New York Journal of Mathematics
New York J. Math. 14 (2008) 1?31.
On the Ktheory of higher rank graph
C
?
algebras
D. Gwion Evans
Abstract. Given a rowfinite kgraph ? with no sources we investigate
the Ktheory of the higher rank graph C
?
algebra, C
?
(?). When k =
2 we are able to give explicit formulae to calculate the Kgroups of
C
?
(?). The Kgroups of C
?
(?) for k>2 can be calculated under
certain circumstances and we consider the case k =3. Weprovethat
for arbitrary k, the torsionfree rank of K
0
(C
?
(?)) and K
1
(C
?
(?)) are
equal when C
?
(?) is unital, and for k = 2 we determine the position of
the class of the unit of C
?
(?) in K
0
(C
?
(?)).
Contents
1. Introduction 1
2. Preliminaries 3
3. The Kgroups of kgraph C
?
algebras 6
4. The Kgroups of unital kgraph C
?
algebras 21
5. Applications and examples 24
References 30
1. Introduction
In [31] Spielberg realised that a crossed product algebra C(?) multicloseright ?, where
? is the boundary of a certain tree and ? is a free group, is isomorphic to
a Cuntz?Krieger algebra [6, 5]. Noticing that such a tree may be regarded
as an a?ne building of type
tildewide
A
1
, Robertson and Steger studied the situation
when a group ? acts simply transitively on the vertices of an a?ne building
of type
tildewide
A
2
with boundary ? [23]. They found that the corresponding crossed
Received May 12, 2004. Revised December 18, 2007.
Mathematics Subject Classification. Primary 46L80; Secondary 46L35.
Key words and phrases. Ktheory, C
?
algebra, kgraphs, graph algebra.
Research supported by the European Union Research Training Network in Quantum
Spaces ? Noncommutative Geometry.
ISSN 10769803/08
1
2 D. Gwion Evans
product algebra C(?)multicloseright? is generated by two Cuntz?Krieger algebras. This
led them to define a C
?
algebra A via a finite sequence of finite 01 matrices
(i.e., matrices with entries in {0,1}) M
1
,...,M
r
satisfying certain condi
tions (H0)?(H3), such that A is generated by r Cuntz?Krieger algebras, one
for each M
1
,...,M
r
. Accordingly they named their algebras higher rank
Cuntz?Krieger algebras, the rank being r.
Kumjian and Pask [13] noticed that Robertson and Steger had constructed
their algebras from a set, W of (higher rank) words in a finite alphabet A
? the common index set of the 01 matrices ? and realised that W could
be thought of as a special case of a generalised directed graph ? a higher
rank graph or kgraph, where k is the rank. Subsequently, Kumjian and
Pask associated a C
?
algebra, C
?
(?) to a higher rank graph ? and showed
that A
?
= C
?
(W)[13, Corollary 3.5 (ii)]. Moreover, they derived a number
of results elucidating the structure of higher rank graph C
?
algebras. They
show in [13, Theorem 5.5] that a simple, purely infinite kgraph C
?
algebra
C
?
(?) may be classified by its Ktheory. This is a consequence of C
?
(?)
satsifying the hypotheses of the Kirchberg?Phillips classification theorem
([12, 18]). Moreover, criteria on the underlying kgraph ? were found that
decided when C
?
(?) was simple and purely infinite (see [13,Proposition
4.8, Proposition 4.9] and [28, Proposition 8.8]). Thus a step towards the
classification of kgraph C
?
algebras is the computation of their Kgroups.
In [25, Proposition 4.1] Robertson and Steger proved that the Kgroups
of a rank 2 Cuntz?Krieger algebra is given in terms of the homology of a
certain chain complex, whoose di?erentials are defined in terms of M
1
and
M
2
. Their proof relied on the fact that a rank 2 Cuntz?Krieger algebra is
stably isomorphic to a crossed product of an AFalgebra by Z
2
. We will
generalise their method to provide explicit formulae for the Kgroups of
2graph C
?
algebras and to gain information on the Kgroups of kgraph
C
?
algebras for k>2.
The rest of this paper is organised as follows. We begin in ?2 by recalling
the fundamental definitions relating to higher rank graphs and their C
?

algebras we will need from [13].
In ?3 we use the fact that the C
?
algebra of a rowfinite kgraph ? with
no sources is stably isomorphic to a crossed product of an AF algebra, B,by
Z
k
([13, Theorem 5.5]) to apply a theorem of Kasparov [11,6.10Theorem]
to deduce that there is a homological spectral sequence ([34, Chapter 5])
converging to K
?
(C
?
(?)) with initial term given by E
2
p,q
?
= H
p
(Z
k
,K
q
(B))
(see [3] for the definition of the homology of a group G with coe?cients in
aleftGmodule M,denotedbyH
?
(G, M)). We will see that it su?ces to
compute H
?
(Z
k
,K
0
(B)). It transpires that H
?
(Z
k
,K
0
(B)) is given by the
socalled vertex matrices of ?. These are matrices with nonnegative integer
entries that encode the structure of the category ?. Next we assemble the
results of ?3 and state them in our main theorem, Theorem 3.15.Wethen
specialise to the cases k =2andk =3. Fork = 2 a complete description
On the Ktheory of higher rank graph C
?
algebras 3
of the Kgroups in terms of the vertex matrices can be given. For k =3
we illustrate how Theorem 3.15 can be used to give a description of the
Kgroups of 3graph C
?
algebras under stronger hypotheses.
In section ?4 we consider the Ktheory of unital kgraph C
?
algebras. We
show that the torsionfree rank of K
0
(C
?
(?)) is equal to that of K
1
(C
?
(?))
when C
?
(?) is unital and give formulae for the torsionfree rank and torsion
parts of the Kgroups of 2graph C
?
algebras.
We conclude with ?5, in which we consider some immediate applications to
the classification of kgraph C
?
algebras by means of the Kirchberg?Phillips
classification theorem. We also consider some simple examples of Kgroup
calculations using the results derived in the previous sections.
Acknowledgements. This paper was written while the author was an Eu
ropean Union Network in Quantum Spaces ? Noncommutative Geometry
funded postdoc at the University of Copenhagen. The paper develops a
part of the author?s Ph.D. thesis, which was written under the supervi
sion of David E. Evans at Cardi? University. We would like to take this
opportunity to thank David for his guidance and support, and Johannes
Kellendonk and Ryszard Nest for enlightening discussions on homological
algebra. We would also like to express our gratitude to the members of the
operator algebras groups in both universities, for maintaining stimulating
environments for research. Finally, we thank the referee for their careful
reading, comments and suggestions, which helped to clarify the exposition
of the paper.
2. Preliminaries
By the usual slight abuse of notation we shall let the set of morphisms
of a small category ? be denoted by ? and identify an object of ? with its
corresponding identity morphism. Also note that a monoid M (and hence
a group) can be considered as a category with one object and morphism
set equal to M, with composition given by multiplication in the monoid.
For convenience of notation, we shall denote a monoid and its associated
category by the same symbol.
The following notation will be used throughout this paper. We let N
denote the abelian monoid of nonnegative integers and we let Z be the
group of integers. For a positive integer k,weletN
k
be the product monoid
viewed as a category. Similarly, we let Z
k
be the product group viewed,
where appropriate, as a category. Let {e
i
}
k
i=1
be the canonical generators of
N
k
as a monoid and Z
k
as a group. Moreover, we choose to endow N
k
and
Z
k
with the coordinatewise order induced by the usual order on N and Z,
i.e., for all m,n ? Z
k
m ? n ?? m ? n ? N
k
.
We will denote by K(H)theC
?
algebra of compact operators on a Hilbert
space H. Where the Hilbert space H is separable and of infinite dimension
we write K for K(H).
4 D. Gwion Evans
The concept of a higher rank graph or kgraph (k =1,2,... being the
rank) was introduced by A. Kumjian and D. Pask in [13]. We recall their
definition of a kgraph.
Definition 2.1 ([13, Definitions 1.1]). A kgraph (rank k graph or higher
rank graph) (?,d) consists of a countable small category ? (with range and
source maps r and s respectively) together with a functor d :??? N
k
satisfying the factorisation property: for every ? ? ?andm,n ? N
k
with
d(?)=m + n, there are unique elements ?,? ? ? such that ? = ?? and
d(?)=m, d(?)=n.Forn ? N
k
and v ? ?
0
we write ?
n
:= d
?1
(n), ?(v):=
r
?1
(v)and?
n
(v):={? ? ?
n
r(?)=v}.
Definition 2.2 ([13, Definitions 1.4]). A kgraph ? is rowfinite if for each
n ? N
k
and v ? ?
0
the set ?
n
(v) is finite. We say that ? has no sources if
?
n
(v) negationslash= ? for all v ? ?
0
and n ? N
k
.
Unless stated otherwise, we will assume that each higher rank graph in
this paper is rowfinite with no sources. Furthermore, we shall denote such
a generic higher rank graph by (?,d) (or more succinctly ? with the under
standing that the degree functor will be denoted by d).
We refer to [15] as an appropriate reference on category theory. There is
no need for a detailed knowledge of category theory as we will be interested
in the combinatorial graphlike nature of higher rank graphs. As the name
suggests a higher rank graph can be thought of as a higher rank analogue
of a directed graph. Indeed, every 1graph is isomorphic (in the natural
sense) to the category of finite paths of a directed graph ([13,Example
1.3]). By [13, Remarks 1.2] ?
0
is the set of identity morphisms of ?. Indeed
it is fruitful to view ?
0
as a set of vertices and ? as a set of (coloured)
paths with composition in ? being concatention of paths. This viewpoint is
discussed further in [7, 20].
