Ideals of Steinberg algebras of strongly effective groupoids, with applications to Leavitt path algebras

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by  Lisa Orloff Clark

IDEALS OF STEINBERG ALGEBRAS OF STRONGLY EFFECTIVE
GROUPOIDS, WITH APPLICATIONS TO LEAVITT PATH
ALGEBRAS
LISA ORLOFF CLARK, CAIN EDIE-MICHELL, ASTRID AN HUEF, AND AIDAN SIMS
Abstract. We consider the ideal structure of Steinberg algebras over a commutative
ring with identity. We focus on Hausdorff groupoids that are strongly effective in the
6
1 sense that their reductions to closed subspaces of their unit spaces are all effective.
0 For sucha groupoid,we completely describe the ideal lattice of the associatedSteinberg
2 algebraoveranycommutativeringwithidentity. Ourresultsarenewevenforthespecial

n case of Leavitt path algebras; so we describe explicitly what they say in this context,
a and give two concrete examples.
J

7
2

1. Introduction
]
A
Leavitt path algebras over a field have been studied intensively since their independent
R
introduction, around 2005, by Abrams–Aranda-Pino in [3] and Ara–Moreno–Pardo in [5].
.
h
One of the earliest questions asked about these algebras was what the ideals look like.
t
a
The lattice of ideals is now completely understood, see, for example, [2, Theorem 2.8.10]
m
or [1, Theorem 11]. Work on the ideal structure of the Leavitt path algebra and its
[
irreducible representations continues. See for example the recent papers on the generators

1
of ideals [25], prime and primitive ideals [8, 24, 17], two-sided chain conditions [4], and
v
8 on irreducible representations [11, 6, 18].
3 In 2011, Tomforde went on to consider Leavitt path algebras over commutative rings
2
R with identity in [32], and again considered the ideal structure. Things are more com-
7
0 plicated in this setting because the ideal structure of the ring R has an effect on the ideal
.
1 structure of the Leavitt path algebra. Tomforde sidestepped this issue by considering only
0
the “basic” ideals which are, roughly speaking, the ideals that contain a scalar multiple
6
1 of a generator if and only if they contain the generator itself, and are therefore insensitive
: to the ideal structure of R. The structure of the basic ideals in Leavitt path algebra has
v
i recently been reconsidered by Larki in [21] for more general graphs than were allowed in
X
[32]. Larki also studies the prime and primitive ideals, and this involves non-basic ideals.
r
a In this paper, we investigate the basic and non-basic ideal structure of a large class
of Steinberg algebras. The Steinberg algebras, introduced independently in [30] and in
[13], are associated to ample groupoids. They include the Kumjian–Pask algebras of
higher-rank graphs introduced in [7], which in turn include the Leavitt path algebras.
Date: 26 January 2016.
1991 Mathematics Subject Classification. 16S99 (Primary); 16S10, 22A22 (Secondary).
Key words and phrases. Groupoid; Steinberg algebra; Leavitt path algebra, Kumjian–Pask algebra,
ideal structure.
This research was supported by Marsden grant 15-UOO-071 from the Royal Society of New Zealand
andbyDiscoveryProjectgrantDP150101598fromtheAustralianResearchCouncil. Partofthisresearch
was completed while the fourth author was attending the workshop Classification of operator algebras:
complexity, rigidity, and dynamics at the Mittag-Leffler Institute.
1
2 LISA ORLOFFCLARK,CAIN EDIE-MICHELL,ASTRID ANHUEF, ANDAIDANSIMS
The advantage of working with the more general Steinberg algebras is that this brings
into play, in the algebraic setting, powerful techniques from Renault’s theory of groupoid
C∗-algebras. Indeed, Renault’s theory previously played a fundamental role in the devel-
opment of the theory of graph C∗-algebras and their analogues.
