:=\square(a,b_{j},a^{j},c_{j})$, $B_{2}:=\square(r,s_{j},r^{j},t_{j})$,
$B_{3}:=\square(x,y_{j},x^{j},z_{j})\in\mathcal{S}_{k}$ and suppose, without
loss of generality, that $M_{i}(A,B_{i})=1$ for each $i\in\\{1,2,3\\}$, and
that $p\notin\\{1,2,3\\}$ (these are purely for notational convenience: the
proof is identical for any three distinct $i\in\\{1,\ldots,k\\}$, or if some
$i=p$).

Then by Definition 2.13, we have $(a^{L})^{-1}=u^{L\cup\\{1\\}}$,
$(b_{j}^{L})^{-1}=w_{j}^{L\cup\\{1\\}}$,
$(c_{j}^{L})^{-1}=v_{j}^{L\cup\\{1\\}}$, and so on, for all compatible $L$ and
$j$.

By Lemma 2.14, we can find a $k$-cube $C_{21}\in\mathcal{S}_{k}$ such that
$M_{2}(B_{1},C_{21})=1$. Then

$C_{21}=\square\Big{(}u^{L\cup\\{1\\}\cup\\{2\\}},v_{j}^{L\cup\\{1\\}\cup\\{2\\}},u^{L\cup\\{1\\}\cup\\{2\\}\cup\\{j\\}},w_{j}^{L\cup\\{1\\}\cup\\{2\\}}\Big{)}.$

Similarly, we can find some $C_{12}\in\mathcal{S}_{k}$ such that
$M_{1}(B_{1},C_{12})=1$. But this can be seen to equal $C_{21}$, and so the
matrices $M_{1},M_{2}$ commute. Indeed, we can show in an identical manner
that all of the matrices $M_{1},\ldots,M_{k}$ commute.

Finally, consider a $k$-cube $D\in\mathcal{S}_{k}$ such that
$M_{3}(C_{21},D)=1$; such a $k$-cube exists which satisfies Definition
2.13(iv) by Lemma 2.14. Then

$D=\square\Big{(}\big{(}u^{L\cup\\{1,2,3\\}}\big{)}^{-1},\Big{(}v_{j}^{L\cup\\{1,2,3\\}}\Big{)}^{-1},\Big{(}u^{L\cup\\{1,2,3\\}\cup\\{j\\}}\Big{)}^{-1},\Big{(}w_{j}^{L\cup\\{1,2,3\\}}\Big{)}^{-1}\Big{)},$

and it is clear that $D$ is also the unique $k$-cube such that
$M_{1}(C_{32},D)=M_{2}(C_{31},D)=1$. ∎

##  3 Higher-rank graphs

######  Definition 3.1.

Let $\Lambda$ be a category such that $\operatorname{Ob}(\Lambda)$ and
$\operatorname{Hom}(\Lambda)$ are countable sets (that is, a countable small
category), and identify $\operatorname{Ob}(\Lambda)$ with the identity
morphisms in $\operatorname{Hom}(\Lambda)$. For a morphism
$\lambda\in\operatorname{Hom}_{\Lambda}(u,v)$, we define range and source maps
$r(\lambda):=v$ and $s(\lambda):=u$ respectively.

Let $d:\Lambda\rightarrow\mathbb{N}^{k}$ be a functor, called the degree map,
and let $\lambda\in\operatorname{Hom}(\Lambda)$. We call the pair
$(\Lambda,d)$ a $k$-rank graph (or simply a $k$-graph) if, whenever
$d(\lambda)=\mathbf{m}+\mathbf{n}$ for some
$\mathbf{m},\mathbf{n}\in\mathbb{N}^{k}$, we can find unique elements
$\mu,\nu\in\operatorname{Hom}(\Lambda)$ such that $\lambda=\nu\mu$, and
$d(\mu)=\mathbf{m}$, $d(\nu)=\mathbf{n}$. Note that for $\mu$, $\nu$ to be
composable, we must have $r(\mu)=s(\nu)$.

For $\mathbf{n}\in\mathbb{N}^{k}$, we write
$\Lambda^{\mathbf{n}}:=d^{-1}(\mathbf{n})$; by the above property, we have
that $\Lambda^{\mathbf{0}}=\operatorname{Ob}(\Lambda)$, and we call the
elements of $\Lambda^{\mathbf{0}}$ the vertices of $(\Lambda,d)$ [15].

Let $(\Lambda,d)$ be a $k$-rank graph, let $\mathbf{n}\in\mathbb{N}^{k}$, and
let $v\in\Lambda^{\mathbf{0}}$. Write $\Lambda^{\mathbf{n}}(v)$ for the set of
morphisms in $\Lambda^{\mathbf{n}}$ which map onto the vertex $v$, that is,
$\Lambda^{\mathbf{n}}(v):=\\{\lambda\in\Lambda^{\mathbf{n}}\mid
r(\lambda)=v\\}$. We say that $(\Lambda,d)$ is row-finite if each set
$\Lambda^{\mathbf{n}}(v)$ is finite, and that $(\Lambda,d)$ has no sources if
each $\Lambda^{\mathbf{n}}(v)$ is non-empty.

######  Remark 3.2.

If $E$ is a directed graph on $n$ vertices, we can construct an $n\times n$
incidence matrix $M_{E}(i,j)$ with $ij$-th entry $1$ if there is an edge from
$i$ to $j$, and $0$ otherwise.

If $E_{1},\ldots,E_{k}$ are directed graphs with the same vertex set, and such
that their associated incidence matrices $M_{1},\ldots,M_{k}$ commute, then we
can construct a $k$-rank graph out of the graphs $E_{i}$, as in [10].

######  Theorem 3.3.

Let $\Gamma$ be a $k$-cube group with adjacency structure
$E_{1},\ldots,E_{k}$. Then $\Gamma$ induces a row-finite $k$-rank graph
$\mathcal{G}(\Gamma)$ with no sources.

######  Remark 3.4.

To the reader who has not come across higher-rank graphs in the past, it may
seem counter-intuitive that a $k$-rank graph be a countably-infinite category,
while a $k$-cube group $\Gamma$ comprises finite data.

The adjacency structure on $\Gamma$ induces a family $\mathcal{E}(\Gamma)$ of
distinctly-coloured directed graphs $\mathcal{E}_{1},\ldots,\mathcal{E}_{k}$
on the same vertices. The shared set of vertices is the set of $k$-cubes of
$\Gamma$, and the incidence matrices are given by $M_{1},\ldots,M_{k}$,
respectively. It is imperative to stress that this is not the same as the cube
complex $\mathcal{M}(\Gamma)$ from Example 2.11. In [7], the collection of
graphs $\mathcal{E}(\Gamma)$ is called a $1$-skeleton.

To view $\mathcal{E}(\Gamma)$ as a $k$-rank graph, we must consider the set
$\mathcal{E}(\Gamma)^{*}$ of $k$-dimensional paths in $\mathcal{E}(\Gamma)$.
These can be regarded as equivalence classes of reduced paths in
$\mathcal{E}(\Gamma)$; two paths $\lambda,\mu$ are identified whenever they
begin at the same vertex and end at the same vertex.

The degree $d(\lambda)$ (called the shape in [21]) of a $k$-dimensional path
$\lambda$ is a tuple whose $i$-th entry is the total length of $\lambda$
restricted to colour $i$, that is, to edges labelled by elements of
$\mathcal{E}_{i}$. Together, the pair $(\mathcal{E}(\Gamma)^{*},d)$ is a
$k$-rank graph, since it satisfies the factorisation property of Definition
1.1 in [15] (consult Figure 6 for an illustration), and we denote this by
$\mathcal{G}(\Gamma)$. We prove Theorem 3.3 like so:

###### Proof.

Let $\mathcal{G}(\Gamma)$ be a collection of directed graphs
$\mathcal{E}_{1},\ldots,\mathcal{E}_{k}$, each on the set of vertices
$\mathcal{S}_{k}(\Gamma)$, and with incidence matrices $M_{1},\ldots,M_{k}$
respectively. From Remark 2.3 in [8] and [21, §1], it is sufficient that, for
each $i,j,l\in\\{1,\ldots,k\\}$ with $i<j<l$:

  1. (i)

$M_{i}$ is non-zero,

  2. (ii)

$M_{i}M_{j}=M_{j}M_{i}$,

  3. (iii)

Each of $M_{i}$, $M_{i}M_{j}$, and $M_{i}M_{j}M_{l}$ has entries in
$\\{0,1\\}$.

But each $M_{i}$ is non-zero by Lemma 2.14, and the matrices are finite-
dimensional, commute, and satisfy the Unique Common Extension Property by
Proposition 2.17. Hence $\mathcal{G}(\Gamma)$ can be regarded as a $k$-rank
graph. ∎

######  Remark 3.5.

Robertson and Steger in [21], [22] initially considered those $2$-rank graphs
whose incidence matrices have entries in $\\{0,1\\}$ (though they did not name
them as such). To our knowledge, $k$-cube groups induce the first examples of
$k$-rank graphs with $\\{0,1\\}$-incidence-matrices for arbitrary $k\geq 2$.

We associate a $C^{\star}$-algebra to a $k$-rank graph as follows:

######  Definition 3.6.

Let $\Lambda=(\Lambda,d)$ be a row-finite $k$-rank graph with no sources. We
define the $k$-rank graph $C^{\star}$-algebra $\mathcal{A}(\Lambda)$ to be the
universal $C^{\star}$-algebra generated by a family
$\\{s_{\lambda}\mid\lambda\in\Lambda\\}$ of partial isometries (that is,
operators $s_{\lambda}$ whose restriction to $(\ker s_{\lambda})^{\perp}$ are
isometries) which have the following properties:

  1. (i)

The set $\big{\\{}s_{v}\mid v\in\Lambda^{\mathbf{0}}\big{\\}}$ satisfies
$(s_{v})^{2}=s_{v}=s_{v}^{*}$ and $s_{u}s_{v}=0$ for all $u\neq v$.

  2. (ii)

If $r(\lambda)=s(\mu)$ for some $\lambda,\mu\in\Lambda$, then
$s_{\mu\lambda}=s_{\mu}s_{\lambda}$.

  3. (iii)

For all $\lambda\in\Lambda$, we have
$s_{\lambda}^{*}s_{\lambda}=s_{s(\lambda)}$.

  4. (iv)

For all vertices $v\in\Lambda^{\mathbf{0}}$ and $\mathbf{n}\in\mathbb{N}^{k}$,
we have:

$s_{v}=\sum_{\lambda\in\Lambda^{\mathbf{n}}(v)}s_{\lambda}s_{\lambda}^{*}.$

Note that without the row-finiteness condition, property (iv) is not well-
defined.

##  4 Spectral sequences and K-theory

We make extensive use of Theorem 3.15 from [6], displayed here as Theorem 4.2;
in the examples presented in this paper, we principally consider the special
cases where $k=3$ or $k=4$. For $k=3$, we make use of the relevant work of
[6], and we derive analogous results for $k=4$ and $k=5$ in Propositions 4.5,
4.7 and Corollary 4.6.

The proofs make use of so-called spectral sequences, generalisations of exact
sequences; we direct the unfamiliar reader to [18] for more detailed
background information, but offer an overview here.

######  Definition 4.1.

Let $\mathcal{C}$ be an Abelian category. A spectral sequence (of homological
type) consists of a family $\\{(E^{r},d^{r})\\}$ of bigraded objects

$E^{r}:=\bigoplus_{p,q\in\mathbb{Z}}E_{p,q}^{r}$

in $\mathcal{C}$, and maps

$d^{r}:E_{p,q}^{r}\longrightarrow E_{p-r,q+r-1}^{r},\quad\text{and}\quad
d^{r}:E_{p+r,q-r+1}^{r}\longrightarrow E_{p,q}^{r},$

called differentials, which are of bidegree $(-r,r-1)$, and which satisfy
$d^{r}\circ d^{r}=0$. We insist that

$E_{p,q}^{r+1}\cong
H(E_{p,q}^{r}):=\frac{\ker\big{(}d^{r}:E_{p,q}^{r}\longrightarrow
E_{p-r,q+r-1}^{r}\big{)}}{\operatorname{im}\big{(}d^{r}:E_{p+r,q-r+1}^{r}\longrightarrow
E_{p,q}^{r}\big{)}}.$

The collections $(E_{p,q}^{r})$ for fixed $r$ are known as the sheets of the
spectral sequence. We move to the next sheet by taking the homology $H$,
defined above.