In the sequel we will use the following higher rank graph constructions
devised by Kumjian and Pask. For further examples of kgraphs see, for
example, [13, 20, 21, 29].
Examples 2.3. (1) Let ?
k
be the category with morphism set equal to
{(m,n) ? Z
k
? Z
k
 m ? n}, object set equal to {(m,m)  m ? Z
k
},
structure maps defined by r(m,n)=m, s(m,n)=n and composition
defined by (m,l)(l,n)=(m,n) for all m,l,n ? Z
k
such that m ? l ?
n. One may define a degree functor d :?
k
?? N
k
by d(m,n)=n?m
so that (?
k
,d)isakgraph. Furthermore, it is straightforward to
check that (?
k
,d) is rowfinite and has no sources.
(2) The product higher rank graph ([13, Proposition 1.8]): Let (?
1
,d
1
)
and (?
2
,d
2
)berankk
1
and k
2
graphs respectively. Then their product
higher rank graph (?
1
??
2
,d
1
?d
2
)isa(k
1
+k
2
)graph, where ?
1
??
2
is the product category and the degree functor d
1
?d
2
:?
1
??
2
??
N
k
1
+k
2
is given by d
1
? d
2
(?
1
,?
2
)=(d
1
(?
1
),d
2
(?
2
)) ? N
k
1
? N
k
2
for
?
1
? ?
1
and ?
2
? ?
2
.
On the Ktheory of higher rank graph C
?
algebras 5
(3) The skewproduct higher rank graph ([13, Definition 5.1]): Given a
countable group G,akgraph ? and a functor c :??? G,theskew
product kgraph G?
c
? consists of a category with object set identified
with G ? ?
0
and morphism set identified with G ? ?. The structure
maps are given by: s(g,?)=(gc(?),s(?)) and r(g,?)=(g,r(?)).
If s(?)=r(?)then(g,?)and(gc(?),?) are composable in G ?
c
?
and (g,?)(gc(?),?)=(g,??). Thedegreemapisgivenbyd(g,?)=
d(?). Furthermore, G acts freely on G ?
c
?byg ? (h,?) mapsto? (gh,?)
for all g,h ? G and ? ? ?(see[13, Remark 5.6] and its preceding
paragraph).
To each rowfinite kgraph with no sources, Kumjian and Pask associated
an unique C
?
algebra in the following way.
Definition 2.4. ([13, Definitions 1.5]) Let ? be a rowfinite kgraph with
no sources. Then C
?
(?) is defined to be the universal C
?
algebra generated
by a family {s
?
 ? ? ?} of partial isometries satisfying:
(i) {s
v
 v ? ?
0
} is a family of mutually orthogonal projections.
(ii) s
??
= s
?
s
?
for all ?,? ? ? such that s(?)=r(?).
(iii) s
?
?
s
?
= s
s(?)
for all ? ? ?.
(iv) for all v ? ?
0
and n ? N
k
we have s
v
=
summationtext
???
n
(v)
s
?
s
?
?
.
For ? ? ?, define p
?
:= s
?
s
?
?
. A family of partial isometries satisfying
(i)?(iv) above is a called a ?representation of ?.
We consider the following C
?
algebras associated with the constructions
notedinExamples2.3, which will be useful in the sequel.
Examples 2.5. (1) Let ?
k
be the rowfinite kgraph with no sources
defined in Examples 2.3.1.Lete
m,n
:= s
(m,q)
s
?
(n,q)
, m,n ? Z
k
, q :=
sup{m,n}.Since{e
m,n
 m,n ? Z
k
} is a complete system of matrix
units, we have C
?
(?
k
)
?
= K(lscript
2
(Z
k
)) (cf. [13, Examples 1.7 (ii)]).
(2) Let (?
i
,d
i
) be a rowfinite k
i
graph with no sources for i =1,2. Then
C
?
(?
1
? ?
2
)
?
= C
?
(?
1
)? C
?
(?
2
)
by [13, Corollary 3.5 (iv)].
1
(3) Let G be a countable group, ? a rowfinite kgraph with no sources
and c :??? G a functor. Then the action of G on G?
c
? described in
Examples 2.3.3 induces an action ? : G ?? Aut(C
?
(G?
c
?)) such that
?
g
(s
(h,?)
)=s
(gh,?)
.FurthermoreC
?
(G?
c
?)multicloseright
?
G
?
= C
?
(?)?K(lscript
2
(G))
[13, Theorem 5.7].
1
The C
?
algebra of a rowfinite higher rank graph with no sources is nuclear [13,The
orem 5.5].
6 D. Gwion Evans
3. The Kgroups of kgraph C
?
algebras
For the remainder of this paper we shall denote by B
?
(or simply B when
there is no ambiguity) the C
?
algebra of the skewproduct of a rowfinite
kgraph (?,d), with no sources, by Z
k
via the degree functor regarded as
a functor into Z
k
, i.e., B := C
?
(Z
k
?
d
?), and by ? the action of Z
k
on
B as described in Examples 2.5.3.Notethatby[13, Corollary 5.3 and
Theorem 5.5] and Takesaki?Takai duality [32](or[13, Theorem 5.7], cf.
Examples 2.5.3), C
?
(?) is stably isomorphic to the crossed product of an
AFalgebra, B,byZ
k
, i.e., B multicloseright
?
Z
k
?
= C
?
(?) ? K. Therefore
K
0
(C
?
(?))
?
= K
0
(B multicloseright
?
Z
k
).
It will be useful for us in the sequel to have an explicit description of
how an isomorphism K
0
(C
?
(?)) ?? K
0
(B multicloseright
?
Z
k
)actsontheK
0
class of
a canonical projection p
v
,v? ?
0
in C
?
(?). To this end, we prefer to
investigate how the isomorphism acts by using an alternative approach to
that outlined above. This we do below by using a standard technique in k
graph C
?
algebra theory, namely by utilising the gaugeinvariant uniqueness
theorem for kgraph C
?
algebras [13, Theorem 3.4].
Theorem 3.1. Let ? be a rowfinite kgraph with no sources. Then there
exists a group isomorphism ?
0
: K
0
(B multicloseright
?
Z
k
) ?? K
0
(C
?
(?)) such that
?
0
([i
B
(p
(0,v)
)]) = [p
v
] for all v ? ?
0
(where we adopt the notation used in
[19] for crossed product C
?
algebras).
Proof. Let (Bmulticloseright
?
Z
k
,i
B
,i
Z
k) be a crossed product for the dynamical system
(B, Z
k
,?) in the sense of [19]. One checks that
{t
(?,(m,n))
 (?,(m,n)) ? ? ? ?}
is a ?representation of ???, where ? := ?
k
(see Examples 2.3.1), and for
(?,(m,n)) ? ? ? ?welet
t
(?,(m,n))
:= i
B
(s
(m,?)
)i
Z
k(m + d(?) ? n).
Moreover, C
?
(t
?
 ? ? ? ? ?) = B multicloseright
?
Z
k
. Thus by the universal property
of C
?
(? ? ?), there exists a ?homomorphism ? : C
?
(? ? ?) ?? B multicloseright
?
Z
k
such that ?(s
?
)=t
?
for all ? ? ? ? ?. Let ? : T
k
?? Aut(B)denotethe
canonical gauge action on B (see [13, ?3]) and let
?
? : T
k
?? Aut(B multicloseright
?
Z
k
)
denote the dual action of ?. There exists an action tildewide? of T
k
on B multicloseright
?
Z
k
such
that i
B
?
z
= tildewide?
z
i
B
for all z ? T
k
. It is clear that setting ?
(z
1
,z
2
)
:= tildewide?
z
1
z
2
?
?
z
?1
2
for all (z
1
,z
2
) ? T
k
?T
k
defines an action ? of T
2k
on B multicloseright
?
Z
k
.Moreover,it
satisfies ??
?
z
= ?
z
? for all z ? T
2k
where ?
?
is the canonical gauge action on
C
?
(???). Clearly ?(p
v
) = 0 for all v ? ???, hence by the gaugeinvariant
uniqueness theorem [13, Theorem 3.4] we see that ? : C
?
(???) ?? Bmulticloseright
?
Z
k
is a ?isomorphism.
On the Ktheory of higher rank graph C
?
algebras 7
For the zero element 0 ? Z
k
,weseethatp
(0,0)
is a minimal projection in
C
?
(?)
?
= K (cf. Examples 2.5.1). Therefore, the homomorphism given by
x mapsto? x ? p
(0,0)
induces an isomorphism K
0
(C
?
(?))
?
= K
0
(C
?
(?) ? C
?
(?)).
The latter is isomorphic to K
0
(C
?
(???)) (see 2.5.2)andK
0
(Bmulticloseright
?
Z
k
). Let
?:K
0
(C
?
(?)) ?? K
0
(B multicloseright
?
Z
k
) be the composition of the preceding group
isomorphisms. Then it follows easily that ?([p
v
]) = [i
B
(p
(0,v)
)]. Setting
?
0
:= ?
?1
completes the proof. square
Therefore we may apply [11, 6.10 Theorem] to describe the Kgroups of
C
?
(?) by means of a homological spectral sequence with initial term given
by H
p
(Z
k
,K
q
(B)), i.e., the homology of the group Z
k
with coe?cients in the
left Z
k
module K
q
(B)[3], where the Z
k
action is given by e
i
?m = K
0
(?
e
i
)(m)
for i =1,...,k, (cf. the proof of [25, Proposition 4.1]). First, we recall the
definition of a homology spectral sequence from [34, ?5] and the notion of
convergence (see also [14]).