Groupoid models in C∗-algebra theory are particularly well-suited to answering ques-
tions about ideal structure [27]. We focus on groupoids G which are strongly effective
in the sense that in every reduction of G to a closed invariant subspace of its unit space
G(0), the interior of the isotropy consists only of units. This reduces to Condition (K)
for graphs and to “strong aperiodicity” for higher-rank graphs. (This is folklore, but we
provide a proof in Corollary 6.5.) Our results provide a complete description of the lattice
of ideals in the Steinberg algebra of such a groupoid. Since these results are new even
for Leavitt path algebras, and hence also for Kumjian–Pask algebras, we give an explicit
account of what our main theorem says in these special cases.
We start in Section 3 by analysing the basic ideals of the Steinberg algebra of a strongly
effectivegroupoid. Wefindthattheidealsareindexedbytheopeninvariantsubsetsofthe
unit space as expected. When R is a field, every ideal is a basic ideal. Thus we can draw
some conclusions about the ideals in Steinberg algebras over fields, and the relationship of
these to the ideals of the corresponding groupoid C∗-algebra, at least when the groupoid
G is amenable.
In Section 4 we build on our analysis of basic ideals to describe all the ideals in the
Steinberg algebra. The extra ideals arising from the ideals of the ring R are encoded by
functions π, satisfying a consistency condition relating nesting of ideals in R to nesting
of subsets of the unit space, from the collection of open invariant subsets of G(0) to the
set L(R) of ideals of R.
Containment of ideals in the Steinberg algebra is encoded by a very natural partial
order on the functions π described in the preceding paragraph. So in principle the lattice
structure on the set of ideals is explicitly described in terms of the functions π. However,
it is difficult to describe the join operation on functions π that corresponds to addition of
ideals of the Steinberg algebra. In Section 5, we introduce an alternative characterisation
of the ideals in the Steinberg algebra in terms of functions ρ : G(0) → L(R) that are
continuous with respect to a suitable topology on L(R). This allows us to describe the
join and meet operations quite naturally.
Finally, in Section 6, we translate our results into the language of Leavitt path algebras
and Kumjian–Pask algebras. Here, the ideals are parameterised by functions from the
collection of saturated hereditary subsets of the vertex set of the graph into the set of
idealsofR,againsatisfyingasuitableconsistencycondition; oralternativelybycontinuous
functions from the infinite-path space of the graph to the ideal space L(R). We detail
the content of our theorems for two concrete examples of graphs, each emphasising the
advantages of one of these two parameterisations.
2. Preliminaries
We use the groupoid conventions of [12]. Let G be a groupoid. A subset U of the unit
space G(0) of G is invariant if s(γ) ∈ U implies r(γ) ∈ U; equivalently,
r(s−1(U)) = U = s(r−1(U)).
IDEALS OF STEINBERG ALGEBRAS 3
Given V ⊆ G(0), we write [V] for the smallest invariant subset of G(0) containing V. Thus
[V] = r(s−1(V)) = s(r−1(V)).
We use the standard notation from [26, page 6] where G = {γ ∈ G : s(γ) = u},
u
Gu = {γ ∈ G : r(γ) = u} and Gu = G ∩Gu for each unit u ∈ G(0). The isotropy groupoid
u u
of G is
Iso(G) := {g ∈ G : s(γ) = r(γ)} = Gu.
u
u∈[G(0)
Let U be an invariant subset of G(0). We write G := s−1(U), and then G coincides
U U
with the restriction
G| := {γ ∈ G : s(γ),r(γ) ∈ U}.
U
of G to U. This G is a groupoid with unit space U.
U
For subsets W,V ⊆ G, we define WV := {γη : γ ∈ W,η ∈ V,s(γ) = r(η)}.
Now let G be a topological groupoid. A subset B of G is a bisection if the source
and range maps restrict to homeomorphisms on B; for an open set to be a bisection we
require the source and range maps to restrict to homeomorphisms onto open subsets of
G(0). Then G is called ample if G has a basis of compact open bisections. In this paper,
we only consider ample Hausdorff groupoids.
An ample Hausdorff groupoid G is effective if the interior of Iso(G) is just G(0). It
follows that when G is effective, if B is a nonempty compact open bisection such that
B ⊆ G\G(0), then B \Iso(G) 6= ∅.