We say that the spectral sequence converges if, for every pair $(p,q)$, the
sequence of objects $E_{p,q}^{r}$ stabilises as $r\rightarrow\infty$; we
denote this limit by $E_{p,q}^{\infty}$, and call it the stable value.

Given a general exact sequence $A:=\cdots\rightarrow
A_{i+1}\overset{\partial_{i+1}}{\longrightarrow}A_{i}\overset{\partial_{i}}{\longrightarrow}A_{i-1}\rightarrow\cdots$,
we frequently write $H_{i}(A)$ to denote the $i$-th homology
$\ker(\partial_{i})/\operatorname{im}(\partial_{i+1})$.

######  Theorem 4.2 (Evans 2008).

Define the sets

$N_{l}:=\begin{cases}\big{\\{}\boldsymbol{\mu}:=(\mu_{1},\ldots,\mu_{l})\in\\{1,\ldots,k\\}^{l}\mid\mu_{1}<\cdots<\mu_{l}\big{\\}}&\text{if
}l\in\\{1,\ldots,k\\},\\\ \\{*\\}&\text{if }l=0,\\\
\emptyset&\text{otherwise},\end{cases}$

and let $\Lambda$ be a row-finite $k$-graph with no sources. Then there exists
a spectral sequence $\\{(E^{r},d^{r})\\}$ converging to
$K_{\ast}(\mathcal{A}(\Lambda))$ with $E_{p,q}^{\infty}\cong E_{p,q}^{k+1}$,
and

$E_{p,q}^{2}\cong\begin{cases}H_{p}(\mathcal{D}_{k})&\text{if
}p\in\\{0,1,\ldots,k\\}\text{ and }q\text{ is even,}\\\
0&\text{otherwise,}\end{cases}$

where $\mathcal{D}_{k}$ is the chain complex with

$(\mathcal{D}_{k})_{p}:=\begin{cases}\bigoplus_{\mu\in
N_{p}}\mathbb{Z}\Lambda^{\mathbf{0}}&\text{if }p\in\\{0,1,\ldots,k\\},\\\
0&\text{otherwise,}\end{cases}$

and whose differentials
$\partial_{p}:(\mathcal{D}_{k})_{p}\longrightarrow(\mathcal{D}_{k})_{p-1}$ are
defined as

$\bigoplus_{\mu\in N_{p}}m_{\mu}\longmapsto\bigoplus_{\lambda\in
N_{p-1}}\sum_{\mu\in
N_{p}}\sum_{i=1}^{p}(-1)^{i+1}\delta_{\lambda,\mu^{i}}(I-M_{\mu_{i}}^{T})m_{\mu},$

for $p\in\\{1,\ldots,k\\}$.∎

######  Proposition 4.3 ($k=3$, Evans 2008).

Let $\Lambda$ be a row-finite $3$-graph with no sources, and let
$\mathbb{Z}\Lambda^{\mathbf{0}}$ be the group of all maps
$\Lambda^{\mathbf{0}}\rightarrow\mathbb{Z}$ with finite support under
pointwise addition. Consider the chain complex $\mathcal{D}_{3}$ defined as
follows:

$0\longrightarrow\mathbb{Z}\Lambda^{\mathbf{0}}\overset{\partial_{3}}{\longrightarrow}\bigoplus_{i=1}^{3}\mathbb{Z}\Lambda^{\mathbf{0}}\overset{\partial_{2}}{\longrightarrow}\bigoplus_{i=1}^{3}\mathbb{Z}\Lambda^{\mathbf{0}}\overset{\partial_{1}}{\longrightarrow}\mathbb{Z}\Lambda^{\mathbf{0}}\longrightarrow
0,$

where $\partial_{1}$, $\partial_{2}$, $\partial_{3}$ are defined by the block
matrices

$\displaystyle\partial_{1}$
$\displaystyle:=\begin{bmatrix}I-M_{1}^{T}&I-M_{2}^{T}&I-M_{3}^{T}\end{bmatrix},$
$\displaystyle\partial_{2}$
$\displaystyle:=\begin{bmatrix}M_{2}^{T}-I&M_{3}^{T}-I&0\\\
I-M_{1}^{T}&0&M_{3}^{T}-I\\\ 0&I-M_{1}^{T}&I-M_{2}^{T}\end{bmatrix},$
$\displaystyle\partial_{3}$ $\displaystyle:=\begin{bmatrix}I-M_{3}^{T}\\\
M_{2}^{T}-I\\\ I-M_{1}^{T}\end{bmatrix}.$

Then for some subgroups $G_{0}\subseteq\operatorname{coker}(\partial_{1})$ and
$G_{1}\subseteq\ker(\partial_{3})$, there exists a short exact sequence

$0\longrightarrow\operatorname{coker}(\partial_{1})/G_{0}\longrightarrow
K_{0}(\mathcal{A}(\Lambda))\longrightarrow\ker(\partial_{2})/\operatorname{im}(\partial_{3})\longrightarrow
0,$

and an isomorphism

$K_{1}(\mathcal{A}(\Lambda))\cong\ker(\partial_{1})/\operatorname{im}(\partial_{2})\oplus
G_{1}.$

∎

######  Corollary 4.4 ($k=3$, Evans 2008).

In addition to the hypotheses of Proposition 4.3:

  1. (i)

If $\partial_{1}$ is surjective, then:

    1. (a)

$K_{0}(\mathcal{A}(\Lambda))\cong\ker(\partial_{2})/\operatorname{im}(\partial_{3})$,

    2. (b)

$K_{1}(\mathcal{A}(\Lambda))\cong(\ker(\partial_{1})/\operatorname{im}(\partial_{2}))\oplus\ker(\partial_{3})$.

  2. (ii)

If $\bigcap_{i}\ker\big{(}I-M_{i}^{T}\big{)}=0$, then there exists a short
exact sequence

$0\longrightarrow\operatorname{coker}(\partial_{1})\longrightarrow
K_{0}(\mathcal{A}(\Lambda))\longrightarrow\ker(\partial_{2})/\operatorname{im}(\partial_{3})\longrightarrow
0,$

and an isomorphism

$K_{1}(\mathcal{A}(\Lambda))\cong\ker(\partial_{1})/\operatorname{im}(\partial_{2}).$

∎

######  Proposition 4.5 ($k=4$).

Let $\Lambda$ be a row-finite $4$-graph with no sources, and consider the
chain complex $\mathcal{D}_{4}$:

$0\longrightarrow\mathbb{Z}\Lambda^{\mathbf{0}}\overset{\partial_{4}}{\longrightarrow}\bigoplus_{i=1}^{4}\mathbb{Z}\Lambda^{\mathbf{0}}\overset{\partial_{3}}{\longrightarrow}\bigoplus_{i=1}^{6}\mathbb{Z}\Lambda^{\mathbf{0}}\overset{\partial_{2}}{\longrightarrow}\bigoplus_{i=1}^{4}\mathbb{Z}\Lambda^{\mathbf{0}}\overset{\partial_{1}}{\longrightarrow}\mathbb{Z}\Lambda^{\mathbf{0}}\longrightarrow
0,$

where $\partial_{1},\ldots,\partial_{4}$ are the group homomorphisms defined
by the block matrices

$\displaystyle\partial_{1}$
$\displaystyle:=\begin{bmatrix}I-M_{1}^{T}&I-M_{2}^{T}&I-M_{3}^{T}&I-M_{4}^{T}\end{bmatrix},$
$\displaystyle\partial_{2}$
$\displaystyle:=\begin{bmatrix}M_{2}^{T}-I&M_{3}^{T}-I&M_{4}^{T}-I&0&0&0\\\
I-M_{1}^{T}&0&0&M_{3}^{T}-I&M_{4}^{T}-I&0\\\
0&I-M_{1}^{T}&0&I-M_{2}^{T}&0&M_{4}^{T}-I\\\
0&0&I-M_{1}^{T}&0&I-M_{2}^{T}&I-M_{3}^{T}\end{bmatrix},$
$\displaystyle\partial_{3}$
$\displaystyle:=\begin{bmatrix}I-M_{3}^{T}&I-M_{4}^{T}&0&0\\\
M_{2}^{T}-I&0&I-M_{4}^{T}&0\\\ 0&M_{2}^{T}-I&M_{3}^{T}-I&0\\\
I-M_{1}^{T}&0&0&I-M_{4}^{T}\\\ 0&I-M_{1}^{T}&0&M_{3}^{T}-I\\\
0&0&I-M_{1}^{T}&I-M_{2}^{T}\end{bmatrix},$ $\displaystyle\partial_{4}$
$\displaystyle:=\begin{bmatrix}M_{4}^{T}-I\\\ I-M_{3}^{T}\\\ M_{2}^{T}-I\\\
I-M_{1}^{T}\end{bmatrix}.$

Write
$H_{i}(\mathcal{D}_{4}):=\ker(\partial_{i})/\operatorname{im}(\partial_{i+1})$,
and let $F_{2}$ be a factor in the ascending filtration of the
$C^{\star}$-algebra $\mathcal{A}(\Lambda)$. Then, for some subgroups

$G_{0}\subseteq\operatorname{coker}(\partial_{1}),\quad
G_{1}\subseteq\ker(\partial_{4}),\quad G_{2}\subseteq
H_{1}(\mathcal{D}_{4}),\quad G_{3}\subseteq H_{3}(\mathcal{D}_{4}),$

there exist short exact sequences as follows:

  1. (i)

$0\longrightarrow\operatorname{coker}(\partial_{1})/G_{0}\longrightarrow
K_{0}(\mathcal{A}(\Lambda))\longrightarrow\dfrac{K_{0}(\mathcal{A}(\Lambda))}{\operatorname{coker}(\partial_{1})/G_{0}}\longrightarrow
0$,

  2. (ii)

$0\longrightarrow\operatorname{coker}(\partial_{1})/G_{0}\longrightarrow
F_{2}\longrightarrow\dfrac{\ker(\partial_{2})}{\operatorname{im}(\partial_{3})}\longrightarrow
0$,

  3. (iii)

$0\longrightarrow F_{2}\longrightarrow
K_{0}(\mathcal{A}(\Lambda))\longrightarrow G_{1}\longrightarrow 0$,

  4. (iv)

$0\longrightarrow\dfrac{\ker(\partial_{1})/\operatorname{im}(\partial_{2})}{G_{2}}\longrightarrow
K_{1}(\mathcal{A}(\Lambda))\longrightarrow G_{3}\longrightarrow 0$,

and sequence (iii) splits, such that $K_{0}(\mathcal{A}(\Lambda))\cong
F_{2}\oplus G_{1}$.

###### Proof.

Write $\\{(E^{r},d^{r})\\}$ to denote the Kasparov spectral sequence of
homological type introduced in [6]. We know that $\\{(E^{r},d^{r})\\}$ is
bounded, and that the stable value of $E_{p,q}^{r}$ is $E_{p,q}^{\infty}\cong
E_{p,q}^{5}$. The spectral sequence converges to
$K_{\ast}(\mathcal{A}(\Lambda))$, so for each family of modules
$\\{\mathcal{K}_{n}\\}$, there exists a finite ascending filtration

$0=F_{s}(\mathcal{K}_{n})\subseteq\cdots\subseteq
F_{p-1}(\mathcal{K}_{n})\subseteq F_{p}(\mathcal{K}_{n})\subseteq
F_{p+1}(\mathcal{K}_{n})\subseteq\cdots\subseteq
F_{t}(\mathcal{K}_{n})=\mathcal{K}_{n},$

and an isomorphism

$E_{p,q}^{\infty}\cong F_{p}(\mathcal{K}_{p+q})/F_{p-1}(\mathcal{K}_{p+q}),$
(2)

with $E_{p,q}^{\infty}\cong E_{p,q}^{5}=0$ whenever
$p\in(\mathbb{Z}\setminus\\{0,\ldots,4\\})$ or $q$ is odd.