2
Definition 3.2. A homology spectral sequence (starting at E
a
) consists of
the following data:
(1) a family {E
r
p,q
} of modules defined for all integers p,q and r ? a,
(2) maps d
r
pq
: E
r
p,q
?? E
r
p?r,q+r?1
that are di?erentials in the sense that
d
r
p?r,q+r?1
d
pq
=0,
(3) isomorphisms E
r+1
pq
?? ker(d
r
pq
)/im(d
r
p+r,q?r+1
).
We will denote the above data by {(E
r
,d
r
)}.Thetotal degree of the term
E
r
pq
is n = p + q. The homology spectral sequence is said to be bounded if
for each n there are only finitely many nonzero terms of total degree n in
{E
r
pq
}, in which case, for each p and q there is an r
0
such that E
r
pq
?
= E
r+1
pq
for all r ? r
0
.WewriteE
?
pq
for this stable value of E
r
pq
.
We say that a bounded spectral sequence converges to K
?
if we are given
a family of modules {K
n
}, each having a finite filtration
0=F
s
(K
n
) ?????F
p?1
(K
n
) ? F
p
(K
n
) ? F
p+1
(K
n
) ?????F
t
(K
n
)=K
n
,
and we are given isomorphisms E
?
pq
?? F
p
(K
p+q
)/F
p?1
(K
p+q
).
Lemma 3.3 (cf. [25, Proposition 4.1]). There exists a spectral sequence
{(E
r
,d
r
)} converging to K
?
(C
?
(?)) := {K
n
}
n?Z
where
K
n
:=
braceleftBigg
K
0
(C
?
(?)) if n is even,
K
1
(C
?
(?)) if n is odd.
Moreover, for p,q ? Z,
E
2
p,q
?
=
braceleftBigg
H
p
(Z
k
,K
0
(B)) if p ?{0,1,...,k} and q is even,
0 otherwise,
2
The reader will notice that the definition is presented in a less general form than in
[34, ?5], but is adequate for our purposes.
8 D. Gwion Evans
E
?
p,q
?
= E
k+1
p,q
and E
k+1
p,q
=0if p ? Z\{0,1,...,k} or q is odd.
Proof. The first assertion follows from an application of [11,6.10Theorem]
to Bmulticloseright
?
Z
k
,whichis?isomorphic to C
?
(?)?K by the proof of Theorem 3.1,
after noting that K
?
(B multicloseright
?
Z
k
) coincides with its ??part? since the Baum?
Connes Conjecture with coe?cients in an arbitrary C
?
algebra is true for
the amenable group Z
k
for all k ? 1.
By the proof of [11, 6.10 Theorem], K
?
(B multicloseright
?
Z
k
)
?
= K
?
(D)forsomeC
?

algebra D which has a finite filtration by ideals: 0 ? D
0
? D
1
?????D
k
=
D since the dimension of the universal covering space of the classifying space
of Z
k
is k.
Here, we are considering the spectral sequence {(E
r
,d
r
)} in homology
K
?
associated with the finite filtration 0 ? D
0
? D
1
? ??? ? D
k
= D
of D ([27, ?6]) which has E
1
p,q
= K
(p+q mod 2)
(D
p
/D
p?1
)whereD
n
=0
for n<0andD
n
= D for n ? k. It follows easily that E
r
p,q
=0for
p ? Z\{0,1,...,k}, for all q ? Z and for all r ? 1andE
?
p,q
?
= E
k+1
p,q
(see
also [27, Theorem 2.1]). This combined with Kasparov?s calculation in the
proof of [11, 6.10 Theorem], giving E
2
p,q
?
= H
p
(Z
k
,K
q
(B)), along with the
observation that K
q
(B)=0foroddq,sinceB is an AFalgebra, proves the
second assertion. square
Now we will compute H
?
(Z
k
,K
0
(B)) in terms of the combinatorial data
encoded in ?. First, let us examine the structure of B, and hence K
0
(B),
in a little more detail.
Lemma 3.4. Let ? be a rowfinite kgraph with no sources. Then
B =
uniondisplay
n?Z
k
B
n
,
where
B
n
= span{s
?
s
?
?
 ?,? ? Z
k
?
d
?,s(?)=s(?)=(n,v) for some v ? ?
0
}
?
=
circleplusdisplay
v??
0
B
n
(v),
where
B
n
(v):=span{s
?
s
?
?
 ?,? ? Z
k
?
d
?,s(?)=s(?)=(n,v)}
?
= K(lscript
2
(s
?1
(v)))
for all v ? ?
0
and n ? Z
k
.
Proof. Follows immediately from the proofs of [13, Lemma 5.4, Theorem
5.5] and the observation that for all n ? Z
k
and v ? ?
0
, s
?1
((n,v)) ?
Z
k
?
d
? may be identified with s
?1
(v) ? ?via(n ? d(?),?) mapsto? ? for all
? ? s
?1
(v). square
On the Ktheory of higher rank graph C
?
algebras 9
Definition 3.5. Let Z?
0
be the group of all maps from ?
0
into Z that have
finite support under pointwise addition. For each u ? ?
0
,wedenoteby?
u
the element of Z?
0
defined by ?
u
(v)=?
u,v
(the Kronecker delta) for all
v ? ?
0
.NotethatZ?
0
is a free abelian group with free set of generators
{?
u
 u ? ?
0
}.
Definition 3.6 (cf. [13, ?6]). Define the vertex matrices of ?, M
i
,bythe
following. For u,v ? ?
0
and i =1,2,...,k,
M
i
(u,v):={? ? ?
e
i
r(?)=u,s(?)=v}.
Remarks 3.7. By the factorisation property, the vertex matrices of a k
graph pairwise commute [13, ?6].
Convention 3.8. Given a matrix M with integer entries and index set ?
0
,
by slight abuse of notation we shall, on occasion, regard M as the group
endomorphism Z?
0
?? Z?
0
, defined in the natural way as
(Mf)(u)=
summationdisplay
v??
0
M(u,v)f(v) for all u ? ?
0
,f? Z?
0
.
Lemma 3.9 (cf. [25] Lemma 4.5 and [17] Proposition 4.1.2). For all n,m ?
Z
k
such that m ? n,letA
m
:= Z?
0
. Moreover, define homomorphisms
j
nm
: A
m
?? A
n
by j
mm
(f)=f, j
m+e
i
,m
:= M
t
i
for all f ? A
m
,u? ?
0
,i?
{1,...,k},andj
m+e
i
+e
j
,m
:= j
m+e
i
+e
j
,m+e
i
j
m+e
i
,m
for all j ?{1,...,k}.
Then (A
m
;j
nm
) is a direct system of groups and
K
0
(B)
?
= A := lim
??
(A
m
;j
nm
).
Proof. It follows from Remarks 3.7 that the connecting homomorphisms
are welldefined and that (A
m
;j
nm
) is a direct system. From Lemma 3.4 we
deduce that
K
0
(B)
?
= lim
??
(K
0
(B
n
);K
0
(?
n,m
)),
where, for m,n ? Z
k
with m ? n, ?
n,m
: B
m
?? B
n
are the inclu
sion maps [33, Proposition 6.2.9]. We also deduce that for n ? Z
k
and
v ? ?
0
, K
0
(B
n
(v)) is isomorphic to Z and is generated by the equivalence
class comprising all minimal projections in B
n
(v), of which p
?
is a member
for any ? ? s
?1
(n,v). Therefore, K
0
(B
n
)
?
=
circleplustext
v??
0
K
0
(B
n
(v)) is gener
ated by {[p
(n,v)
]
n
 v ? ?
0
},where[?]
n
denotes the equivalence classes
of K
0
(B
n
) for all n ? Z
k
.Thusthemap?
n
: A
n
?? K
0
(B
n
)givenby
f mapsto?
summationtext
u??
0
f(u)[p
(n,u)
]
n
is a group isomorphism for all n ? Z
k
.
The embedding ?
n,m
: B
m
?? B
n
is given by
?
n,m
(s
(m?d(?),?)
s
?
(m?d(?),?)
)=
summationdisplay
???
n?m
(s(?))
s
(m?d(?),??)
s
?
(m?d(?),??)
10 D. Gwion Evans
for all ?,? ? ?. Therefore,
K
0
(?
n+e
i
,n
)
parenleftbig
[p
(n,v)
]
n
parenrightbig
=
bracketleftbig
?
n+e
i
,n
parenleftbig
p
(n,v)
parenrightbigbracketrightbig
n+e
i
=
?
?
summationdisplay
???
e
i(v)
p
(n,?)
?
?
n+e
i
=
summationdisplay
u??
0
M
i
(v,u)[p
(n+e
i
,u)
]
n+e
i
and
K
0
(?
n+e
i
,n
)
?
?
summationdisplay
v??
0
f(v)[p
(n,v)
]
n
?
?
=
summationdisplay
u??
0
?
?
summationdisplay
v??
0
M
i
(v,u)f(v)
?
?
[p
(n+e
i
,u)
]
n+e
i
=
summationdisplay
u??
0
j
n+e
i
,n
(f)(u)[p
(n+e
i
,u)
]
n+e
i
.
Thus the following square commutes for all i ?{1,2,...,k}:
A
n
j
n+e
i
,n
????? A
n+e
i
?
n
?
?
arrowbt
?
?
arrowbt
?
n+e
i
K
0
(B
n
)
K
0
(?
n+e
i
,n
)
???????? K
0
(B
n+e
i
).
The result follows. square
Henceforth, we shall follow the notation introduced in Lemma 3.9 and its
proof.
Now we begin to examine the action of Z
k
on K
0
(B)intermsofthe
description of K
0
(B) provided by Lemma 3.9.