When G is second countable, G is effective if and only if it is topologically principal
in the sense that {u ∈ G(0) : Gu = {u}} is dense in G(0) (see [28, Proposition 3.6]). Our
u
results apply to groupoids G that are not second countable, so for us the two conditions
are, in general, different.
Definition 2.1. A groupoid G is strongly effective if for every nonempty closed invariant
subset V of G(0), the groupoid G is effective.
V
If G is strongly effective, then it is effective because G(0) is a closed invariant set. If
G is second countable, then so is G for every invariant subset V of G(0), and so G is
V
strongly effective if and only if it is essentially principal in the sense of [26, Chapter 2,
Definition 4.3].
Let G be an ample Hausdorff groupoid and R a commutative ring with identity. We
write A (G) for the Steinberg algebra of all locally constant, compactly supported func-
R
tions f : G → R, equipped with the convolution product. As a set, A (G) is the R-linear
R
span
span {1 : B is a compact open bisection}.
R B
For f ∈ A (G), the set {γ ∈ G : f(γ) 6= 0} is a finite union of compact open sets, and
R
so is itself compact and open. Since compact subsets of a Hausdorff space are closed, we
have
supp(f) := {γ ∈ G : f(γ) 6= 0} = {γ ∈ G : f(γ) 6= 0}.
Under the convolution product on C (G), for f,g ∈ A (G) we have supp(f ∗ g) ⊆
c R
supp(f)supp(g).
4 LISA ORLOFFCLARK,CAIN EDIE-MICHELL,ASTRID ANHUEF, ANDAIDANSIMS
3. Basic ideal structure
Throughout, G is an ample Hausdorff groupoid and R is a commutative ring with
identity.
When the coefficient ring R is not a field, the ideal structure of A (G) depends on the
R
ideal structure of R. For example, if G = {e} is the trivial group, then the Steinberg
algebra A (G) is isomorphic to R as an R-algebra, and then the ideals of A (G) are
R R
precisely the ideals of R. An ideal I of A (G) is a basic ideal if
R
K a compact open subset of G(0),0 6= r ∈ R and r1 ∈ I =⇒ 1 ∈ I.
K K
When G = {e}, the only nonzero basic ideal is R itself. In general, the basic ideals are the
ones that reflect the structure of G alone, and do not reflect the structure of R; we expect
the basic-ideal structure to be independent of R. Basic ideals of A (G) were introduced
R
by the first two authors in [12], and they generalise the basic ideals of a Leavitt path
algebra studied by Tomforde in [32].
The first step in studying the ideal structure of A (G) is to study the basic ideals. By
R
[12, Theorem 4.1], if G is an ample Hausdorff groupoid, then A (G) has no proper basic
R
ideals if and only if G is effective and minimal. In this paper we consider groupoids that
are strongly effective (hence effective) but not minimal. The main result of this section
is the following.
Theorem 3.1. Let G be an ample Hausdorff groupoid, and let R be a commutative ring
with identity. Then G is strongly effective if and only if
U 7→ I := {f ∈ A (G) : suppf ⊆ G }
U R U
is a lattice isomorphism from the open invariant subsets of G(0) onto the basic ideals of
A (G).
R
Before proving Theorem 3.1, we need to establish some helper results.
Lemma 3.2. Let G be an ample Hausdorff groupoid, let R be a commutative ring with
identity and let U ⊆ G(0) be an open invariant subset. Then I is a basic ideal in A (G).
U R
Proof. The set I is closed under addition and scalar multiplication. To see that I is an
U U
ideal, fix f ∈ I and g ∈ A (G). Let α ∈/ G and β ∈ Gr(α). Then s(β−1α) = s(α) 6∈ U,
U R U
and hence s(β) 6∈ U because U is invariant. Hence f(β) = 0. Thus
(f ∗g)(α) = f(β)g(β−1α) = 0.