Firstly, we turn our attention to $K_{0}(\mathcal{A}(\Lambda))$. Write
$K_{0}=K_{0}(\mathcal{A}(\Lambda))=\mathcal{K}_{p+q}$, as in Lemma 3.3 of [6].

Case I: Fix the total degree, $p+q$, to be zero.

We have that $E_{p,q}^{5}=0$ unless $p\in\\{0,2,4\\}$, since if $p$ is odd and
$p+q=0$, then $q$ is odd. Suppose, then, that $p\notin\\{0,2,4\\}$, such that
$0=E_{p,q}^{5}=F_{p}(K_{0})/F_{p-1}(K_{0})$, and hence $F_{p}(K_{0})\neq
F_{p-1}(K_{0})$. We can deduce that, in our filtration, we have
$F_{1}(K_{0})=F_{0}(K_{0})$, and $F_{i+1}(K_{0})=F_{i}(K_{0})$ for all $i\geq
2$.

By the same argument, it follows that $F_{i}(K_{0})=0$ for all $i<0$, and so
the filtration becomes

$0\subseteq F_{0}(K_{0})\subseteq F_{2}(K_{0})\subseteq K_{0}.$

Next, we consider the non-zero $E_{p,q}^{5}$ terms. From (2), we have:

  * •

$E_{0,0}^{5}\cong F_{0}(K_{0})$,

  * •

$E_{2,-2}^{5}\cong F_{2}(K_{0})/F_{1}(K_{0})\cong F_{2}(K_{0})/F_{0}(K_{0})$,

  * •

$E_{4,-4}^{5}\cong F_{4}(K_{0})/F_{3}(K_{0})\cong K_{0}/F_{2}(K_{0})$.

It then follows that we have short exact sequences:

  1. (i’)

$0\longrightarrow E_{0,0}^{5}\longrightarrow K_{0}\longrightarrow
K_{0}/E_{0,0}^{5}\longrightarrow 0$,

  2. (ii’)

$0\longrightarrow E_{0,0}^{5}\longrightarrow F_{2}(K_{0})\longrightarrow
E_{2,-2}^{5}\longrightarrow 0$,

  3. (iii’)

$0\longrightarrow F_{2}(K_{0})\longrightarrow K_{0}\longrightarrow
E_{4,-4}^{5}\longrightarrow 0$.

Case II: $p+q=1$.

We consider $K_{1}(\mathcal{A}(\Lambda))$. Note that in order for
$E_{p,q}^{5}$ to be non-zero, we must have $p\in\\{0,\ldots,4\\}$ and $q$
even. But, the only pairs $(p,q)$ of total degree $1$ are $(1,0)$ and
$(3,-2)$. Thus, it follows analogously from Proposition 3.17 in [6] that there
is a short exact sequence

$0\longrightarrow E_{1,0}^{5}\longrightarrow
K_{1}(\mathcal{A}(\Lambda))\longrightarrow E_{3,-2}^{5}\longrightarrow 0.$

* * *

The final step of the proof is to compute the following:

  1. (a)

For $K_{1}(\mathcal{A}(\Lambda))$, the terms $E_{1,0}^{5}$ and $E_{3,-2}^{5}$,

  2. (b)

For $K_{0}(\mathcal{A}(\Lambda))$, the terms $E_{0,0}^{5}$, $E_{2,-2}^{5}$,
and $E_{4,-4}^{5}$.

Step (a): $E_{1,0}^{5}$ and $E_{3,-2}^{5}$.

We proceed by considering the differentials $d^{4}$, $d^{3}$, and $d^{2}$.
Since $E_{p,q}^{4}=0$ whenever $p\in(\mathbb{Z}\setminus\\{1,\ldots,4\\})$, we
necessarily have:

$d^{4}:E_{p,q}^{4}\longrightarrow E_{p-4,q+3}^{4},\qquad
d^{4}:E_{p+4,q-3}^{4}\longrightarrow E_{p,q}^{4},$

for $p\in\\{0,4\\}$. However, in either case we must have $q,q+3$ or $q,q-3$
both even: a contradiction. Hence $d^{4}$ is the zero map.

Similarly, it follows that the only non-zero components of the $d^{3}$
differential are

$d^{3}:E_{3,q}^{3}\longrightarrow E_{0,q+2},\qquad
d^{3}:E_{4,q}^{3}\longrightarrow E_{1,q+2}^{3},$

for $q$ even. Furthermore, we can deduce that $d^{2}$ must also be the zero
map, as in Proposition 3.16 of [6]. Thus, we have:

$\displaystyle E_{1,0}^{5}$ $\displaystyle\cong
H\big{(}E_{1,0}^{4}\big{)}=\frac{\ker\big{(}d^{4}:E_{1,0}^{4}\rightarrow
E_{-3,3}^{4}\big{)}}{\operatorname{im}\big{(}d^{4}:E_{5,-3}^{4}\rightarrow
E_{1,0}^{4}\big{)}}=E_{1,0}^{4},$ $\displaystyle E_{1,0}^{4}$
$\displaystyle\cong
H\big{(}E_{1,0}^{3}\big{)}=\frac{\ker\big{(}d^{3}:E_{1,0}^{3}\rightarrow
E_{-2,2}^{3}\big{)}}{\operatorname{im}\big{(}d^{3}:E_{4,-2}^{3}\rightarrow
E_{1,0}^{3}\big{)}}=E_{1,0}^{3}/\operatorname{im}\big{(}d^{3}:E_{4,-2}^{3}\rightarrow
E_{1,0}^{3}\big{)}.$

Now, let $G_{2}$ be a subgroup of $E_{1,0}^{3}=H_{1}(\mathcal{D}_{4})$, namely
$G_{2}:=\operatorname{im}\big{(}d^{3}:E_{4,-2}^{3}\rightarrow
E_{1,0}^{3}\big{)}$. Then we have

$E_{1,0}^{3}\cong
H\big{(}E_{1,0}^{2}\big{)}=E_{1,0}^{2}=H_{1}(\mathcal{D}_{4}),$

and so
$E_{1,0}^{5}\cong(\ker(\partial_{1})/\operatorname{im}(\partial_{2}))/G_{2}$.

It remains to compute $E_{3,-2}^{5}$. We have:

$\displaystyle E_{3,-2}^{5}$ $\displaystyle\cong
H\big{(}E_{3,-2}^{4}\big{)}=\frac{\ker\big{(}d^{4}:E_{3,-2}^{4}\rightarrow
E_{-1,1}^{4}\big{)}}{\operatorname{im}\big{(}d^{4}:E_{7,-5}^{4}\rightarrow
E_{3,-2}^{4}\big{)}}=E_{3,-2}^{4},$ $\displaystyle E_{3,-2}^{4}$
$\displaystyle\cong
H\big{(}E_{3,-2}^{3}\big{)}=\frac{\ker\big{(}d^{3}:E_{3,-2}^{3}\rightarrow
E_{0,0}^{3}\big{)}}{\operatorname{im}\big{(}d^{3}:E_{6,-4}^{3}\rightarrow
E_{3,-2}^{3}\big{)}}=\ker\big{(}d_{3,-2}^{3}\big{)}\subseteq E_{3,-2}^{3}.$

Now, $E_{3,-2}^{3}\cong H(E_{3,-2}^{3})=E_{3,-2}^{3}=H_{3}(\mathcal{D}_{4})$,
and hence

$E_{3,-2}^{5}\cong\ker\big{(}d_{3,-2}^{3}\big{)}\subseteq
E_{3,-2}^{3}=H_{3}(\mathcal{D}_{4}).$

Write $G_{3}:=\ker(d_{3,-2}^{3})$, which is a subgroup of
$H_{3}(\mathcal{D}_{4})$. Then we have the short exact sequence (iv):

$0\longrightarrow H_{1}(\mathcal{D}_{4})/G_{2}\longrightarrow
K_{1}(\mathcal{A}(\Lambda))\longrightarrow G_{3}\longrightarrow 0.$

Step (b): $E_{0,0}^{5}$, $E_{2,-2}^{5}$, and $E_{4,-4}^{5}$.

Firstly, consider $E_{0,0}^{5}$. We know that $E_{0,0}^{5}\cong
H(E_{0,0}^{4})=E_{0,0}^{4}$, since the differential $d^{4}$ is the zero map.
We also have

$E_{0,0}^{4}\cong
H\big{(}E_{0,0}^{3}\big{)}=\frac{\ker\big{(}d^{3}:E_{0,0}^{3}\rightarrow
E_{-3,0}^{3}\big{)}}{\operatorname{im}\big{(}d^{3}:E_{3,-2}^{3}\rightarrow
E_{0,0}^{3}\big{)}}=E_{0,0}^{3}/\operatorname{im}\big{(}d_{3,-2}^{3}\big{)}.$

Note that $E_{0,0}^{3}\cong
H(E_{0,0}^{2})=E_{0,0}^{2}=H_{0}(\mathcal{D}_{4})=\operatorname{coker}(\partial_{1})$,
so that if we write $G_{0}:=\operatorname{im}(d_{3,-2}^{3})$, we obtain
$E_{0,0}^{5}=\operatorname{coker}(\partial_{1})/G_{0}$. This, together with
the sequence (i’) above, gives us the sequence (i).

Now, we turn our attention to $E_{2,-2}^{5}$ and $E_{4,-4}^{5}$. We know that
$E_{2,-2}^{5}\cong H(E_{2,-2}^{4})=E_{2,-2}^{4}$, by virtue of $d^{4}$ being
the zero map. We also have

$E_{2,-2}^{4}\cong
H\big{(}E_{2,-2}^{3}\big{)}=\frac{\ker\big{(}d^{3}:E_{2,-2}^{3}\rightarrow
E_{-1,0}^{3}\big{)}}{\operatorname{im}\big{(}d^{3}:E_{5,-4}^{3}\rightarrow
E_{2,-2}^{3}\big{)}}=E_{2,-2}^{3}\cong
H\big{(}E_{2,-2}^{2}\big{)}=H_{2}(\mathcal{D}_{4}),$

and so $E_{2,-2}^{5}\cong H_{2}(\mathcal{D}_{4})$. Together with (ii’) and the
above, this gives us (ii). We also know that $E_{4,-4}^{5}\cong
H(E_{4,-4}^{4})=E_{4,-4}^{5}$, and

$E_{4,-4}^{4}\cong
H\big{(}E_{4,-4}^{3}\big{)}=\frac{\ker\big{(}d^{3}:E_{4,-4}^{3}\rightarrow
E_{1,-2}^{3}\big{)}}{\operatorname{im}\big{(}d^{3}:E_{7,-6}^{3}\rightarrow
E_{4,-4}^{3}\big{)}}=\ker\big{(}d_{4,-4}^{3}\big{)}\subseteq E_{4,-4}^{3},$

and so $E_{4,-4}^{3}\cong
H(E_{4,-4}^{2})=E_{4,-4}^{2}=H_{4}(\mathcal{D}_{4})$. Writing
$G_{1}:=\ker(d_{4,-4}^{3})$, and putting this together with (iii’), we obtain
the sequence (iii). We know that $H_{4}(\mathcal{D}_{4})$ is a free Abelian
group, and since subgroups of such groups are also free Abelian, it follows
that $G_{1}$ is free Abelian, and sequence (iii) splits. ∎

######  Corollary 4.6 ($k=4$).

In addition to the hypotheses of Proposition 4.5:

  1. (i)

If $\partial_{1}$ is surjective, then there exists an isomorphism
$F_{2}\cong\ker(\partial_{2})/\operatorname{im}(\partial_{3})$, and the short
exact sequences reduce to:

    1. (a)

$K_{0}(\mathcal{A}(\Lambda))\cong\dfrac{\ker(\partial_{2})}{\operatorname{im}(\partial_{3})}\oplus
G_{1}$,

    2. (b)

$0\longrightarrow\dfrac{\ker(\partial_{1})/\operatorname{im}(\partial_{2})}{G_{2}}\longrightarrow
K_{1}(\mathcal{A}(\Lambda))\longrightarrow\ker(\partial_{3})/\operatorname{im}(\partial_{4})\longrightarrow
0$.