Lemma 3.10 (cf. [25] Lemma 4.10). Fix i ?{1,...,k} and define a homo
morphism ?
i,n
: A
n
?? A
n
by ?
i,n
:= M
t
i
for all n ? Z
k
.Let?
i
: A ?? A
be the homomorphism induced by the system of homomorphisms {?
i,n
 n ?
Z
k
}.Then??
i
= K
0
(?
e
i
)?,where? : A ?? K
0
(B) is the isomorphism
constructed in Lemma 3.9.
Proof. It follows from Remarks 3.7 that ?
i,n
j
nm
= j
nm
?
i,m
for all m,n ? Z
k
so that ?
i
is welldefined for all i ?{1,...,k}.
Now, we let
tildewide
? : A ?? lim
??
(K
0
(B
m
);K
0
(?
n,m
)) be the unique isomorphism
such that K
0
(?
n
)? ?
n
=
tildewide
? ? j
n
for all n ? Z
k
where
?
n
: A
n
?
=
?? K
0
(B
n
)
f mapsto?
summationdisplay
u??
0
f(u)[p
(n,u)
]
n
(proof of Lemma 3.9). Then ? : A ?? K
0
(B) is the composition of
tildewide
? with
the canonical isomorphism lim
??
(K
0
(B
n
);K
0
(?
n,m
))
?
= K
0
(B).
On the Ktheory of higher rank graph C
?
algebras 11
We will show that
K
0
(B
n
)
K
0
(?
n
)
????? K
0
(B)
e
?
i,n
?
?
arrowbt
?
?
arrowbt
K
0
(?
e
i
)
K
0
(B
n
)
K
0
(?
n
)
????? K
0
(B)
commutes for all i =1,2,...,k and n ? Z
k
where ?
n
: B
n
?? B is the
inclusion map for all n ? Z
k
and
tildewide
?
i,n
= ?
n
? ?
i,n
? ?
?1
n
. The lemma then
follows from the universal properties of direct limits.
Fix i ?{1,...,k} and n ? Z
k
.SinceK
0
(B
n
) is generated by
{[p
(n,v)
]
n
 v ? ?
0
},
it su?ces to show that K
0
(?
e
i
) ? K
0
(?
n
)([p
(n,v)
]
n
)=K
0
(?
n
) ?
tildewide
?
i,n
([p
(n,v)
]
n
)
for all v ? ?
0
. To see that this holds let v ? ?
0
.Then
K
0
(?
e
i
)? K
0
(?
n
)([p
(n,v)
]
n
)=K
0
(?
e
i
)([p
(n,v)
]) = [p
(n+e
i
,v)
],
while
K
0
(?
n
)?
tildewide
?
i,n
([p
(n,v)
]
n
)=
summationdisplay
u??
0
M
i
(v,u)[p
(n,u)
]
=
summationdisplay
???
e
i(v)
[p
(n+e
i
,?)
]=[p
(n+e
i
,v)
]. square
Having established a description of K
0
(B)asaleftZ
k
module in terms of
the structure of ?, we are almost in a position to describe H
?
(Z
k
,K
0
(B)).
First we recall some relevant notions from homological algebra.
It will be convenient to use multiplicative notation for the free abelian
group Z
k
, generated by k generators. Thus we set
G := ?s
i
 s
i
s
j
= s
j
s
i
for all i,j ?{1,...,k}?
and R := ZG, the group ring of G [3]. We shall compute H
?
(Z
k
,K
0
(B)) by
means of a Koszul resolution K(x) for an appropriate regular sequence x
on R,[34, Corollary 4.5.5].
By a regular sequence on R we mean a sequence x = {x
i
}
n
i=1
of elements
of R such that:
(a) (x
1
,...,x
n
) negationslash= R.
(b) For i =1,...,n, x
i
negationslash?Z(R/(x
1
,...,x
i?1
)).
Here, we regard R as an Rmodule, write (x
1
,...,x
j
) for the ideal of R
generated by {x
i
}
j
i=1
,andwriteZ(M) for the set of zerodivisors on a R
module M, i.e., Z(M):={r ? R  r ?m = 0 for some nonzero m ? M} (see
[10, ?3.1] for more details).
It is straightforward to check that for any finite set {t
1
,...,t
k
} of gener
ators of G, the subset x = {x
i
}
k
i=1
,wherex
i
:= 1 ? t
i
for i =1,...,k,isa
regular sequence on R.
12 D. Gwion Evans
Following [34, ?4.5] we will describe the Koszul complex, K(x), in terms
of the exterior algebra of a free Rmodule [2]. It will be convenient for us to
describe the terms of the exterior algebra as follows.
Definition 3.11. For any nonnegative integer l,letE
l
(R
k
)denotethel
th
term of the exterior algebra of the free Rmodule R
k
:=
circleplustext
k
i=1
R
i
,overR,
where R
i
= R for i =1,...,k. Moreover, for any negative integer l,let
E
l
(R
k
)={0}.Forl ? Z let
N
l
:=
?
?
?
?
?
{(?
1
,...,?
l
) ?{1,...,k}
l
 ?
1
< ???n.
Using the above notation we may describe the l
th
term (l ? Z)ofthe
exterior algebra over R
k
, E
l
(R
k
), as being generated by the set N
l
as a free
Rmodule and having rank
parenleftbig
k
l
parenrightbig
.
Now let K(x) be the chain complex
0 ?? E
0
(R
k
) ?? E
1
(R
k
) ?? ? ? ? ?? E
k
(R
k
) ?? 0
where the di?erentials E
l
(R
k
) ?? E
l?1
(R
k
) are given by mapping
? mapsto?
l
summationdisplay
j=1
(?1)
j+1
x
?
j
?
j
for all ? =(?
1
,...,?
l
) ? N
l
,
if l ?{1,...,k} and the zero map otherwise. By [34, Corollary 4.5.5] K(x)
is a free resolution of R/I over R where I is the ideal of R generated by
x. It is wellknown (see, e.g., [34, Chapter 6],[3, ?I.2]) that I =kerepsilon1 where
epsilon1 : R ?? Z : g mapsto? 1 is the augmentation map of the group ring R = ZG.
Thus we have a free (and hence projective) resolution of Z over ZG,which
we may use to compute H
?
(G, K
0
(B)) (see [3, Chapter III]).
Lemma 3.12. Following the above notation, we have H
?
(G, K
0
(B)) iso
morphic to the homology of the chain complex
B :0?? K
0
(B) ?? ? ? ? ??
circleplusdisplay
N
l
K
0
(B) ?? ? ? ? ?? K
0
(B) ?? 0,
On the Ktheory of higher rank graph C
?
algebras 13
where the di?erentials
tildewide
?
l
:
circleplustext
N
l
K
0
(B) ??
circleplustext
N
l?1
K
0
(B)(l ?{1,...,k})
are defined by
circleplusdisplay
??N
l
m
?
mapsto?
circleplusdisplay
??N
l?1
summationdisplay
??N
l
l
summationdisplay
i=1
(?1)
i+1
?
?,?
i(m
?
? K
0
(?
?
i
)(m
?
)).
(Recall that the Gaction on K
0
(B) is given by s
i
? m = K
0
(?
e
i
)(m) for all
m ? K
0
(B),i=1,...,k.)
Proof. By definition H
?
(G, K
0
(B))
?
= H
?
(K(x)?
G
K
0
(B)), where the latter
chain complex is obtained by applying the functor ??
G
K
0
(B)termwiseto
the chain complex K(x). The lemma follows via the canonical isomorphism
between E
l
(R
k
)?
G
K
0
(B)and
circleplustext
N
l
K
0
(B)(l ?{0,...,k}), setting t
i
:= s
?1
i
(i =1,...,k) as our generators of G to obtain x as described above. square
For m,n ? Z
k
with m ? n,letA
(n)
be the chain complex
0 ?? A
n
?? ? ? ? ??
circleplusdisplay
N
l
A
n
?? ? ? ? ?? A
n
?? 0,
where l =0,...,k, A
n
= Z?
0
and for l>0 the di?erentials
?
(n)
l
:
circleplusdisplay
N
l
A
n
??
circleplusdisplay
N
l?1
A
n
are defined by
circleplusdisplay
??N
l
m
?
mapsto?
circleplusdisplay
??N
l?1
summationdisplay
??N
l
l
summationdisplay
i=1
(?1)
i+1
?
?,?
i(m
?
? ?
?
i
,n
(m
?
)),
where for i =1,...,k and n ? Z
k
, ?
i,n
is the homomorphism defined in
Lemma 3.10. Furthermore, let (?
n
m
)
p
: A
(m)
p
?? A
(n)
p
be the homomorphism
defined by
(?
n
m
)
p
?
?
circleplusdisplay
??N
p
m
?
?
?
=
circleplusdisplay
??N
p
j
nm
(m
?
)
for all p ?{0,1,...,k} and the trivial map for p ? Z\{0,1,...,k} (see
Lemma 3.9).
Following [30, Chapter 4, ?1], by a chain map ? : C??C
prime
we mean
a collection {?
p
: C
p
?? C
prime
p
} of homomorphisms that commute with the
di?erentials in the sense that commutativity holds in each square:
C
p
?????C
p?1
?
p
?
?
arrowbt
?
?
arrowbt
?
p?1
C
prime
p
?????C
prime
p?1
14 D. Gwion Evans
Recall that there is a category of chain complexes whose objects are chain
complexes and whose morphisms are chain maps. Moreover, the category of
chain complexes admits direct limits.
Lemma 3.13. Following the above notation, for each m,n ? Z
k
the system
of homomorphisms {(?
n
m
)
p
 p ? Z} defines a chain map ?
n
m
: A
(m)
?? A
(n)
such that (A
(n)
;?
n
m
) is a direct system in the category of chain complexes.