β∈XGr(α)
So f ∗g ∈ I. A similar argument gives g ∗f ∈ I. That I is basic follows immediately
U
from its definition. (cid:3)
Proposition 3.3. Let G be an ample Hausdorff groupoid and let R be a commutative
ring with identity. Then U 7→ I is an injective lattice morphism from the open invariant
U
subsets of G(0) to the basic ideals of A (G).
R
Proof. We first prove that for open invariant subsets U,V of G(0), we have I ⊆ I if
U V
and only if U ⊆ V. Suppose that I ⊆ I , and fix u ∈ U. Choose a compact open
U V
neighbourhood K of u such that K ⊆ U. Then 1 ∈ I ⊆ I , giving u ∈ K ⊆ V. Hence
K U V
U ⊆ V. Conversely, if U ⊆ V, then G ⊆ G , and hence I ⊆ I .
U V U V
It follows immediately that I = I implies U = V, so U 7→ I is injective.
U V U
IDEALS OF STEINBERG ALGEBRAS 5
We will show that
I = I ∩I and I = I +I
U∩V U V U∪V U V
Since the set of open invariant subsets of G(0) (with set inclusion, intersection and union)
forms a lattice, it will then follow that {I : U is an open invariant subset of G(0)} is a
U
lattice (with set inclusion, intersection and +), and that U 7→ I is a lattice morphism.
U
Since G ∩G = s−1(U)∩s−1(V) = s−1(U ∩V) = G , we have
U V U∩V
I ∩I = {f : supp(f) ∈ G ∩G } = {f : supp(f) ∈ G } = I .
U V U V U∩V U∩V
If f ∈ I +I , say f = f +f , then
U V U V
supp(f) ⊆ supp(f )∪supp(f ) ⊆ G ∪G = s−1(U)∪s−1(V) = s−1(U ∪V) = G .
U V U V U∪V
This gives I +I ⊆ I .
U V U∪V
For the reverse containment, suppose that f ∈ I . So supp(f) ⊆ s−1(U ∪V). The
U∪V
set K := supp(f)\s−1(V) ⊆ s−1(U) is compact because it is a closed subset of supp(f),
U
and similarly K := supp(f)\s−1(U) is a compact subset of s−1(V). Let u ∈ K . Since
V U
supp(f)∩s−1(U)isopen, andsinceGisample, wecanfindacompactopenneighbourhood
N of u ∈ N ⊆ supp(f)∩s−1(U). By taking the union of a finite subcover of the cover
u u
{N : u ∈ K } of K , we obtain a compact open subset K′ of supp(f) such that K ⊆
u U U U U
K′ ⊆ s−1(U). Let K′ := supp(f)\K′ . Then supp(f) = K′ ⊔K′ with K′ ⊆ s−1(U)
U V U U V U
and K′ ⊆ s−1(V). Since K′ and K′ are compact and open, we obtain locally constant
V U V
functions fU and fV by setting fU(γ) := 1K′ (γ)f(γ) and fV(γ) := 1K′ (γ)f(γ). By
U V
construction, f ∈ I and f ∈ I , and so f = f +f ∈ I +I . Thus I +I = I .
U U V V U V U V U V U∪V
As discussed above, this proves the proposition. (cid:3)
Lemma 3.4. Let G be an ample Hausdorff groupoid and let R be a commutative ring with
identity. Suppose that G is not effective. Then there is a nonzero basic ideal I of A (G)
R
such that I ∩A (G(0)) = {0}.
R
Proof. Let F (G(0)) denote the free R-module generated by a copy of G(0); to reduce
R
confusion, we shall write δ for the spanning element of F (G(0)) corresponding to u ∈
u R
G(0). LetEnd(F (G(0)))denotetheR-algebraofendomorphisms ofF (G(0)). Byapplying
R R
[12, Proposition 4.2(2)] to the G-invariant set G(0) there is a homomorphism π : A (G) →
R
End(F (G(0))) such that
R
π(f)δ = f(γ)δ .
u r(γ)
γX∈Gu
Since G is not effective, Int(Iso(G))\G(0) is nonempty. Since Int(Iso(G))\G(0) is open,
there exists a compact open bisection B ⊆ Int(Iso(G))\G(0). If u ∈ s(B) then π(1 )δ =
B u
δ = π(1 )δ , and both are 0 otherwise. Now 0 6= 1 −1 ∈ kerπ. Thus kerπ is a
u s(B) u B s(B)
nonzero ideal, and it is basic by [12, Lemma 4.5].