  2. (ii)

If $\bigcap_{i}\ker\big{(}I-M_{i}^{T}\big{)}=0$, then
$K_{0}(\mathcal{A}(\Lambda))\cong F_{2}$, and the sequences reduce to:

    1. (a)

$0\longrightarrow\operatorname{coker}(\partial_{1})/G_{0}\longrightarrow
K_{0}(\mathcal{A}(\Lambda))\longrightarrow\ker(\partial_{3})/\operatorname{im}(\partial_{2})\longrightarrow
0$,

    2. (b)

$0\longrightarrow\ker(\partial_{1})/\operatorname{im}(\partial_{2})\longrightarrow
K_{1}(\mathcal{A}(\Lambda))\longrightarrow G_{3}\longrightarrow 0$.

###### Proof.

To show (i), suppose that $\partial_{1}$ is surjective, such that
$\operatorname{coker}(\partial_{1})=0$, and
$F_{2}(K_{0})\cong\ker(\partial_{2})/\operatorname{im}(\partial_{3})$. Then
the split exact sequence (iii) from Proposition 4.5 gives us (i)(a).

Now, we have
$0=\operatorname{coker}(\partial_{1})=H_{0}(\mathcal{D}_{4})=E_{0,0}^{3}$, and
so $d^{3}:E_{3,-2}^{3}\rightarrow E_{0,0}^{3}$ is the zero map. Hence
$\ker(d_{3,-2}^{3})=E_{3,-2}^{3}=H_{3}(\mathcal{D}_{4})$, and we obtain (i)(b)
from Proposition 4.5(iv).

To show (ii), suppose that $\bigcap_{i}\big{(}I-M_{i}^{T}\big{)}=0$. Then
$\ker(\partial_{4})=0$, and hence $G_{1}=0$ and $K_{0}\cong F_{2}$. This gives
us (ii)(a).

Now, from Proposition 4.5(iv), we have the sequence

$0\longrightarrow\frac{\ker(\partial_{1})/\operatorname{im}(\partial_{2})}{G_{2}}\longrightarrow
K_{1}\longrightarrow G_{3}\longrightarrow 0,$

where

$\displaystyle G_{2}$
$\displaystyle:=\operatorname{im}\big{(}d_{4,-2}^{3}:E_{4,-2}^{3}\longrightarrow
E_{1,0}^{3}=H_{1}(\mathcal{D}_{4})\big{)},$ $\displaystyle G_{3}$
$\displaystyle:=\ker\big{(}d_{3,-2}^{3}\big{)}\subseteq\ker(\partial_{3})/\operatorname{im}(\partial_{4}).$

However, we also have that $E_{4,-2}^{3}\cong
H(E_{4,-2}^{2})=E_{4,-2}^{2}=H_{4}(\mathcal{D}_{4})=\ker(\partial_{4})$. Since
$\ker(\partial_{4})=0$, it follows that the differential $\partial_{4,-2}^{3}$
has domain $0$, and is hence the zero map. Therefore $G_{2}=0$, and the result
follows. ∎

We have computed similar short exact sequences in the case where $k=5$, and
the proof, omitted, is broadly similar to the above.

######  Proposition 4.7 ($k=5$).

Let $\Lambda$ be a row-finite $5$-graph with no sources, and let
$\mathcal{D}_{5}$ be the corresponding chain complex with differentials
$\partial_{1},\ldots,\partial_{5}$ defined in Theorem 4.2. Let $F_{2}$,
$F_{3}$ be factors in the ascending filtration of the $C^{\star}$-algebra
$\mathcal{A}(\Lambda)$. Then, for some subgroups

$\begin{array}[]{ll}G_{0}\subseteq\operatorname{coker}(\partial_{1})=H_{0}(\mathcal{D}_{5}),&G_{4}\cong\ker\big{(}d_{5,-4}^{5}\big{)}\subseteq\ker\big{(}d_{5,-4}^{3}\big{)}\subseteq
H_{5}(\mathcal{D}_{5}),\\\
G_{1}\cong\operatorname{im}\big{(}d_{5,-4}^{5}\big{)}\subseteq
H_{0}(\mathcal{D}_{5})/G_{0},&G_{5}\cong\operatorname{im}\big{(}d_{4,-2}^{3}\big{)}\subseteq
H_{1}(\mathcal{D}_{5}),\\\
G_{2}\cong\operatorname{im}\big{(}d_{5,-4}^{3}\big{)}\subseteq
H_{2}(\mathcal{D}_{5}),&G_{6}\cong\ker\big{(}d_{3,-2}^{3}\big{)}\subseteq
H_{3}(\mathcal{D}_{5}),\\\ G_{3}\cong\ker\big{(}d_{4,-4}^{3}\big{)}\subseteq
H_{4}(\mathcal{D}_{5}),&\end{array}$

there exist short exact sequences as follows:

  1. (i)

$0\longrightarrow
A:=\dfrac{\operatorname{coker}(\partial_{1})/G_{0}}{G_{1}}\longrightarrow
K_{0}(\mathcal{A}(\Lambda))\longrightarrow
K_{0}(\mathcal{A}(\Lambda))/A\longrightarrow 0$,

  2. (ii)

$0\longrightarrow A\longrightarrow
F_{2}\longrightarrow\dfrac{\ker(\partial_{2})/\operatorname{im}(\partial_{3})}{G_{2}}\longrightarrow
0$,

  3. (iii)

$0\longrightarrow F_{2}\longrightarrow
K_{0}(\mathcal{A}(\Lambda))\longrightarrow G_{3}\longrightarrow 0$,

and there is an isomorphism $K_{1}(\mathcal{A}(\Lambda))\cong F_{3}\oplus
G_{4}$, where $F_{3}$ satisfies

$0\longrightarrow\frac{\ker(\partial_{1})/\operatorname{im}(\partial_{2})}{G_{5}}\longrightarrow
F_{3}\longrightarrow G_{6}\longrightarrow 0.$

∎

By the Kirchberg–Phillips Classification ([13], [20]), any separable, nuclear,
unital, purely-infinite, simple $C^{\star}$-algebra which satisfies the
Rosenberg–Schochet Universal Coefficient Theorem [23] is completely determined
by its K-groups and the class of the identity in $K_{0}$.

######  Lemma 4.8.

Let $\Gamma$ be a $k$-cube group with adjacency structure
$E_{1},\ldots,E_{k}$. Then the induced $k$-rank graph $\mathcal{G}(\Gamma)$
satisfies the _Aperiodicity Condition_ ([15], [19]).

###### Proof.

The result can be obtained with a slight adaptation to that of Lemma 4.2 in
[19], as a result of the observations in §2 of [21]. Briefly, if $\varphi$ is
an infinite path in $\mathcal{G}(\Gamma)$, then the Aperiodicity Condition is
satisfied if $\varphi$ does not enter or exit an infinite cycle. Since
$|E_{i}|\geq 4$ for each $i$, there are always at least two $k$-cubes which
are $E_{i}$-adjacent to some $k$-cube $S$. Hence, we can always exit some
cycle by diverting our path down another edge in $\mathcal{G}(\Gamma)$ at any
vertex $S$.

Likewise, given a $k$-cube $T$, there are always at least two $k$-cubes to
which $T$ is $E_{i}$-adjacent. Thus we may always find an infinite aperiodic
path, and $\mathcal{G}(\Gamma)$ satisfies the Aperiodicity Condition. ∎

######  Lemma 4.9.

Let $\Gamma$ be a $k$-cube group, and let $\mathcal{G}(\Gamma)$ be its induced
$k$-rank graph. Then $\mathcal{G}(\Gamma)$ is connected in the sense that, for
any two vertices $S,T\in\mathcal{G}(\Gamma)^{\mathbf{0}}$, there is a path
from $S$ to $T$.

We give a geometric proof, based on the cube complex $\mathcal{M}(\Gamma)$,
although we point out that this can also be proved in the manner of Lemma 4.2
in [12].

###### Proof.

Let $\Gamma$ be a $k$-cube group with adjacency structure
$E_{1},\ldots,E_{k}$, and consider a pointed $k$-cube $S$ in the cube complex
$\mathcal{M}(\Gamma)$. Let $S_{H}$ be the $k$-cube obtained by reflecting $S$
through the edges labelled by elements of $E_{1}$, leaving the basepoint and
orientation the same as in $S$ (Figure 1).

Firstly, we aim to show that there is a sequence of $k$-cubes
$S=T_{0},T_{1},\ldots,T_{n}=S_{H}$ such that $M_{1}(T_{j},T_{j+1})=1$ for all
$j$, that is, such that each $k$-cube is $E_{1}$-adjacent to the next.

Each $k$-cube $X$ in $\mathcal{M}(\Gamma)$ contains two $(k-1)$-faces (that
is, $(k-1)$-sub-cubes) labelled by elements of $E_{2},\ldots,E_{k}$. Since the
$k$-cubes have a predetermined orientation, we may label these faces $X^{L}$
and $X^{R}$, such that $M_{1}(X,Y)=1$ if and only if $Y^{L}=X^{R}$ and $Y\neq
X_{H}$. We may therefore assign to each $k$-cube $X$ the pair $(X^{L},X^{R})$
such that, in the above sequence, $T_{j+1}^{L}=T_{i}^{R}$, and
$T_{j+1}\neq(T_{j})_{H}$, for all $i$.

Observe that each $(k-1)$-cube appears as $X^{L}$ (resp. $X^{R}$) for some
$X\in\mathcal{S}_{k}$ precisely $|E_{1}|$ times, and that, by assumption,
$|E_{1}|\geq 4$ and $|E_{1}|$ is even.

Write $A_{0}:=S^{R}$, and let $T_{1}$ be a $k$-cube $E_{1}$-adjacent to $S$;
such a $k$-cube exists by the above observation. If $T_{1}^{R}=A_{0}$, then
$M_{1}(T_{1},S_{H})=1$ and we are done. Assume then that $T_{1}^{R}=A_{1}\neq
A_{0}$, and let $T_{2}$ be $E_{1}$-adjacent to $T_{1}$. If $T_{2}^{R}=A_{0}$,
then $M_{1}(T_{2},S_{H})=1$ , and if $T_{2}^{R}=A_{1}$, then
$M_{1}(T_{2},(T_{1})_{H})=1$, and $M_{1}((T_{1})_{H},S_{H})=1$. In both cases,
we have obtained a sequence of adjacent $k$-cubes linking $S$ to $S_{H}$.

By induction, we see that if $T_{q}^{R}=T_{p}^{R}$ for any $p<q$, we obtain
the sequence we desire. But also, by the fact that each $(k-1)$-cube appears
as $X^{R}$ for some $X\in\mathcal{S}_{k}$ an even number of times, there must
be some $q>p$ for which $T_{q}^{R}=T_{p}^{R}$. Hence such a sequence exists,
and there is a path connecting the vertices labelled $S$ and $S_{H}$ in
$\mathcal{G}(\Gamma)$.

In the same manner, we may show that there is a sequence of adjacent $k$-cubes
connecting each $S\in\mathcal{S}_{k}$ to each of its symmetries, i.e., the
$k$-cubes which belong to the same orbit as $S$ under action by the symmetry
group of the $k$-dimensional cube.

* * *

Now, we construct the set $\mathcal{P}$ of all $k$-cubes which can be reached
by a sequence of adjacent $k$-cubes (in any sequence of directions) from an
initial $k$-cube $S$. Certainly $S_{H}$ is in $\mathcal{P}$, by the above.
Moreover, by construction, $\mathcal{P}$ contains $(|E_{1}|-1)$ more distinct
$k$-cubes which are $E_{1}$-adjacent to $S$, to total $|E_{1}|$ distinct
$k$-cubes. Each of these $k$-cubes is $E_{2}$-adjacent to $|E_{2}|$ $k$-cubes
by the same argument. These are distinct from each other by the uniqueness
property of C3.