Furthermore, (B;?
n
) is a direct limit for (A
(n)
;?
n
m
),where?
n
: A
(n)
?? B
is given by (?
n
)
p
(
circleplustext
??N
p
m
?
)=
circleplustext
??N
p
?j
n
(m
?
) for all p ?{0,1,...,k}
and the trivial map otherwise.
Proof. That ?
n
m
: A
(m)
?? A
(n)
is a chain map for all m,n ? Z
k
, follows
immediately from the fact that ?
i,n
j
nm
= j
nm
?
i,m
for all i ?{1,...,k} and
m,n ? Z
k
(cf.proofofLemma3.10). That (A
(n)
;?
n
m
) is a direct system of
chain complexes follows immediately from the fact that (A
m
;j
nm
) is a direct
system of groups (by Lemma 3.9).
Note that since K
0
(?
e
i
)? = ??
i
for all i ?{1,...,k} by Lemma 3.10,
a direct calculation shows that
tildewide
?
p
(?
n
)
p
=(?
n
)
p?1
?
(n)
p
for all p ? Z
k
,thus
?
n
: A
(n)
?? B is a chain map for all n ? Z
k
.Thefactthat?
m
= ?
n
?
nm
for all m,n ? Z
k
follows immediately by construction of the maps.
Now suppose that (A;?
n
) is a direct limit for (A
(n)
;?
n
m
). Then by the
above and the universal property of direct limits, there exists a morphism
? : A??Bsuch that ??
n
m
= ?
n
for all m,n ? Z
k
. In order to show that
? is an isomorphism it su?ces to show that each (?)
p
: A
p
?? B
p
is an
isomorphism for all p ? Z. This follows immediately from the fact that
? : A ?? K
0
(B) is an isomorphism (Lemma 3.9) and that direct limits
commute with (finite) direct sums in the category of abelian groups in the
obvious way. We have shown, therefore, that (B;?
n
) is a direct limit for
(A
(n)
;?
n
m
) in the category of chain complexes. square
Note that each chain complex A
(n)
does not actually depend on n ? Z
k
,
thus for ease of notation we let D denote this common chain complex with
di?erentials ?
p
:= ?
(n)
p
for all p ? Z.
Theorem 3.14. Using the above notation, the homology of Z
k
with coef
ficients in the left Z
k
module K
0
(B) is given by the homology of the chain
complex D, i.e., we have H
?
(G, K
0
(B))
?
= H
?
(D).
Proof. The homology functor commutes with direct limits ([30, Chapter 4,
?1, Theorem 7]), therefore it follows that
H
?
(G, K
0
(B))
?
= lim
??
(H
?
(A
(n)
);H
?
(?
n
m
)).
Thus, it su?ces to prove that H
p
(?
m+e
j
m
) is the identity map for all p ?
Z,m? Z
k
,j?{1,...,k}. To prove that this is true, we fix j ?{1,...,k}
and show that
(3.1) y ? (?
m+e
j
m
)
p
y ? im ?
p+1
for y ? ker ?
p
and p ?{0,1,...,k}.
On the Ktheory of higher rank graph C
?
algebras 15
Here,werecallthatify =
circleplustext
??N
p
y
?
,then
(?
m+e
j
m
)
p
y =
circleplusdisplay
??N
p
M
t
j
y
?
,
so for a given p ?{0,1,...,k},(3.1)says
(3.2)
circleplusdisplay
??N
p
(1 ? M
t
j
)y
?
? im ?
p+1
for y =
circleplusdisplay
??N
p
y
?
? ker ?
p
.
The case p = 0 is obvious, so we concentrate on the case p ?{1,...,k}.
For ? ? N
p
,let?(?):=
summationtext
p
i=1
?
j,?
i
i, i.e., if j is a component of ?,then
?(?) denotes the unique i ?{1,...,p} such that ?
i
= j;otherwise?(?)=0.
Now fix a ?
prime
=(?
prime
1
,...,?
prime
p
) ? N
p
and let y =
circleplustext
??N
p
y
?
be in ker?
p
.
First, suppose that ?(?
prime
) > 0andlet? =(?
prime
)
?(?
prime
)
.Then
0=?
p
(y)
?
=
summationdisplay
??N
p
p
summationdisplay
i=1
(?1)
i+1
?
?,?
i(1 ? M
t
?
i
)y
?
=
p
summationdisplay
i=1
(?1)
i+1
?
?,(?
prime
)
i(1 ? M
t
?
i
)y
?
prime +
summationdisplay
??N
p
?negationslash=?
prime
p
summationdisplay
i=1
(?1)
i+1
?
?,?
i(1 ? M
t
?
i
)y
?
=(?1)
?(?
prime
)+1
(1 ? M
t
j
)y
?
prime +
summationdisplay
??N
p
?negationslash=?
prime
p
summationdisplay
i=1
(?1)
i+1
?
?,?
i(1 ? M
t
?
i
)y
?
,
so that
(1 ? M
t
j
)y
?
prime =
summationdisplay
??N
p
?negationslash=?
prime
p
summationdisplay
i=1
(?1)
i+?(?
prime
)+1
?
?,?
i(1 ? M
t
?
i
)y
?
.
Note that the case p = k (where ?(?) is necessarily greater than zero)
follows immediately from the above. Therefore, we may assume that p ?
{1,...,k? 1}, in which case we claim that
circleplusdisplay
??N
p
(1 ? M
t
j
)y
?
= ?
p+1
?
?
circleplusdisplay
??N
p+1
z
?
?
?
where z
?
=
summationtext
p+1
i=1
(?1)
i+1
?
?
i
,j
y
?
i for all ? ? N
p+1
.
Now
?
p+1
?
?
circleplusdisplay
??N
p+1
z
?
?
?
?
prime
=
summationdisplay
??N
p+1
p+1
summationdisplay
i,r=1
(?1)
i+r+2
?
?
prime
,?
i?
j,?
r
(1 ? M
t
?
i
)y
?
r
16 D. Gwion Evans
=
summationdisplay
??N
p+1
?(?)>0
p+1
summationdisplay
i=1
(?1)
i+?(?)
?
?
prime
,?
i(1 ? M
t
?
i
)y
?
?(?)
=
summationdisplay
??N
p+1
?(?)>0
?
?
?
?(?)?1
summationdisplay
i=1
(?1)
i+?(?
prime
)+1
?
?
prime
,?
i(1 ? M
t
?
i
)y
?
?(?)
+
p+1
summationdisplay
i=?(?)+1
(?1)
i+?(?
prime
)
?
?
prime
,?
i(1 ? M
t
?
i
)y
?
?(?)
?
?
?
=
summationdisplay
??N
p
?(?)=0
p
summationdisplay
i=1
(?1)
i+?(?
prime
)+1
?
?,?
i(1 ? M
t
?
i
)y
?
since for every ? ? N
p+1
such that ?(?) > 0andforeveryi ?{1,...,
p +1}\{?(?)} we have:
(1) ?
?
prime
,?
?(?) =0,
(2) ?(?)=
braceleftBigg
?(?
prime
)+1 if?
prime
= ?
i
with i 0and?
?(?)
= ?
prime
. square
Combining the results of this section we get the following theorem.
On the Ktheory of higher rank graph C
?
algebras 17
Theorem 3.15. Let ? be a rowfinite kgraph with no sources. Then there
exists a spectral sequence {(E
r
,d
r
)} converging to K
?
(C
?
(?)) with E
?
p,q
?
=
E
k+1
p,q
and
E
2
p,q
?
=
braceleftBigg
H
p
(D) if p ?{0,1,...,k} and q is even,
0 otherwise,
where D is the chain complex with
D
p
:=
braceleftBigg
circleplustext
??N
p
Z?
0
if p ?{0,1,...,k},
0 otherwise,
and di?erentials
?
p
: D
p
?? D
p?1
:
circleplusdisplay
??N
p
m
?
mapsto?
circleplusdisplay
??N
p?1
summationdisplay
??N
p
p
summationdisplay
i=1
(?1)
i+1
?
?,?
i(1 ? M
t
?
i
)m
?
for p ?{1,...,k}.
Specialising Theorem 3.15 to the case when k = 2 gives us explicit formu
lae to compute the Kgroups of the C
?
algebras of rowfinite 2graphs with
no sources.
Proposition 3.16. Let ? be a rowfinite 2graph with no sources and vertex
matrices M
1
and M
2
. Then there is an isomorphism
?:coker(1? M
t
1
,1 ? M
t
2
) ? ker
parenleftbigg
M
t
2
? 1
1 ? M
t
1
parenrightbigg
?? K
0
(C
?
(?))
such that ?((?
u
+im?
1
) ? 0) = [p
u
] for all u ? ?
0
(see Definition 3.5) and
whereweregard
(1 ? M
t
1
,1 ?M
t
2
):Z?
0
? Z?
0
?? Z?
0
,
parenleftbigg
M
t
2
? 1
1? M
t
1
parenrightbigg
: Z?
0
?? Z?
0
? Z?
0
,
as group homomorphisms defined in the natural way.
Moreover, we have
K
1
(C
?
(?))
?
= ker(1 ? M
t
1
,1 ? M
t
2
)/im
parenleftbigg
M
t
2
? 1
1 ?M
t
1
parenrightbigg
.
Proof. The Kasparov spectral sequence converging to K
?
(C
?
(?)) of Theo
rem 3.15 has E
?
p,q
?
= E
3
p,q
for all p,q ? Z. However, it follows from E
2
p,q
=0
for odd q that the di?erential d
2
is the zero map and E
3
p,q
?