We will show that kerπ ∩ A (G(0)) = {0}, and this proves the lemma. Let f ∈
R
A (G(0)) \ {0}. Fix u ∈ G(0) such that f(u) 6= 0. Then 0 6= f(u)δ = π(f)δ , and so
R u u
f ∈/ kerπ. Thus kerπ ∩A (G(0)) = {0}. (cid:3)
R
If G is an effective ample Hausdorff groupoid, then every nonzero ideal I of A (G)
R
has nonzero intersection with A (G(0)) by [31, Proposition 3.3]. Combining this with
R
Lemma 3.4 gives the following corollary.
6 LISA ORLOFFCLARK,CAIN EDIE-MICHELL,ASTRID ANHUEF, ANDAIDANSIMS
Corollary 3.5. Let G be an ample Hausdorff groupoid and let R be a commutative ring
with identity. Then G is effective if and only if every nonzero ideal I of A (G) has
R
nonzero intersection with A (G(0)).
R
Suppose that U is an open invariant subset of G(0) and let D := G(0) \U. Since U is
open, there is a function i : A (G ) → A (G) such that
U R U R
f(γ) if γ ∈ U
i (f)(γ) =
U
0 otherwise.
(
Likewise, since D, and hence G , is closed, restriction of functions gives a function q :
D U
A (G) → A (G ).
R R D
Lemma 3.6. Let G be an ample Hausdorff groupoid and let R be a commutative ring with
identity. Let U be an open invariant subset of G(0), and D := G(0) \ U. The functions
i : A (G ) → A (G) and q : A (G) → A (G ) are ∗-homomorphisms, and the
U R U R U R R D
sequence
0 −→ A (G ) −i→U A (G) −q→U A (G ) −→ 0
R U R R D
is exact. Further, I = i (A (G )), and
U U R U
I = span {1 : B ⊆ G is a compact open bisection with s(B) ⊆ U}.
U R B
Proof. Since U is invariant, i : A (G ) → A (G) is a homomorphism, and since D is
U R U R
invariant, q is also a homomorphism. It is clear that i is injective. To see that q is
U U U
surjective, fix a compact open subset K of G . Since K is also compact in G, and G is
D
ample, we can find a finite cover L of K by mutually disjoint compact open subsets
L∈F
of G. Then 1 = q ( 1 ). Since A (G ) is spanned by the 1 it follows that q
K U B∈F L S R V K U
is surjective. By definition of i and q it is clear that imi ⊆ kerq . For the reverse
U U U U
P
containment, take f ∈ kerq . Write f = r 1 where F is a collection of mutually
U B∈F B B
disjoint bisections of G and the r are all nonzero. Since q (f) = 0, each B ∈ F is
B U
P
contained in G , and so is a compact open subset of G . So we can define f ∈ A (G )
U U 0 R U
by f = r 1 , and we have i (f ) = f by construction. Finally, we have
0 B∈F B B U 0
I = i (A (G )) = span {i (1 ) : B is a compact open bisection of G }
UP U R U R U B U
= span {1 : B ⊆ G is a compact open bisection with s(B) ⊆ U}. (cid:3)
R B
Proof of Theorem 3.1. By Proposition 3.3, U 7→ I is an injective lattice morphism. So
U
it remains to prove that G is strongly effective if and only if U 7→ I is surjective.
U
Suppose that G is not strongly effective. There exists a nonempty closed invariant
subset V of G(0) such that G is not effective. By Lemma 3.4, there is a nonzero basic
V
(0)
ideal I of A (G ) which has zero intersection with A (G ). Let
R V R V
J := {f ∈ A (G) : f| ∈ I}.