We may proceed inductively to find that $\mathcal{P}$ must contain at least
$\prod_{i=1}^{k}|E_{i}|$ distinct $k$-cubes, but this is precisely
$|\mathcal{S}_{k}|=|\mathcal{G}(\Gamma)^{\mathbf{0}}|$. Hence each $k$-cube in
$\mathcal{S}_{k}$ can be reached from some $k$-cube $S$ by a sequence of
adjacent $k$-cubes. Equivalently, given any vertex labelled by $S$ in
$\mathcal{G}(\Gamma)^{\mathbf{0}}$, there is a path from $S$ to every other
vertex. ∎

######  Theorem 4.10.

Let $\Gamma$ be a $k$-cube group with adjacency structure
$E_{1},\ldots,E_{k}$. Then
$\mathcal{A}(\Gamma):=\mathcal{A}(\mathcal{G}(\Gamma))$ is separable, nuclear,
purely-infinite, simple, and satisfies the Universal Coefficient Theorem.
Hence $\mathcal{A}(\Gamma)$ is completely determined by its K-groups and the
class of the identity of $\mathcal{A}(\Gamma)$ in $K_{0}$, up to isomorphism.

###### Proof.

By Lemma 4.9 and Proposition 4.8 in [15], it follows that
$\mathcal{A}(\Gamma)$ is simple. Also by Lemma 4.9, together with the fact
that $|E_{i}|\geq 4$ for all $i$, it follows that for every
$S\in\mathcal{G}(\Gamma)^{\mathbf{0}}$ we can find
$\lambda,\mu\in\mathcal{G}(\Gamma)$ such that $d(\mu)\neq\mathbf{0}$,
$r(\lambda)=S$, and $s(\lambda)=r(\mu)=s(\mu)$. Hence by Proposition 4.9 in
[15] it follows that $\mathcal{A}(\Gamma)$ is purely-infinite.

From Theorem 3.3 we know that $\mathcal{G}(\Gamma)$ is a row-finite $k$-graph
with no sources, and in [6] it is shown that such a $k$-graph has a
corresponding $C^{\star}$-algebra which is separable, nuclear, unital, and
satisfies the Universal Coefficient Theorem, hence we are done. ∎

######  Proposition 4.11.

Let $\Gamma$ be a $k$-cube group with adjacency structure
$E_{1},\ldots,E_{k}$, where $|E_{i}|=m_{i}$, and define
$\rho:=\gcd\\{(m_{i}/2)-1\mid 1\leq i\leq k\\}$. Factorise $\rho$ as $2^{q}r$,
where $r$ is an odd number: if $\rho$ is odd then $q=0$. Then the order of the
class of $\operatorname{id}\in\mathcal{A}(\Gamma)$ in
$K_{0}(\mathcal{A}(\Gamma))$ is at most $\rho$, and is:

  1. (i)

Equal to $\rho$ if $\rho$ is odd,

  2. (ii)

Divisible by $\rho/(2^{q})$ if $1\leq q<(k-1)$,

  3. (iii)

Divisible by $\rho/(2^{k-1})$ if $q\geq(k-1)$.

###### Proof.

From a straightforward generalisation of Proposition 5.4 in [12], we know that
the order of the class of $\operatorname{id}\in\mathcal{A}(\Gamma)$ in
$K_{0}(\mathcal{A}(\Gamma))$ divides $\rho$.

Now, given a higher-rank graph $\Lambda$, the sum of all elements of
$\mathcal{A}(\Lambda)$ of the form $s_{v}$, where $v\in\Lambda^{\mathbf{0}}$,
is an identity for $\mathcal{A}(\Lambda)$ (e.g. [21, Remark 3.4]). Hence for a
$k$-cube group $\Gamma$, the sum $\sum_{S\in\mathcal{S}(\Gamma)}s_{S}$ forms
an identity in $\mathcal{A}(\Gamma)$. Recall the map

$\partial_{1}:\mathbb{Z}\Gamma^{\mathbf{0}}\longrightarrow\bigoplus_{i=1}^{k}\mathbb{Z}\Gamma^{\mathbf{0}}$

defined by the matrix $\big{[}I-M_{1}^{T},\ldots,I-M_{k}^{T}\big{]}$. The
Covariance Relation of [12, §5] generalises to $k$-graphs, and so from [22] it
follows that the map

$\varphi:\operatorname{coker}(\partial_{1})=\bigg{\langle}S\in\mathcal{S}_{k}\biggm{|}\sum_{T\in\mathcal{S}}M_{i}(S,T)\cdot
S\bigg{\rangle}\longrightarrow K_{0}(\mathcal{A}(\Gamma))$

which takes $S$ to its class $[S]$ is injective. But each column of $M_{i}$
has exactly $(m_{i}-1)$ ones, the rest of the entries being zero, and so
$\Sigma=(m_{i}-1)\Sigma$ for each $i\in\\{1,\ldots,k\\}$, where
$\Sigma:=\sum_{S\in\mathcal{S}_{k}}$. Since $\sum_{S\in\mathcal{S}_{k}}s_{S}$
is an identity in $\mathcal{A}(\Gamma)$, the class $[\operatorname{id}]\in
K_{0}$ is the image of $\Sigma$ under $\varphi$; By the above, we also know
that $(m_{i}-2)\Sigma$ is zero for each $i$.

Write $2\rho=\gcd\\{m_{i}-2\\}$, and define the map
$\psi:\operatorname{coker}(\partial_{1})\rightarrow\mathbb{Z}/2\rho$ by
$\psi(S)=1\mod 2\rho$, as in the proof of [12, Prop. 5.4]. Now,

$\prod_{i}(m_{i}-2)=\bigg{(}\prod_{i}m_{i}\bigg{)}-2^{k}\mod 2\rho,$

and since $(m_{i}-2)=0\mod 2\rho$, this means that $\psi(\Sigma)=2^{k}\mod
2\rho$, and so $\rho\cdot\psi(\Sigma)=0\mod 2\rho$. If $\rho$ is odd, then
$\psi(\Sigma)$ has order $\rho$ in $\mathbb{Z}/2\rho$. If $\rho$ is even, then
$\rho=2^{q}r$ for some odd number $r$, and
$\rho\cdot\psi(\Sigma)=2^{k+q}r\mod(2^{q+1}r)$. Hence the order of $\Sigma$ in
$\operatorname{coker}(\partial_{1})$ is divisible by $\rho$ in the former
case, and by $\max\\{\rho/(2^{q}),\rho/(2^{k-1})\\}$ in the latter. ∎

##  5 Examples for $k=3$ and $k=4$

######  Example 5.1.

Consider the product of three free groups, each with two generators, defined
as follows:

$\mathbb{F}_{2}^{3}:=\langle
a_{1},a_{2},b_{1},b_{2},c_{1},c_{2}\mid[a_{i},b_{j}],[a_{i},c_{j}],[b_{i},c_{j}],\text{
for all }i,j\in\\{1,2\\}\rangle,$

where $[x,y]$ denotes the commutator $xyx^{-1}y^{-1}$. This is a $3$-cube
group with adjacency structure
$\\{a_{i},a_{i}^{-1}\\},\\{b_{i},b_{i}^{-1}\\},\\{c_{i},c_{i}^{-1}\\}$. We
construct the chain complex from Proposition 4.3 using the three corresponding
adjacency matrices, to find that
$\operatorname{coker}(\partial_{1})\cong\mathbb{Z}^{8}$, and
$\ker(\partial_{2})/\operatorname{im}(\partial_{3})\cong\ker(\partial_{1})/\operatorname{im}(\partial_{2})\cong\mathbb{Z}^{24}$.

We can use the Künneth Theorem for tensor products [26, §9.3] to calculate the
K-theory explicitly up to isomorphism; we have
$K_{0}(\mathcal{A}(\mathbb{F}_{2}^{3}))\cong
K_{1}(\mathcal{A}(\mathbb{F}_{2}^{3}))\cong\mathbb{Z}^{32}$. Then, since all
of the groups from Proposition 4.3 are free Abelian, we are able to deduce
that $G_{0}=0$, and $G_{1}\cong\mathbb{Z}^{8}$. This complies with Matui’s HK-
Conjecture [17].

######  Example 5.2.

Now consider the product $\Gamma$ of three free groups, each with three
generators; this is a $3$-cube group whose corresponding cube complex has as
universal cover $T(6)\times T(6)\times T(6)$. We again construct the chain
complex $\mathcal{D}_{3}$ from Proposition 4.3 using the three corresponding
adjacency matrices, to find that:

  * •

$\operatorname{coker}(\partial_{1})\cong\mathbb{Z}^{27}\oplus(\mathbb{Z}/2)^{37}$,

  * •

$\ker(\partial_{2})/\operatorname{im}(\partial_{3})\cong\mathbb{Z}^{81}\oplus(\mathbb{Z}/2)^{37}$,

  * •

$\ker(\partial_{1})/\operatorname{im}(\partial_{2})\cong\mathbb{Z}^{81}\oplus(\mathbb{Z}/2)^{74}$,

  * •

$\ker(\partial_{3})\cong\mathbb{Z}^{27}$.

Hence we have a short exact sequence

$0\longrightarrow\frac{\mathbb{Z}^{27}\oplus(\mathbb{Z}/2)^{37}}{G_{0}}\longrightarrow
K_{0}(\mathcal{A}(\Gamma))\longrightarrow\mathbb{Z}^{81}\oplus(\mathbb{Z}/2)^{37}\longrightarrow
0,$ (3)

and an isomorphism
$K_{1}(\mathcal{A}(\Gamma))\cong\mathbb{Z}^{r}\oplus(\mathbb{Z}/2)^{74}$, for
some $G_{0}\subseteq\mathbb{Z}^{27}\oplus(\mathbb{Z}/2)^{37}$ and $81\leq
r\leq 108$. We deduce from (3) and the fact that $K_{0}$ and $K_{1}$ must have
the same torsion-free rank [6, Prop. 4.1] that the torsion-free part of
$K_{0}$ is isomorphic to $\mathbb{Z}^{r}$.

Write $A,B,C$ for the adjacency structure of $\Gamma$. By Lemma 2.6, the three
subgroups of $\Gamma$ isomorphic to $\mathbb{F}_{3}^{2}$, obtained by removing
one of $A,B,C$ from the generating set, are each $2$-cube groups (or BMW-
groups). The $3$-group $\Gamma$ is a free product with amalgamation of these
three groups (Proposition 2.7). The K-theory of their induced $k$-rank graph
algebras is given by

$K_{0}(\mathbb{F}_{3}^{2})\cong
K_{1}(\mathbb{F}_{3}^{2})\cong\mathbb{Z}^{18}\oplus(\mathbb{Z}/2)^{7}.$

Compare this to the K-theory of the $k$-rank graph algebra induced by
$\Gamma$, calculated above. We discern no obvious structure inherited by the
K-theory of $\mathcal{A}(\Gamma)$ from the K-theory induced by its $2$-cube
subgroups.

######  Example 5.3.

Recall the group $\Gamma=\Gamma_{\\{3,5,7\\}}$ from Example 2.12. It is
readily verifiable that this is a $3$-cube group, and its corresponding cube
complex comprises one vertex, $26$ squares labelled by the relators in $R$,
and $24$ cubes. We can construct three $192\times 192$ adjacency matrices
$M_{1},M_{2},M_{3}$ based on adjacency of pointed cubes in the $\\{a_{i}\\}$,
$\\{b_{i}\\}$, and $\\{c_{i}\\}$ directions respectively. We know that
$\mathcal{G}(\Gamma)$ is a $3$-rank graph by Theorem 3.3, and so we can input
the matrices $M_{1},M_{2},M_{3}$ into Proposition 4.3 to garner information
about the algebra $\mathcal{A}(\Gamma)$.