= E
2
p,q
?
= H
p
(D)
for all p ?{0,1,...,k} and even q,whereD is the chain complex
0 ?? Z?
0
?
1
?? Z?
0
? Z?
0
?
2
?? Z?
0
?? 0
with ?
1
=(1? M
t
1
,1 ? M
t
2
)and?
2
=
parenleftbigg
M
t
2
? 1
1? M
t
1
parenrightbigg
for a suitable choice of
bases.
18 D. Gwion Evans
Convergence of the spectral sequence to K
?
(C
?
(?)) and the above means
that we have the following finite filtration of K
1
= K
1
(C
?
(?)):
0=F
0
(K
1
) ? F
1
(K
1
)=F
2
(K
1
)=K
1
,
with F
1
(K
1
)
?
= H
1
(D). Hence, K
1
(C
?
(?))
?
= H
1
(D) as required.
Now, we could proceed to obtain an isomorphism of K
0
(C
?
(?)) by use
of the spectral sequence; however we choose to use the Pimsner?Voiculescu
sequence in order to deduce relatively easily the action of the isomorphism
on basis elements.
By applying the Pimsner?Voiculescu sequence to B multicloserightZand
B multicloserightZ
2
?
= (B multicloserightZ) multicloserightZ
in succession, we may deduce that K
0
(i
B
):K
0
(B) ?? K
0
(B multicloserightZ
2
)factors
through an injection ?
1
: H
0
(B) ?? K
0
(B multicloserightZ
2
), where i
B
is the canon
ical injection i
B
: B ?? B multicloserightZ
2
and B is the chain complex defined in
Lemma 3.12. We may also deduce that there is an exact sequence that
constitutes the first row of the following commutative diagram:
0 ????? H
0
(B)
?
1
????? K
0
(B multicloserightZ
2
) ????? H
2
(B) ????? 0
?
2
?
?
arrowbt
?
?
arrowbt
?
0
?
?
arrowbt
0 ????? H
0
(D) ????? K
0
(C
?
(?)) ????? H
2
(D) ????? 0.
where all downward arrows represent isomorphisms. In particular,
?
0
: K
0
(C
?
(?)) ?? K
0
(B multicloserightZ
2
)
is the isomorphism constructed in Theorem 3.1 and
?
2
: H
0
(B) ?? H
0
(D)
is one of the isomorphisms in Theorem 3.14.
Now H
2
(D) ? Z?
0
is a free abelian group, thus the exact sequences split
andwehaveanisomorphism?:H
0
(D) ? H
2
(D) ?? C
?
(?) such that
?(g ? 0) = ?
0
?
1
?
?1
2
(g) for all g ? H
0
(D). It is straightforward to check
that ?(?
u
+im?
1
?0) = [p
u
] for all u ? ?
0
. Thus the theorem is proved. square
Evidently complications arise when k>2, however it is worth noting
that under some extra assumptions on the vertex matrices it is possible
to determine a fair amount about the Kgroups of higher rank graph C
?

algebras. For example, the case k = 3 is considered below.
On the Ktheory of higher rank graph C
?
algebras 19
Proposition 3.17. Let ? be a rowfinite 3graph with no sources. Consider
the following group homomorphisms defined by block matrices:
?
1
=(1? M
t
1
1 ?M
t
2
1? M
t
3
):
3
circleplusdisplay
i=1
Z?
0
?? Z?
0
,
?
2
=
?
?
M
t
2
? 1 M
t
3
?10
1 ? M
t
1
0 M
t
3
? 1
01? M
t
1
1 ? M
t
2
?
?
:
3
circleplusdisplay
i=1
Z?
0
??
3
circleplusdisplay
i=1
Z?
0
,
?
3
=
?
?
1 ? M
t
3
M
t
2
? 1
1 ? M
t
1
?
?
: Z?
0
??
3
circleplusdisplay
i=1
Z?
0
.
There exists a short exact sequence:
0 ?? coker ?
1
/G
0
?? K
0
(C
?
(?)) ?? ker ?
2
/im ?
3
?? 0,
and
K
1
(C
?
(?))
?
= ker ?
1
/im ?
2
? G
1
,
where G
0
is a subgroup of coker ?
1
and G
1
is a subgroup of ker ?
3
.
Proof. By Theorem 3.15, there exist short exact sequences
0 ?? E
4
0,0
?? K
0
(C
?
(?)) ?? E
4
2,?2
?? 0,
0 ?? E
4
1,0
?? K
1
(C
?
(?)) ?? E
4
3,?2
?? 0.
However, since E
4
p,q
=0ifp ? Z\{0,1,2,3}, the only nonzero components
of the di?erential d
3
are d
3
3,q
: E
3
3,q
?? E
3
0,q+2
,whereq ? 2Z. Moreover, as
in the proof of Proposition 3.16, the di?erential d
2
is the zero map. Thus
we have
E
4
1,0
?
= E
3
1,0
?
= E
2
1,0
?
= H
1
(D),E
3
0,0
?
= E
2
0,0
?
= H
0
(D),
E
4
2,?2
?
= E
3
2,?2
?
= E
2
2,?2
?
= H
2
(D),E
3
3,?2
?
= E
2
3,?2
?
= H
3
(D).
Also note that E
4
3,?2
is isomorphic to ker d
3
3,?2
? E
3
3,?2
, which is a subgroup
of the free abelian group H
3
(D)
?
= ker ?
3
.ThusE
4
3,?2
is itself a free abelian
group, from which we deduce that the exact sequence for K
1
(C
?
(?)) splits.
Hence the result follows by setting G
0
to be the image of im d
3
3,?2
? E
3
0,0
under the isomorphism E
3
0,0
?? E
2
0,0
?? H
0
(D), and G
1
to be the image of
ker d
3
3,?2
? E
3
3,?2
under the isomorphism E
3
3,?2
?? E
2
3,?2
?? H
3
(D). square
Now we consider two cases for which we can describe the Kgroups of the
C
?
algebra of a row finite 3graph with no sources in terms of its vertex
matrices by the immediate application of Proposition 3.17.
20 D. Gwion Evans
Corollary 3.18. In addition to the hypothesis of Proposition 3.17:
(1) If ?
1
is surjective then
K
0
(C
?
(?))
?
= ker ?
2
/im ?
3
,
K
1
(C
?
(?))
?
= ker ?
1
/im ?
2
? ker ?
3
.
(2) If ?
3
i=1
ker(1 ? M
t
i
)=0then
K
1
(C
?
(?))
?
= ker ?
1
/im ?
2
and there exists a short exact sequence
0 ?? coker ?
1
?? K
0
(C
?
(?)) ?? ker ?
2
/im ?
3
?? 0.
Proof. To prove (1), note that we have 0 = coker ?
1
thus the exact sequence
for K
0
(C
?
(?)) collapses to give the result for K
0
(C
?
(?)). Also note that
0=coker?
1
= H
0
(D)
?
= E
3
0,0
. Therefore d
3
3,?2
: E
3
3,?2
?? E
3
3,0
is the zero
map and kerd
3
3,?2
= E
3
3,?2
?
= E
2
3,?2
?
= H
3
(D)=ker?
3
. Therefore, G
1
in
Proposition 3.17 is ker ?
3
and (1)isproved.
To prove (2), if
intersectiontext
3
i=1
ker(1 ?M
t
i
)=0thenker?
3
= 0, which implies that
G
1
in Proposition 3.17 is the trivial group. It also follows that E
3
3,?2
=0so
that im d
3
3,?2
=0andG
0
in Proposition 3.17 is the trivial group. Whence
(2) follows immediately from Proposition 3.17. square
Remarks 3.19. (i) One may recover [16, Theorem 3.1] from Theorem
3.15 by setting k equal to 1.
(ii) By [13, Corollary 3.5 (ii)] a rank k Cuntz?Krieger algebra ([24, 25]) is
isomorphic to a kgraph C
?
algebra. Thus, Propsition 3.16 generalises
[25, Proposition 4.1], the proof of which inspired the methods used
throughout this paper.
(iii) By showing that the C
?
algebra of a rowfinite 2graph, ?, with no
sources and finite vertex set, satisfying some further conditions, is
isomorphic to a rank 2 Cuntz?Krieger algebra, Allen, Pask and Sims
used Robertson and Steger?s [25, Proposition 4.1] result to calculate
the Kgroups of C
?
(?) [1, Theorem 4.1]. Moreover, in [1,Remark4.7.
(1)] they note that their formulae for the Kgroups holds for more
general 2graph C
?
algebras, namely the C
?
algebras of rowfinite 2
graphs, ?, with no sinks (i.e., s
?1
(v)??
n
negationslash= ? for all n ? N
k
,v? ?
0
)
nor sources and finite vertex set.
(iv) The notion of associating a C
?
algebra, C
?
(?), to a kgraph ? was
generalised by Raeburn, Sims and Yeend [20] to include the case where
? is finitelyaligned; a property identified by them to enable an appro
priate C
?
algebra to be constructed. The family of finitelyaligned k
graphs and their associated C
?
algebras admit kgraphs with sources
and those that are not rowfinite. In [8], Farthing devised a method
of constructing, from an arbitrary finitelyaligned kgraph ? with
sources, a rowfinite kgraph with no sources,
?
?, which contains ?
On the Ktheory of higher rank graph C
?
algebras 21
as a subgraph. If, in addition, ? is rowfinite then Farthing showed
that C
?
(
?
?) is strong Morita equivalent to C
?
(?) and thus has isomor
phic Kgroups to those of C
?
(?). Therefore, in principle, the results
in this paper could be extended to the case where ? is rowfinite but
with sources.