R GV
If V = G(0), then J = I. If V 6= G(0), then I is a nonzero ideal of A (G) contained
G(0)\V R
in J. In either case, J is a nonzero ideal of A (G).
R
To see that J is a basic ideal, suppose that K ⊆ G(0) is a compact open subset of
G(0) and 0 6= r ∈ R with r1 ∈ J. Then r1 | = r1 ∈ I. Since I is basic,
K K GV K∩V
1 | = 1 ∈ I, and hence 1 ∈ J. Thus J is basic.
K GV K∩V K
To see that J is not of the form I , fix a nonempty open invariant U ⊆ G(0). First
U
suppose that U ∩ V = ∅. Fix a nonzero element g ∈ I. Lemma 3.6 shows that q :
V
IDEALS OF STEINBERG ALGEBRAS 7
A (G) → A (G ) is surjective, so there exists f ∈ A (G) such that f| = g, and so
R R V R GV
f ∈ J. Since U∩V = ∅, we have h| = 0 for all h ∈ I , and we conclude that f ∈ J\I .
GV U U
On the other hand, if U ∩V 6= ∅, then any nonempty compact open subset K ⊆ U ∩V
(0) (0)
satisfies 1 ∈ I ∩A (G ). Since J∩A (G ) = {0}, this implies J 6= I . Thus U 7→ I
K U R V R V U U
is not surjective.
Conversely, suppose that G is strongly effective. We just have to show that U 7→ I is
U
surjective. Let I be a nonzero basic ideal in A (G). Define
R
D := f−1(0) and U := G(0) \D = f−1(R\{0}).
f∈I,sup\pf⊆G(0) f∈I,sup[pf⊆G(0)
We will show that U is an open invariant subset of G(0) and that I = I .
U
To see that U is invariant, let u ∈ U and choose γ such that s(γ) = u. We must show
that r(γ) ∈ U. Fix f ∈ I such that suppf ⊆ G(0) and f(s(γ)) 6= 0. Let B be a compact
open bisection containing γ. A calculation shows that
1 ∗f ∗1 (ζ) = f(s(γ))1 (η−1ζ)
B B−1 B−1
{η∈B:r(η)=r(ζ)}
X
f(s(η)) if there exists unique η ∈ B such that ζ = r(η)
=
0 else.
(
In particular, 1 ∗ f ∗ 1 (r(γ)) = f(s(γ)) 6= 0 and supp(1 ∗ f ∗ 1 ) ⊆ G(0). Thus
B B−1 B B−1
r(γ) ∈/ D, and hence r(γ) ∈ U. Thus U is invariant, and it is open because it is a union
of open sets f−1(R\{0}).
Now we will show that I = I . For the ⊆ direction, recall from Lemma 3.6 that
U
I = kerq , and so q induces an isomorphism q˜ : A (G)/I → A (G ). By definition
U U U U R U R D
of I , we have I ∩A (G(0)) = {f ∈ A (G(0)) : supp(f) ⊆ U} = I ∩A (G(0)). Thus
U R R U R
q˜ (I +I )∩A ((G )(0)) = q˜ (I ∩A (G(0)))+I
U U R D U R U
= qU(cid:0)(I ∩AR(G(0))) (cid:1)
= q (I ∩A (G(0))) = {0}.
U U R
Since G is strongly effective, G is effective, and [31, Proposition 3.3] gives q˜ (I +I ) =
D U U
{0}. Thus I ⊆ I .