For $\operatorname{coker}(\partial_{1})$, it suffices to compute the
elementary divisors (that is, the diagonal elements of the Smith normal form)
of $\partial_{1}$. This is because the cokernel of a linear map is equal to
the cokernel of its Smith normal form. The Smith normal form $S(\partial_{1})$
is a $576\times 576$ diagonal matrix with entries

$\underbrace{1,\ldots,1}_{181\text{
times}},4,4,12,\underbrace{0,\ldots,0}_{7\text{ times}}.$

Hence we have
$\operatorname{coker}(\partial_{1})\cong\mathbb{Z}^{7}\oplus(\mathbb{Z}/4)^{2}\oplus(\mathbb{Z}/12)$,
and we are able to work out similarly that
$\ker(\partial_{3})\cong\mathbb{Z}^{7}$. We also verify using MAGMA that:

  * •

$\ker(\partial_{2})/\operatorname{im}(\partial_{3})\cong\mathbb{Z}^{21}\oplus(\mathbb{Z}/4)^{2}\oplus(\mathbb{Z}/12)$,
and

  * •

$\ker(\partial_{1})/\operatorname{im}(\partial_{2})\cong\mathbb{Z}^{21}\oplus(\mathbb{Z}/2)^{6}\oplus(\mathbb{Z}/4)^{2}\oplus(\mathbb{Z}/12)^{2}$.

We therefore have a short exact sequence

$0\longrightarrow\frac{\mathbb{Z}^{7}\oplus(\mathbb{Z}/4)^{2}\oplus(\mathbb{Z}/12)}{G_{0}}\longrightarrow
K_{0}(\mathcal{A}(\Gamma))\longrightarrow\mathbb{Z}^{21}\oplus(\mathbb{Z}/4)^{2}\oplus(\mathbb{Z}/12)\longrightarrow
0,$

and an isomorphism

$K_{1}(\mathcal{A}(\Gamma))\cong\mathbb{Z}^{21}\oplus(\mathbb{Z}/2)^{6}\oplus(\mathbb{Z}/4)^{2}\oplus(\mathbb{Z}/12)^{2}\oplus
G_{1},$

for some
$G_{0}\subseteq\mathbb{Z}^{7}\oplus(\mathbb{Z}/4)^{2}\oplus(\mathbb{Z}/12)$
and $G_{1}\subseteq\mathbb{Z}^{7}$. From this and Proposition 4.1 in [6], we
can deduce that the torsion-free part of $K_{0}$ is isomorphic to
$\mathbb{Z}^{r}$, and that
$K_{1}\cong\mathbb{Z}^{r}\oplus(\mathbb{Z}/2)^{6}\oplus(\mathbb{Z}/4)^{2}\oplus(\mathbb{Z}/12)^{2}$,
for some $21\leq r\leq 28$.

We may also compute the cellular homology of the cube complex $\Gamma$;
firstly by determining the relevant boundary map matrices. We consider the
barycentric subdivision of $\Gamma$, which has the same cellular homology as
$\Gamma$, and whose edge set contains no loops. Then, from the Smith normal
forms of the boundary maps, we discover that

$H_{i}(\Gamma)\cong\begin{cases}\mathbb{Z}&\text{ when }i=0,\\\
(\mathbb{Z}/2)^{2}\oplus(\mathbb{Z}/4)^{2}&\text{ when }i=1,\\\
(\mathbb{Z}/2)^{2}\oplus(\mathbb{Z}/12)&\text{ when }i=2,\\\ 0&\text{ when
}i\geq 3.\end{cases}$

######  Example 5.4.

Consider the group $\Gamma=\Gamma_{\\{2,3,4\\}}$ from Example 2.36 of [24],
defined as follows:

$\Gamma_{\\{2,3,4\\}}:=\langle
a_{1},a_{5},a_{9},b_{2},b_{6},b_{10},c_{3},c_{7},c_{11}\mid R\rangle,$

where

$\displaystyle R:=\big{\\{}$ $\displaystyle
a_{1}b_{2}a_{5}^{-1}b_{10}^{-1},a_{1}b_{6}a_{9}b_{10},a_{1}b_{10}a_{9}b_{6},a_{1}b_{2}^{-1}a_{9}^{-1}b_{2}^{-1},$
$\displaystyle
a_{1}b_{6}^{-1}a_{5}b_{6}^{-1},a_{1}b_{10}^{-1}a_{5}^{-1}b_{2},a_{5}b_{2}a_{9}^{-1}b_{6},a_{5}b_{6}a_{9}^{-1}b_{2},a_{5}b_{10}^{-1}a_{9}b_{10}^{-1},$
$\displaystyle
a_{1}c_{3}a_{5}^{-1}c_{3},a_{1}c_{7}a_{1}^{-1}c_{7}^{-1},a_{1}c_{11}a_{9}c_{11},a_{1}c_{3}^{-1}a_{1}c_{11}^{-1},$
$\displaystyle
a_{5}c_{3}a_{5}c_{7}^{-1},a_{5}c_{7}a_{9}^{-1}c_{7},a_{5}c_{11}a_{5}^{-1}c_{11}^{-1},a_{9}c_{3}a_{9}^{-1}c_{3}^{-1},a_{9}c_{7}a_{9}c_{11}^{-1},$
$\displaystyle
b_{2}c_{3}b_{6}^{-1}c_{11}^{-1},b_{2}c_{7}b_{10}c_{11},b_{2}c_{11}b_{10}c_{7},b_{2}c_{3}^{-1}b_{10}^{-1}c_{3}^{-1},$
$\displaystyle
b_{2}c_{7}^{-1}b_{6}c_{7}^{-1},b_{2}c_{11}^{-1}b_{6}^{-1}c_{3},b_{6}c_{3}b_{10}^{-1}c_{7},b_{6}c_{7}b_{10}^{-1}c_{3},b_{6}c_{11}^{-1}b_{10}c_{11}^{-1}\big{\\}}.$

This is a $3$-cube group with adjacency structure
$\\{a_{i},a_{i}^{-1}\\},\\{b_{i},b_{i}^{-1}\\},\\{c_{i},c_{i}^{-1}\\}$, and
which acts freely and transitively on $T(6)\times T(6)\times T(6)$, as in
Example 5.2. The corresponding cube complex $\mathcal{M}$ has one vertex, $27$
squares labelled with the relators in $R$, and $27$ cubes.

We can construct three $216\times 216$ adjacency matrices, and use Proposition
4.3 to reveal information about the K-groups $K_{0},K_{1}$ of the $3$-rank
graph algebra $\mathcal{A}(\Gamma)$.

As in the previous example, we use MAGMA to compute the relevant kernels and
cokernels, culminating with:

  * •

$\operatorname{coker}(\partial_{1})\cong\mathbb{Z}^{9}\oplus(\mathbb{Z}/2)\oplus(\mathbb{Z}/20)\oplus(\mathbb{Z}/80)$,

  * •

$\ker(\partial_{3})\cong\mathbb{Z}^{9}$,

  * •

$\ker(\partial_{2})/\operatorname{im}(\partial_{3})\cong\mathbb{Z}^{27}\oplus(\mathbb{Z}/2)\oplus(\mathbb{Z}/20)\oplus(\mathbb{Z}/80)$,

  * •

$\ker(\partial_{1})/\operatorname{im}(\partial_{2})\cong\mathbb{Z}^{27}\oplus(\mathbb{Z}/2)^{4}\oplus(\mathbb{Z}/4)^{2}\oplus(\mathbb{Z}/8)^{2}$.

Then, from Proposition 4.3 we obtain the short exact sequence

$0\rightarrow\frac{\mathbb{Z}^{9}\oplus(\mathbb{Z}/2)\oplus(\mathbb{Z}/20)\oplus(\mathbb{Z}/80)}{G_{0}}\longrightarrow
K_{0}\longrightarrow\mathbb{Z}^{27}\oplus(\mathbb{Z}/2)\oplus(\mathbb{Z}/20)\oplus(\mathbb{Z}/80)\rightarrow
0,$

and the isomorphism

$K_{1}\cong\mathbb{Z}^{27}\oplus(\mathbb{Z}/2)^{4}\oplus(\mathbb{Z}/4)^{2}\oplus(\mathbb{Z}/8)^{2}\oplus
G_{1},$

where
$G_{0}\subseteq\mathbb{Z}^{9}\oplus(\mathbb{Z}/2)\oplus(\mathbb{Z}/20)\oplus(\mathbb{Z}/80)$
and $G_{1}\subseteq\mathbb{Z}^{9}$. Hence the torsion-free part of $K_{0}$ is
isomorphic to $\mathbb{Z}^{r}$, and
$K_{1}\cong\mathbb{Z}^{r}\oplus(\mathbb{Z}/2)^{4}\oplus(\mathbb{Z}/4)^{2}\oplus(\mathbb{Z}/8)^{2}$,
for some $27\leq r\leq 36$. The $K_{1}$ group in particular is distinct from
those of Examples 5.2 and 5.3, so we may conclude by Theorem 4.10 that the
$C^{\star}$-algebras induced by each of the cube complexes are different.

######  Remark 5.5.

In each of the examples above, the torsion-free rank $r$ of $K_{\ast}$ lies
within a range of values. If Matui’s HK-Conjecture [17] is true, then
$K_{1}\cong H_{1}(\mathcal{D}_{k})\oplus H_{3}(\mathcal{D}_{k})$, and so $r$
must be maximal in this range.

######  Example 5.6.

Consider the product of four free groups, each with two generators, defined as
follows:

$\mathbb{F}_{2}^{4}:=\langle
a_{1},a_{2},b_{1},b_{2},c_{1},c_{2},d_{1},d_{2}\mid[a_{i},b_{j}],[a_{i},c_{j}],[a_{i},d_{j}],\\\
[b_{i},c_{j}],[b_{i},d_{j}],[c_{i},d_{j}]\text{ for all
}i,j\in\\{1,2\\}\rangle.$

This is a $4$-cube group with adjacency structure
$\\{a_{i},a_{i}^{-1}\\},\ldots,\\{d_{i},d_{i}^{-1}\\}$. We construct the chain
complex from Proposition 4.5 using the four corresponding adjacency matrices,
and obtain short exact sequences:

  1. (i)

$0\longrightarrow\mathbb{Z}^{16}/G_{0}\longrightarrow
K_{0}(\mathcal{A}(\mathbb{F}_{2}^{4}))\longrightarrow\dfrac{K_{0}(\mathcal{A}(\mathbb{F}_{2}^{4}))}{\mathbb{Z}^{16}/G_{0}}\longrightarrow
0$,

  2. (ii)

$0\longrightarrow\mathbb{Z}^{16}/G_{0}\longrightarrow
F_{2}\longrightarrow\mathbb{Z}^{96}\longrightarrow 0$,

  3. (iii)

$0\longrightarrow F_{2}\longrightarrow
K_{0}(\mathcal{A}(\mathbb{F}_{2}^{4}))\longrightarrow G_{1}\longrightarrow 0$,

  4. (iv)

$0\longrightarrow\mathbb{Z}^{64}/G_{2}\longrightarrow
K_{1}(\mathcal{A}(\mathbb{F}_{2}^{4}))\longrightarrow G_{3}\longrightarrow 0$,

for some subgroups $G_{0},G_{1}\subseteq\mathbb{Z}^{16}$, and
$G_{2},G_{3}\subseteq\mathbb{Z}^{64}$. Compare this with the values obtained
for $K_{0}$ and $K_{1}$ by the Künneth Theorem for tensor products:
$K_{0}(\mathcal{A}(\mathbb{F}_{2}^{4}))\cong
K_{1}(\mathcal{A}(\mathbb{F}_{2}^{4}))\cong\mathbb{Z}^{128}$.

Each of the groups in the sequences above is free Abelian, so we can use this
information to deduce from (iv) that $G_{2}=0$, $G_{3}\cong\mathbb{Z}^{64}$,
and $G_{0}\oplus G_{1}\cong\mathbb{Z}^{16}$.