4. The Kgroups of unital kgraph C
?
algebras
Recall that if ? is a rowfinite higher rank graph with no sources then ?
0
finite is equivalent to C
?
(?) being unital ([13, Remarks 1.6 (v)]). Thus in
this section we specialise in the case where the vertex set of our higher rank
graph, hence each vertex matrix, is finite. We will continue to denote the
Kasparov spectral sequence converging to K
?
(C
?
(?)) of the previous section
by {(E
r
,d
r
)} and we shall denote the torsionfree rank of an abelian group
G by r
0
(G) (see, e.g., [9]).
Proposition 4.1. If ? is a rowfinite higher rank graph with no sources and
?
0
finite then K
0
(C
?
(?)) and K
1
(C
?
(?)) have equal torsionfree rank.
Proof. Let the rank of the given higher rank graph ? be k and let ?
0
 = n.
Since E
?
p,q
?
= E
k+1
p,q
for all p,q ? Z and E
k+1
p,q
=0ifp ? Z\{0,1,...,k} or q
odd by Lemma 3.3, it follows from the definition of convergence of {(E
r
,d
r
)}
(Definition 3.2) that there exist finite filtrations,
0=F
?1
(K
0
) ? E
k+1
0,0
?
= F
0
(K
0
) ? F
1
(K
0
) ?????F
k?1
(K
0
) ? F
k
(K
0
)=K
0
,
and
0=F
0
(K
1
) ? E
k+1
1,0
?
= F
1
(K
1
) ? F
2
(K
1
) ?????F
k?1
(K
1
) ? F
k
(K
1
)=K
1
of K
0
= K
0
(C
?
(?)) and K
1
= K
1
(C
?
(?)) respectively, such that
E
k+1
p,q
?
= F
p
(K
p+q
)/F
p?1
(K
p+q
).
Thus,
r
0
(K
0
(C
?
(?))) = r
0
(F
k
(K
0
)) = r
0
(F
k?1
(K
0
)) + r
0
(E
k+1
k,?k
)=???
= r
0
(F
0
(K
0
)) +
summationdisplay
s?1
r
0
(E
k+1
s,?s
)=
summationdisplay
s?Z
r
0
(E
k+1
s,?s
),
and
r
0
(K
1
(C
?
(?))) = r
0
(F
k
(K
1
)) = r
0
(F
k?1
(K
1
)) + r
0
(E
k+1
k,?k+1
)=???
= r
0
(F
1
(K
1
)) +
summationdisplay
s?2
r
0
(E
k+1
s,?s+1
)=
summationdisplay
s?Z
r
0
(E
k+1
s,?s+1
).
Now we claim that
summationdisplay
s?Z
r
0
(E
k+1
s,?s
) ?r
0
(E
k+1
s,?s+1
)=
summationdisplay
s?Z
r
0
(E
2
s,?s
) ? r
0
(E
2
s,?s+1
).
22 D. Gwion Evans
To see that this holds it is su?cient to prove that for all r ? 2wehave
summationdisplay
s?Z
r
0
(E
r+1
s,?s
) ?r
0
(E
r+1
s,?s+1
)=
summationdisplay
s?Z
r
0
(E
r
s,?s
) ? r
0
(E
r
s,?s+1
).
Recall that for all r ? 1,p,q? Z,E
r+1
p,q
?
= Z(E
r
)
p,q
/B(E
r
)
p,q
where
Z(E
r
)
p,q
=kerd
r
p,q
and B(E
r
)
p,q
=imd
r
p+r,q?r+1
.Thus
r
0
(E
r+1
p,q
)=r
0
(Z(E
r
)
p,q
)? r
0
(B(E
r
)
p,q
)
= r
0
(Z(E
r
)
p,q
)? r
0
(E
r
p+r,q?r+1
)+r
0
(Z(E
r
)
p+r,q?r+1
)
for all r ? 1,p,q? Z. Moreover, it follows from the definition of the
Kasparov spectral sequence that given any r ? 1andp,q,q
prime
? Z with
q = q
prime
mod 2 there exist isomorphisms
? : E
r
p,q
?? E
r
p,q
prime,
? : E
r
p?r,q+r?1
?? E
r
p?r,q
prime
+r?1
,
such that d
r
p,q
prime
?? = ? ? d
r
p,q
. Therefore,
summationdisplay
s?Z
r
0
(E
r+1
s,?s
)? r
0
(E
r+1
s,?s+1
)=
summationdisplay
s?Z
{r
0
(Z(E
r
)
s,?s
)? r
0
(Z(E
r
)
s+r,?s?r+2
)
? r
0
(Z(E
r
)
s,?s+1
)+r
0
(Z(E
r
)
s+r,?s?r+1
)
+ r
0
(E
r
s+r,?s?r+2
)? r
0
(E
r
s+r,?s?r+1
)
bracerightbig
=
summationdisplay
s?Z
r
0
(E
r
s,?s
) ?r
0
(E
r
s,?s+1
)
for all r ? 1. Combining the above gives
r
0
(K
0
(C
?
(?))) ? r
0
(K
1
(C
?
(?))) =
summationdisplay
s?Z
r
0
(E
2
s,?s
) ?r
0
(E
2
s,?s+1
).
Now, recall that for all p ? Z and q ? 2Z,
E
2
p,q
?
= H
p
(Z
k
,K
0
(B))
?
= ker ?
p
/im ?
p+1
by Theorem 3.14. Therefore,
On the Ktheory of higher rank graph C
?
algebras 23
r
0
(K
0
(C
?
(?))) ? r
0
(K
1
(C
?
(?))) =
summationdisplay
s?Z
r
0
(E
2
2s,?2s
)? r
0
(E
2
2s+1,?2s
)
=
summationdisplay
s?Z
r
0
(ker ?
2s
)? r
0
(im ?
2s+1
)? r
0
(ker ?
2s+1
)+r
0
(im ?
2s+2
)
=
summationdisplay
s?Z
r
0
(ker ?
2s
)? r
0
?
?
?
?
circleplusdisplay
N
2s+1
Z?
0
?
?
/ker ?
2s+1
?
?
? r
0
(ker ?
2s+1
)
+ r
0
?
?
?
?
circleplusdisplay
N
2s+2
Z?
0
?
?
/ker ?
2s+2
?
?
=
summationdisplay
s?Z
r
0
(ker ?
2s
)?
parenleftbigg
k
2s +1
parenrightbigg
n + r
0
(ker ?
2s+1
) ? r
0
(ker ?
2s+1
)
+
parenleftbigg
k
2s +2
parenrightbigg
n? r
0
(ker ?
2s+2
)
=
summationdisplay
s?Z
braceleftbiggparenleftbigg
k
2s
parenrightbigg
?
parenleftbigg
k
2s ? 1
parenrightbiggbracerightbigg
n
=
summationdisplay
s?Z
braceleftbiggparenleftbigg
k ? 1
2s
parenrightbigg
+
parenleftbigg
k ? 1
2s ? 1
parenrightbigg
?
parenleftbigg
k ?1
2s ? 1
parenrightbigg
?
parenleftbigg
k ? 1
2s ? 2
parenrightbiggbracerightbigg
n =0. square
Corollary 4.2. If ? is a rowfinite higher rank graph with no sources and
?
0
is finite then there exists a nonnegative integer r such that for i =0,1,
K
i
(C
?
(?))
?
= Z
r
? T
i
for some finite group T
i
,whereZ
0
:= {0}.
Proof. It is wellknown that if B is a finitely generated subgroup of an
abelian group A such that A/B is also finitely generated then A must be
finitely generated too [9]. Now, for all p,q ? Z, E
k+1
p,q
is isomorphic to a
subquotient of the finitely generated abelian group E
2
p,q
?
= H
p
(Z
k
,K
q
(B)),
so E
k+1
p,?p
is also finitely generated. Moreover, E
k+1
0,i
?
= F
0
(K
i
(C
?
(?))) and
for p ?{1,2,...,k},E
k+1
p,?p+i
?
= F
p
(K
i
(C
?
(?)))/F
p?1
(K
i
(C
?
(?))), which
implies that K
i
(C
?
(?)) = F
k
(K
i
(C
?
(?))) is finitely generated. The result
follows from Proposition 4.1 by noting that every finitely generated abelian
group A is isomorphic to the direct sum of a finite group with Z
r
,where
r = r
0
(A)(see,e.g.,[9, Theorem 15.5]). square
Remarks 4.3. Note that it is wellknown that when k = 1 we always have
T
1
= 0 in the above, i.e., K
1
(C
?
(?)) is torsionfree. However, for k>1,
K
1
(C
?
(?)) may contain torsion elements.
Formulae for the torsionfree rank and torsion parts of the Kgroups of
unital C
?
algebras of rowfinite 2graphs with no sources can be given in
24 D. Gwion Evans
terms of the vertex matrices (cf. [25, Proposition 4.13]). This we do in
Proposition 4.4 below.
Proposition 4.4 (cf. [25, Proposition 4.13]). Let ? be a rowfinite 2graph
with no sources and finite vertex set. Then
r
0
(K
0
(C
?
(?))) = r
0
(K
1
(C
?
(?)))
= r
0
(coker(1 ? M
t
1
,1 ? M
t
2
)) + r
0
(coker(1 ? M
1
,1 ? M
2
)),
tor(K
0
(C
?
(?)))
?
= tor(coker(1 ? M
t
1
,1 ? M
t
2
)),
tor(K
1
(C
?
(?)))
?
= tor(coker(1 ? M
1
,1 ? M
2
)).
Proof. We have already seen in Proposition 4.1 that the torsionfree rank of
the K
0
group and K
1
group of a kgraph are equal so it is su?cient to calcu
late the torsionfree rank of K
0
(C
?
(?)). Let n := ?
0
.ByProposition3.16
we have
r
0
(K
0
(C
?
(?))) = r
0
(coker(1 ? M
t
1
,1 ? M
t
2
)) + r
0
parenleftbigg
ker
parenleftbigg
1 ? M
t
1
1 ? M
t
2
parenrightbiggparenrightbigg
= r
0
(coker(1 ? M
t
1
,1 ? M
t
2
)) + n ? r
0
(im(1 ? M
1
,1 ? M
2
))
= r
0
(coker(1 ? M
t
1
,1 ? M
t
2
)) + r
0
(coker(1 ? M
1
,1 ? M
2
)).
Furthermore, the assertion about the torsion part of K
0
(C
?
(?)) is obvious.
The torsion part of K
1
(C
?
(?)) is given by
tor(K
1
(C
?
(?)))
?
= tor
parenleftbigg
ker(1 ? M
t
1
,1 ? M
t
2
)/im
parenleftbigg
M
t
2
?1
1? M
t
1
parenrightbiggparenrightbigg
,
which is clearly isomorphic to tor(coker
parenleftbig
M
t
2
?1
1?M
t
1
parenrightbig
). However, by reduction to
Smith normal forms, coker
parenleftbig
M
t
2
?1
1?M
t
1
parenrightbig
is isomorphic to coker(1?M
1
,1?M
2
). square
Remarks 4.5. We note that, in the case where ? is a rowfinite 3graph
with no sources and finite vertex set, and with ?
1
,?
2
defined as in Proposi
tion 3.17, it is straightforward to show that if ?
1
is surjective then
K
0
(C
?
(?))
?
= K
1
(C
?
(?))
?
= Z
m
,
where m := r
0
(ker ?
2
)??
0
 = r
0
(coker ?
2
)??
0
 (with Z
0
:= 0).
5. Applications and examples
We begin this section with two corollaries to the results in the preceding
section, which facilitate the classification of the C
?
algebras of rowfinite 2
graphs with no sources. We then end the paper with some simple illustrative
examples.
On the Ktheory of higher rank graph C
?
algebras 25
Corollary 5.1. Let ? be a rowfinite 2graph with no sources, finite vertex
set and vertex matrices M
1
and M
2
. Then there exists an isomorphism
?:coker(1? M
t
1
,1 ? M
t
2
) ? ker
parenleftbigg
M
t
2
? 1
1 ? M
t
1
parenrightbigg
?? K
0
(C
?
(?))
such that ?(e +im?
0
)=[1],wheree(v)=1for all v ? ?
0
.
Proof. Follows from Proposition 3.16 and that
summationtext
u??
0
p
u
=1. square
Remarks 5.2. We note that the C
?
algebra of a rowfinite kgraph ?, with
no sources, is separable, nuclear and satisfies the UCT [26]. If in addition the
C
?
algebra is simple and purely infinite we say that it is a Kirchberg algebra,
and note that by the Kirchberg?Phillips classification theorem ([12, 18]) it
is classifiable by its Ktheory (see [13, Theorem 5.5]). We also note that
conditions on the underlying kgraph have been identified, which determine
whether the C
?
algebra is simple ([13, Proposition 4.8], [22, Theorem 3.2])
and purely infinite ([28, Proposition 8.8]).
Corollary 5.3. Let ? and ? be two rowfinite 2graphs with no sources.
Furthermore, suppose that C
?
(?) and C
?
(?) are both simple and purely
infinite, and that ? and ? share the same vertex matrices. Then
C
?
(?)
?
= C
?
(?).
Proof. Let ? and ? be two 2graphs satisfying the hypothesis. Then, by
Proposition 3.16 their Kgroups are isomorphic.
Suppose that the vertex set of ? (and hence that of ?) is infinite. Then
C
?
(?) and C
?
(?) are both nonunital, and thus stable, Kirchberg algebras
with isomorphic Kgroups. Thus by the Kirchberg?Phillips classification
theorem C
?
(?)
?
= C
?
(?).
In the case where the vertex set of ? (and hence that of ?) is finite,
C
?
(?) and C
?
(?) are both unital Kirchberg algebras with isomorphic K
groups. Furthermore, by Corollary 5.1 we see that the isomorphism of K
groups maps the K
0
class of the unit of one of the C
?
algebras onto that
of the other. Therefore, by the Kirchberg?Phillips classification theorem we
conclude that C
?
(?)
?
= C
?
(?) and the result is proved. square
Examples 5.4. (1) Let ? be a rowfinite 2graph with no sources. Sup
pose that the vertex matrices of ? are both equal to M,say. By
Proposition 3.16 and [16, Theorem 3.1] we have:
K
i
(C
?
(?))
?
= Z?
0
/im(1 ? M
t
)? ker(1 ? M
t
)
?
= K
0
(C
?
(E)) ? K
1
(C
?
(E)),
for i =1,2, where E is the 1graph with vertex matrix M.Theiso
morphism for K
0
(C
?
(?)) is immediately obvious. That for K
1
(C
?
(?))
is given by
parenleftbigg
x
y
parenrightbigg
+im
parenleftbigg
M
t
? 1
1 ? M
t
parenrightbigg
mapsto? (x +im(1? M
t
)) ? (x + y)
26 D. Gwion Evans
for all
parenleftbig
x
y
parenrightbig
? ker(1 ? M
t
,1 ? M
t
).
(2) Fix nonzero n
1
,n
2
? N
2
and let ? be a 2graph with one vertex
and vertex matrices M
1
=(n
1
)andM
2
=(n
2
) respectively. By
Proposition 3.16,wehaveK
0
(C
?
(?))
?
= K
1
(C
?
(?))
?
= Z/gZ,whereg
is the greatest common divisor of n
1
? 1andn
2
? 1.
Note that we recover the Kgroups of tensor products of Cuntz
algebras [4] by letting ? be the product 2graph of two 1graphs each
with one vertex and a finite number of edges (i.e., morphisms of degree
1). We also note that tensor products of two Cuntz algebras are not
the only examples of C
?
algebras of such 2graphs ? (cf. [13, ?6]).
However, by Corollary 5.3,theyare,upto?isomorphism, the only
examples of Kirchberg algebras arising from rowfinite 2graphs with
one vertex.
(3) For each positive integer n,letO
n
be the 1graph with 1 vertex, star,and
n edges, ?
1
,?
2
,...,?
n
.Letc : O
3
?O
n
?? Z be the unique functor
that satisfies c(?
i
,star)=?
i,1
(i =1,2,3) and c(star, ?
i
)=1(i =1,...,n).
Define ? to be the 2graph Z?
c
(O
3
?O
n
). Let T
i
:= 1?M
t
i
,where
M
1
and M
2
are the vertex matrices of ?. Then
T
1
?
u
= ??
u
? ?
u+1
and(5.1)
T
2
?
u
= ?
u
? n?
u+1
,(5.2)
where we identify ?
0
with Z.
Clearly, ker
parenleftbig
?T
2
T
1
parenrightbig
= 0. Now consider coker(T
1
,T
2
)andforeach
g ? Z?
0
let [g] be the image of g under the natural homomorphism
Z?
0
?? coker(T
1
,T
2
). By (5.1)and(5.2)wehave(n +1)[?
u
]=0.
Therefore, coker(T
1
,T
2
) is a cyclic group, generated by [?
0
]say,whose
order divides n + 1. We claim that ?[?
0
] negationslash=0foreach? =1,...,n.
Suppose the contrary, then we have
??
0
= T
1
x + T
2
y
for some x,y ? Z?
0
and ? ?{1,...,n}.Thus,foreachu ? ?
0
we
have
??
0
(u)=?x(u)? x(u ? 1) + y(u)? ny(u ? 1).
Since x,y ? Z?
0
,thereexistsN such that x(u)=y(u)=0ifu >N,
which we assume, without loss of generality, to be greater than zero.
It follows that
y(u)=
?
?
?
?
?
x(u),u= ?N,
x(u)+(n +1)
summationtext
N?1+u
j=0
n
N?1+u?j
x(?N + j), ?N +1? u ??1,
x(u)+(n +1)
summationtext
N?1+u
j=0
n
N?1+u?j
x(?N + j)+?n
u
,u? 0.
On the Ktheory of higher rank graph C
?
algebras 27
Setting u = N + 1, we arrive at the contradiction (n +1)?n
N+1
.
Therefore, by Proposition 3.16 K
0
(C
?
(?))
?
= Z/(n +1)Z.
Now we turn our attention to ker(T
1
,T
2
)/im
parenleftbig
?T
2
T
1
parenrightbig
. Suppose that
x ? y ? ker(T
1
,T
2
), then there exists N such that x(u)=y(u)=0if
u >N,
y(?N)=x(?N)and(5.3)
y(u)=x(u)+(n +1)
u?1
summationdisplay
j=?N
n
u?1?i
x(j) for all u ??N +1.(5.4)
Let P : Z?
0
? Z?
0
?? Z?
0
be the projection onto the second com
ponent, i.e., P(x ? y)=y.From(5.3)and(5.4)weseethatP is
injective on ker(T
1
,T
2
) and thus induces an isomorphism
ker(T
1
,T
2
)/im
parenleftbigg
?T
2
T
1
parenrightbigg
?
= P(ker(T
1
,T
2
))/im T
1
.
Moreover, P(ker(T
1
,T
2
)) = {y ? Z?
0

summationtext
j?Z
(?1)
j
y(j)=0}.Now
given y ? P(ker(T
1
,T
2
)), define z :?
0
?? Z by
z(u)=0ifu