######  Example 5.7.

Consider the group $\Gamma=\Gamma_{\\{1,2,3,4\\}}$, found as a result of [24],
and defined as follows:

$\Gamma_{\\{1,2,3,4\\}}:=\langle
a_{1},a_{2},a_{3},b_{1},b_{2},b_{3},c_{1},c_{2},c_{3},d_{1},d_{2},d_{3}\mid
R\rangle,$

where

$\displaystyle R:=\big{\\{}$ $\displaystyle
a_{1}b_{1}a_{3}^{-1}b_{1},a_{1}b_{1}^{-1}a_{2}^{-1}b_{3},a_{1}b_{2}a_{2}b_{2},a_{1}b_{2}^{-1}a_{3}b_{3}^{-1},$
$\displaystyle
a_{1}b_{3}a_{2}^{-1}b_{1}^{-1},a_{1}b_{3}^{-1}a_{3}b_{2}^{-1},a_{2}b_{3}a_{3}b_{3},a_{3}b_{1}a_{2}^{-1}b_{2},a_{3}b_{2}a_{2}^{-1}b_{1}$
$\displaystyle
a_{1}c_{1}a_{2}^{-1}c_{1},a_{1}c_{1}^{-1}a_{1}c_{3}^{-1},a_{1}c_{2}a_{1}^{-1}c_{2}^{-1},a_{1}c_{3}a_{3}c_{3},$
$\displaystyle
a_{2}c_{1}a_{2}c_{2}^{-1},a_{2}c_{2}a_{3}^{-1}c_{2},a_{2}c_{3}a_{2}^{-1}c_{3}^{-1},a_{3}c_{1}^{-1}a_{3}^{-1}c_{1},a_{3}c_{2}a_{3}c_{3}^{-1}$
$\displaystyle
a_{1}d_{1}a_{3}^{-1}d_{3},a_{1}d_{1}^{-1}a_{2}d_{2},a_{1}d_{2}a_{2}d_{1}^{-1},a_{1}d_{2}^{-1}a_{1}d_{3}^{-1},$
$\displaystyle
a_{1}d_{3}a_{3}^{-1}d_{1},a_{2}d_{1}a_{2}d_{3}^{-1},a_{2}d_{2}^{-1}a_{3}d_{3},a_{2}d_{3}a_{3}d_{2}^{-1},a_{3}d_{1}a_{3}d_{2}$
$\displaystyle
b_{1}c_{1}b_{3}^{-1}c_{1},b_{1}c_{1}^{-1}b_{2}^{-1}c_{3},b_{1}c_{2}b_{2}c_{2},b_{1}c_{2}^{-1}b_{3}c_{3}^{-1},$
$\displaystyle
b_{1}c_{3}b_{2}^{-1}c_{1}^{-1},b_{1}c_{3}^{-1}b_{3}c_{2}^{-1},b_{2}c_{3}b_{3}c_{3},b_{3}c_{1}b_{2}^{-1}c_{2},b_{3}c_{2}b_{2}^{-1}c_{1},$
$\displaystyle
b_{1}d_{1}b_{2}^{-1}d_{1},b_{1}d_{1}^{-1}b_{1}d_{3}^{-1},b_{1}d_{2}b_{1}^{-1}d_{2}^{-1},b_{1}d_{3}b_{3}d_{3},$
$\displaystyle
b_{2}d_{1}b_{2}d_{2}^{-1},b_{2}d_{2}b_{3}^{-1}d_{2},b_{2}d_{3}b_{2}^{-1}d_{3}^{-1},b_{3}d_{1}^{-1}b_{3}^{-1}d_{1},b_{3}d_{2}b_{3}d_{3}^{-1}$
$\displaystyle
c_{1}d_{1}c_{3}^{-1}d_{1},c_{1}d_{1}^{-1}c_{2}^{-1}d_{3},c_{1}d_{2}c_{2}d_{2},c_{1}d_{2}^{-1}c_{3}d_{3}^{-1},$
$\displaystyle
c_{1}d_{3}c_{2}^{-1}d_{1}^{-1},c_{1}d_{3}^{-1}c_{3}d_{2}^{-1},c_{2}d_{3}c_{3}d_{3},c_{3}d_{1}c_{2}^{-1}d_{2},c_{3}d_{2}c_{2}^{-1}d_{1}\big{\\}}.$

We have written a program in Python which determines whether a group is a
$4$-cube group, and if so, outputs four adjacency matrices. In this example,
$\Gamma$ is a $4$-cube group, but the adjacency matrices are very large. MAGMA
has a limit on the dimensions of the input, so we are currently exploring
other languages and ways around the bounds of the software.

##  6 Higher-rank graphs arising as double covers of cube complexes

In this section, we deduce information about the K-theory of a certain class
of $k$-rank graphs with two vertices, which arise as double covers of the cube
complexes discussed above. Whereas the $k$-rank graphs $\mathcal{G}(\Gamma)$
had vertices labelled by the $k$-cubes of $\Gamma$, these $k$-rank graphs
$\Lambda$ have vertices labelled by the vertices of $\Gamma$. For further
detail on how these graphs arise, we direct the reader to [16, §8], but we
outline the process here.

Recall the cube complex $\mathcal{M}(\Gamma)$, which has one vertex $v$, and
has as cover the product of $k$ trees by Proposition 2.9 and Example 2.11. Let
$\ell:\Gamma\rightarrow\mathbb{Z}/2$ be a labelling of the elements of
$\Gamma$. Then we obtain a cover $\tilde{\mathcal{M}}^{2}$ of
$\mathcal{M}(\Gamma)$ with vertex set $\\{v\\}\times(\mathbb{Z}/2)$, and edge
set $\Gamma\times(\mathbb{Z}/2)$. In $\tilde{\mathcal{M}}^{2}$, for a given
element $a\in\Gamma$, either $(a,0)$ and $(a,1)$ are loops based at $(v,0)$,
$(v,1)$ respectively, or $(a,0)$ goes from $(v,1)$ to $(v,0)$ and $(a,1)$ from
$(v,0)$ to $(v,1)$.

We therefore have the following construction. For some non-negative integers
$m_{1},\ldots,m_{k}$ and $n_{1},\ldots,n_{k}$, not necessarily distinct,
consider the matrices

$D_{i}:=\begin{bmatrix}2m_{i}&0\\\ 0&2m_{i}\end{bmatrix},\qquad
T_{i}:=\begin{bmatrix}0&2n_{i}\\\ 2n_{i}&0\end{bmatrix}.$

For each $i\in\\{1,\ldots,k\\}$, we let $M_{i}$ equal one of $D_{i}$ or
$T_{i}$. Provided that we have at least one $T_{i}$, we can construct a
$k$-rank graph $\Lambda$ with our choices of $M_{i}$ as incidence matrices
(Figure 7). Such $k$-rank graphs are clearly cofinal and satisfy the
Aperiodicity Condition, so their graph $C^{\star}$-algebras are uniquely
determined by their K-theory.

Our aim is to apply Evans’ result (Proposition 4.3 above) to compute the
K-theory of the corresponding $3$-rank graph algebras $\Lambda$ in each case,
for all $m_{i},n_{i}\geq 2$. To do this, we must investigate the Smith normal
forms of the boundary maps $\partial_{i}$. We do so firstly for $k=3$, in
order to illustrate the method for arbitrary $k$.

######  Lemma 6.1.

Let $k=3$, and let $\Lambda$ be one of the three possible $3$-rank graphs
constructed as above, namely one with incidence matrices

  1. (a)

$M_{1}=T_{1}$, $M_{2}=D_{2}$, $M_{3}=D_{3}$,

  2. (b)

$M_{1}=T_{1}$, $M_{2}=T_{2}$, $M_{3}=D_{3}$,

  3. (c)

$M_{1}=T_{1}$, $M_{2}=T_{2}$, $M_{3}=T_{3}$,

up to reordering. Write

$a_{i}:=\begin{cases}1-2m_{i}&\text{ whenever }M_{i}=D_{i},\\\
1-4n_{i}^{2}&\text{ whenever }M_{i}=T_{i}.\end{cases}$

Recall the matrices $\partial_{1},\partial_{2},\partial_{3}$ from Proposition
4.3, and write $I_{2},0_{2}$ to denote the $2\times 2$ identity and zero
matrix, respectively. Then the Smith normal forms of the $\partial_{i}$ are
given by

$S(\partial_{1})=S(\partial_{3})^{T}=\begin{bmatrix}1&0&0&0&0&0\\\
0&g&0&0&0&0\end{bmatrix},\quad\text{and}\quad
S(\partial_{2})=\begin{bmatrix}I_{2}&0_{2}&0_{2}\\\ 0_{2}&gI_{2}&0_{2}\\\
0_{2}&0_{2}&0_{2}\end{bmatrix},$

where $g:=\gcd(a_{1},a_{2},a_{3})$.

Figure 7: The ($1$-skeleton of the) $3$-graph with two vertices and incidence
matrices $\big{[}\begin{smallmatrix}4&0\\\ 0&4\end{smallmatrix}\big{]}$,
$\big{[}\begin{smallmatrix}0&2\\\ 2&0\end{smallmatrix}\big{]}$, and
$\big{[}\begin{smallmatrix}0&2\\\ 2&0\end{smallmatrix}\big{]}$, respectively
represented by blue, magenta, and yellow arrows.

The proof relies on an argument based on the Cauchy–Binet Theorem: for a
matrix $\partial$ of rank $r$, the product of the invariant factors of the
Smith normal form $S(\partial)$ is equal to the greatest common divisor of all
of the determinants of the $r\times r$ minors of $\partial$.

###### Proof.

Consider the matrices $\partial_{1},\partial_{2},\partial_{3}$ in each of the
three cases. In case (a), we have

$\partial_{1}=\begin{bmatrix}1&-2n_{1}&1-2m_{2}&0&1-2m_{3}&0\\\
-2n_{1}&1&0&1-2m_{2}&0&1-2m_{3}\end{bmatrix}.$

It is known that $\partial_{1}$ and $\partial_{3}^{T}$ have the same Smith
normal form, so it suffices to check $S(\partial_{1})$. The matrix
$\partial_{1}$ has rank $2$, and the non-zero determinants of its $2\times 2$
minors are given by

$\begin{gathered}1-4n_{1}^{2},\,(1-2m_{2})^{2},\,(1-2m_{3})^{2},\,1-2m_{2},\,\\\
1-2m_{3},\,2n_{1}(1-2m_{2}),\,2n_{1}(1-2m_{3}),\,(1-2m_{2})(1-2m_{3}).\end{gathered}$

Let $g$ be the greatest common divisor of these determinants. Then, after some
relatively harmless algebra, we find that
$g=\gcd(1-4n_{1}^{2},1-2m_{2},m_{2}-m_{3})$.

For $\partial_{2}$, which has rank $4$, we compute the determinants of all
$4\times 4$ minors:

$\begin{gathered}(1-4n_{1}^{2})^{2},\,(1-2n_{2})^{4},\,(1-2m_{3})^{4},\,\\\
(1-4n_{1}^{2})(1-2n_{2})^{2},\,(1-4n_{1}^{2})(1-2m_{3})^{2},\,(1-2n_{2})^{2}(1-2m_{3})^{2},\end{gathered}$

and find that the greatest common divisor of these is equal to $g^{2}$. Hence

$S(\partial_{1})=S(\partial_{3})^{T}=\begin{bmatrix}1&0&0&0&0&0\\\
0&g&0&0&0&0\end{bmatrix},\quad\text{and}\quad
S(\partial_{2})=\begin{bmatrix}I_{2}&0_{2}&0_{2}\\\ 0_{2}&gI_{2}&0_{2}\\\
0_{2}&0_{2}&0_{2}\end{bmatrix},$

where $I_{2},0_{2}$ are the $2\times 2$ identity matrix and zero matrix,
respectively. Cases (b) and (c) are shown in a similar manner. ∎

######  Proposition 6.2.

Let $k=3$, let $\Lambda$ be one of the three possible $3$-rank graphs
constructed as above, and let $g$ be the corresponding value as defined in
Lemma 6.1. Then

  1. (i)

If $g=1$, then $K_{0}(\mathcal{A}(\Lambda))\cong
K_{1}(\mathcal{A}(\Lambda))\cong 0$,

  2. (ii)

If $g\geq 2$, then
$K_{1}(\mathcal{A}(\Lambda))\cong(\mathbb{Z}/g)\oplus(\mathbb{Z}/g)$, and
$K_{0}(\mathcal{A}(\Lambda))$ is isomorphic to a group of order $g^{2}$.

###### Proof.

If $g=1$, then $\operatorname{coker}(\partial_{1})\cong\ker(\partial_{3})\cong
0$, and we may apply Corollary 4.4(i). Using the Smith normal form to
calculate the homologies, we find that $K_{0}(\mathcal{A}(\Lambda))\cong
K_{1}(\mathcal{A}(\Lambda))\cong 0$.

If $g\geq 2$, then $\operatorname{coker}(\partial_{1})\cong(\mathbb{Z}/g)$,
and $\ker(\partial_{3})\cong 0$. Once again, we apply Proposition 4.3 to
obtain the short exact sequence

$0\longrightarrow(\mathbb{Z}/35)\longrightarrow
K_{0}(\mathcal{A}(\Lambda))\longrightarrow\ker(\partial_{2})/\operatorname{im}(\partial_{1})\longrightarrow
0,$

and the isomorphism
$K_{1}(\mathcal{A}(\Lambda))\cong\ker(\partial_{1})/\operatorname{im}(\partial_{2})$.
It is well known that, when the differentials are finitely-generated, the
homologies of a chain complex can be computed via the formula

$\ker(\partial_{i})/\operatorname{im}(\partial_{i+1})\cong\mathbb{Z}^{c-r-s}\oplus\bigoplus_{j=1}^{r}(\mathbb{Z}/a_{j}),$

where $c$ is the number of columns of $\partial_{i}$,
$r:=\operatorname{rk}(\partial_{i+1})$, $s:=\operatorname{rk}(\partial_{i})$,
and $a_{j}$ are the non-zero entries of $S(\partial_{i+1})$. In our example,
this gives

$\ker(\partial_{1})/\operatorname{im}(\partial_{2})\cong(\mathbb{Z}/g)\oplus(\mathbb{Z}/g),\qquad\text{and}\qquad\ker(\partial_{2})/\operatorname{im}(\partial_{3})\cong(\mathbb{Z}/g),$

and the result follows. ∎

######  Proposition 6.3.

Let $\Lambda$ be one of the $k$ possible $k$-rank graphs constructed as above,
with incidence matrices $M_{1},\ldots,M_{k}$. Write

$a_{i}:=\begin{cases}1-2m_{i}&\text{ whenever }M_{i}=D_{i},\\\
1-4n_{i}^{2}&\text{ whenever }M_{i}=T_{i},\end{cases}$

and $g:=\gcd(a_{1},\ldots,a_{k})$. Recall the matrices
$\partial_{1},\ldots,\partial_{k}$ from Theorem 4.2, and write $I_{s}$ to
denote the $s\times s$ identity matrix. Then the Smith normal forms of the
matrices $\partial_{i}$ are given by

$S(\partial_{1})=S(\partial_{k})^{T}=\begin{bmatrix}1&0&\mathbf{0}\\\
0&g&\mathbf{0}\end{bmatrix}_{2\times 2k},$

and for $2\leq i\leq k-1$, the rank of $\partial_{i}$ is equal to
$2R:=2{k-1\choose i-1}$, and

$S(\partial_{i})=\begin{bmatrix}I_{R}&0&\mathbf{0}\\\ 0&gI_{R}&\mathbf{0}\\\
\mathbf{0}&\mathbf{0}&\mathbf{0}\end{bmatrix}_{2{k\choose i-1}\times
2{k\choose i}}.$

###### Proof.

The main argument of the proof is that, knowing the non-trivial invariant
factor of $S(\partial_{1})=S(\partial_{k})^{T}$, along with the rank of each
$\partial_{i}$, suffices to completely determine the Smith normal form of
these matrices.

Firstly, we compute $S(\partial_{1})=S(\partial_{k})^{T}$ in the same manner
as Lemma 6.1. Secondly, we determine the shape (and hence the rank) of the
matrices $\partial_{2},\ldots,\partial_{k-1}$. Recall from Theorem 4.2 the
sets

$N_{l}:=\begin{cases}\big{\\{}\boldsymbol{\mu}:=(\mu_{1},\ldots,\mu_{l})\in\\{1,\ldots,k\\}^{l}\mid\mu_{1}<\cdots<\mu_{l}\big{\\}}&\text{if
}l\in\\{1,\ldots,k\\},\\\ \\{*\\}&\text{if }l=0,\\\
\emptyset&\text{otherwise}.\end{cases}$

From now on, we treat $\partial_{i}$ exclusively as block matrices, with shape
$|N_{i-1}|\times|N_{i}|$, and which comprise $2\times 2$ blocks which are
either zero or $\pm(I_{2}-M_{j}^{T})$ for $j\in\\{1,\ldots,k\\}$. Notice that
$N_{i}$ has $(\begin{smallmatrix}k\\\ i\end{smallmatrix})$ elements, each a
strictly-increasing tuple of length $i$ whose entries belong to
$\\{1,\ldots,k\\}$. Thus we consider $N_{i}$ and $N_{i-1}$ as ordered sets,
with $\boldsymbol{\mu}>\boldsymbol{\mu}^{\prime}$ if there is some $r$ such
that $\mu_{r}>\mu_{r}^{\prime}$ and $\mu_{t}=\mu_{t}^{\prime}$ for $t<r$. We
label the (blocks of the) rows and columns of each matrix $\partial_{i}$ with
the elements, in order, of $N_{i-1}$ and $N_{i}$ respectively, writing
$\boldsymbol{\mu}(p)\in N_{i-1}$ for the label of row $p$, and
$\boldsymbol{\nu}(q)\in N_{i}$ for the label of column $q$.

Denote by $\partial_{i}(p,q)$ a block of $\partial_{i}$ in position $(p,q)$.
Then $\partial_{i}(p,q)$ is either $\pm(I_{2}-M_{j}^{T})$ for some $j$, or the
$2\times 2$ zero matrix. It is non-zero if and only if $\boldsymbol{\mu}(p)$
can be obtained by deleting one element, say $\nu(q)_{t}$, of
$\boldsymbol{\nu}(q)$. Furthermore, the value of $\nu(q)_{t}$ determines the
index of the matrix $M_{j}$, and the position $t$ determines the sign of
$\partial_{i}(p,q)$ as follows:

$\partial_{i}(p,q)=\begin{cases}I_{2}-M_{j}^{T}&\text{if $t$ is odd,}\\\
M_{j}^{T}-I_{2}&\text{if $t$ is even.}\end{cases}$

Using this interpretation of the blocks of $\partial_{i}$, we conclude that
the number of non-zero blocks is equal to
$|N_{i-1}|\cdot(k-(i-1))=|N_{i}|\cdot(k-i)$.

It remains to deduce the rank of each matrix $\partial_{i}$. Firstly, observe
that

$\operatorname{rk}(\partial_{i})\leq\min 2\bigg{\\{}{k\choose i-1},{k\choose
i}\bigg{\\}}.$

By construction of $\partial_{i}$, there exists a $2(\begin{smallmatrix}k-1\\\
i-1\end{smallmatrix})\times 2(\begin{smallmatrix}k-1\\\ i-1\end{smallmatrix})$
minor of $\partial_{i}$, with block-elements $I_{2}-M_{1}^{T}$ on the
diagonal, obtained by considering those column-blocks labelled by elements
$\boldsymbol{\nu}\in N_{i}$ with $\nu_{1}=1$, and row-blocks labelled by
elements $\boldsymbol{\mu}\in N_{i-1}$ with $\mu_{1}\neq 1$. There are

${k\choose i}-{k-1\choose i}={k-1\choose i-1},\quad\text{and}\quad{k\choose
i-1}-{k-1\choose i-2}={k-1\choose i-1}$

such column- and row-blocks, respectively, and so

$\operatorname{rk}(\partial_{i})\geq 2{k-1\choose i-1}.$

We claim that this is in fact an equality, and we display here an outline of
the proof. Suppose that there exists a non-vanishing minor $A$ which is larger
than the one above. Given the number of non-zero blocks, and the fact that
minors are square matrices, the diagonal (or when relevant, the anti-diagonal)
of $A$ must not contain any zero blocks, else $A$ vanishes. Using elementary
row operations, we take the last $(\begin{smallmatrix}k-1\\\
i-1\end{smallmatrix})$ row-blocks of $\partial_{i}$ and move them to the top,
such that the elements $\boldsymbol{\mu}\in N_{i-1}$ with $\mu_{1}=1$ now
label the rows at the bottom of $\partial_{i}$. Hence we must be able to find
a column labelled by an element $\boldsymbol{\nu}\in N_{i}$ such that
$\nu_{1}\neq 1$, and from which we can find the $\boldsymbol{\mu}$, but this
is a contradiction.

Indeed, by some careful considerations, it turns out that regardless of the
row operations performed, we will arrive at such a contradiction. Thus
$\operatorname{rk}(\partial_{i})=2{k-1\choose i-1}$, and from this the result
is readily verified. ∎

######  Proposition 6.4 ($k=4$).

Let $\Lambda$ be one of the four possible $4$-rank graphs constructed as
above, and let $g$ be the corresponding value as defined in Lemma 6.3. Then:

  1. (i)

If $g=1$, then $K_{0}(\mathcal{A}(\Lambda))\cong
K_{1}(\mathcal{A}(\Lambda))\cong 0$,

  2. (ii)

If $g\geq 2$, then we can find short exact sequences

    1. (a)

$0\longrightarrow(\mathbb{Z}/g)/G_{0}\longrightarrow
K_{0}(\mathcal{A}(\Lambda))\longrightarrow(\mathbb{Z}/g)^{3}\longrightarrow
0$,

    2. (b)

$0\longrightarrow(\mathbb{Z}/g)^{3}/G_{2}\longrightarrow
K_{1}(\mathcal{A}(\Lambda))\longrightarrow
G_{3}\subseteq(\mathbb{Z}/g)\longrightarrow 0$,

where $G_{0}$ and $G_{2}$ are as in Proposition 4.5.

###### Proof.

Firstly, if $g=1$, it follows that $\ker(\partial_{4})=0$ and we may apply
Corollary 4.6(i). But $H_{3}(\mathcal{D}_{4})$, $H_{2}(\mathcal{D}_{4})$, and
$H_{1}(\mathcal{D}_{4})$ are all trivial, and so the K-theory of
$\mathcal{A}(\Lambda)$ is trivial.

If $g\geq 2$, then $\ker(\partial_{4})=0$,
$\operatorname{coker}(\partial_{1})\cong(\mathbb{Z}/g)$, and Proposition 4.5
gives us:

  1. (a)

$0\longrightarrow\operatorname{coker}(\partial_{1})/G_{0}\longrightarrow
K_{0}(\mathcal{A}(\Lambda))\longrightarrow\ker(\partial_{2})/\operatorname{im}(\partial_{3})\longrightarrow
0$,

  2. (b)

$0\longrightarrow\dfrac{\ker(\partial_{1})/\operatorname{im}(\partial_{2})}{G_{2}}\longrightarrow
K_{1}(\mathcal{A}(\Lambda))\longrightarrow G_{3}\longrightarrow 0$.

Again, we use the Smith normal forms to compute the homologies
$\ker(\partial_{i})/\operatorname{im}(\partial_{i+1})$ of the chain complex
$\mathcal{D}_{4}$, which reveal that

$\ker(\partial_{1})/\operatorname{im}(\partial_{2})\cong\ker(\partial_{2})/\operatorname{im}(\partial_{3})\cong(\mathbb{Z}/g)^{3},\qquad\text{and}\qquad\ker(\partial_{3})/\operatorname{im}(\partial_{4})\cong(\mathbb{Z}/g),$

and the above sequences reduce to those we desire. ∎

## Acknowledgments

The authors would like to thank David Evans and Gwion Evans for discussing the
connections between the theory of buildings, higher-rank graphs, and the
K-theory of their corresponding $C^{\star}$-algebras at the Newton Institute,
Cambridge in Spring 2017.

The authors express their gratitude to Newcastle University for providing an
excellent research environment, and to the EPSRC for funding part of this
research.

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School of Mathematics, Statistics and Physics,
Newcastle University, Newcastle upon Tyne NE1 7RU, UK

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