U
For the ⊇ direction, we first claim that if B is a compact open bisection and s(B) ⊆ U,
then 1 ∈ I. To see this, observe that s(B) isa compact opensubset ofU. Bydefinition
s(B)
of U, for each u ∈ U there is an element f ∈ I such that supp(f ) ⊆ G(0) and f (u) 6= 0.
u u u
This f is locally constant, so V := f−1(f(u)) is a compact open neighbourhood of u in
u u
G(0) and f| = f(u)1 . Since I is an ideal, we deduce that f(u)1 = f ∗1 belongs
Vu Vu Vu Vu
to I. Since I is a basic ideal, we deduce that 1 ∈ I. Now the V cover s(B), which
Vu u
is compact, so we can write s(B) as a finite union s(B) = V ∪ ··· ∪ V . Putting
u1 un
W := V \ i−1 V we obtain pairwise disjoint compact open sets that cover s(B), and
i ui j=1 uj
each 1 = 1 ∗1 ∈ I because I is an ideal. Thus 1 = 1 ∈ I as claimed. So
Wi SWi Vui s(B) Wi
the final statement of Lemma 3.6 implies that I ⊆ I. So I = I and hence the map
U U
P
U 7→ I is surjective. (cid:3)
U
InthesituationwhereR = Fisafield, allidealsarebasicandthefollowingisimmediate.
8 LISA ORLOFFCLARK,CAIN EDIE-MICHELL,ASTRID ANHUEF, ANDAIDANSIMS
Corollary 3.7. Let G be an ample Hausdorff groupoid and let F be a field. Suppose that
G is strongly effective. Then
U 7→ I := {f ∈ A (G) : suppf ⊆ G }
U F U
is an isomorphism from the lattice of open invariant subsets of G(0) onto the lattice of
ideals of AF(G).
If in additionthe groupoidG is second countable andamenable, then [10, Corollary 5.9]
shows that there is a lattice isomorphism between the open invariant subsets of G(0) and
the closed ideals of the C∗-algebra C∗(G) = C∗(G). Combining this with Corollary 3.7
r
gives the following.
Corollary 3.8. Let G be an ample Hausdorff groupoid. Suppose that G is second count-
able, amenable and strongly effective. If we regard AC(G) as a ∗-subalgebra of C∗(G),
then the closure operation is a lattice isomorphism from the ideals of AC(G) to the closed
ideals of C∗(G).
4. Nonbasic ideal structure
Proposition 4.1. Let G be an ample Hausdorff groupoid and let R be a commutative ring
with identity. Suppose that G is strongly effective, and let I be an ideal in A (G). Then
R
I = span{r1 : B is a compact open bisection and r1 ∈ I}.
B s(B)
The proof of this proposition uses two technical lemmas. The first is about compact
open bisections in the complement of the unit space of a strongly effective groupoid. The
second shows that restriction of functions to the unit space in such groupoids respects
ideal structure.
Lemma 4.2. Let G be an ample Hausdorff groupoid which is strongly effective. Suppose
that U ⊆ G(0) is a compact open set and B ⊆ G\G(0) is a compact open bisection such that
s(B),r(B) ⊆ U. Then there is a finite collection of compact open bisections {N ,...,N }
1 m
such that
(a) for each i, r(N ) ⊆ U;
i
(b) for each i, N−1BN = ∅;
i i
(c) s(N )∩s(N ) = ∅ for i 6= j; and
i j
(d) s(B) = s(N ).
i i
Proof. Since GFis strongly effective it is effective, and hence B \ Iso(G) 6= ∅. For each
γ ∈ B\Iso(G), we can apply [10, Claim 3.2]to obtaina compact openbisection V ⊆ s(B)
γ
such that γ ∈ BV and V BV = ∅. Let
γ γ γ
C := G(0) \ [V ].
γ
γ∈B\Iso(G)
[
Since each [V ] is an open invariant set, C is closed and invariant. We have
γ
B \Iso(G) ⊆ BV ⊆ G\G ,
γ C
γ∈B\Iso(G)
[
and hence B ∩ G ⊆ Iso(G). We claim that B ∩ G = ∅. Suppose, aiming for a
C C
contradiction, that B ∩ G 6= ∅. Then B ∩ G is a nonempty open compact bisection
C C
IDEALS OF STEINBERG ALGEBRAS 9
of G . Since G is strongly effective, G is effective. Thus (B ∩ G ) \ Iso(G ) 6= ∅,
C C C C
contradicting that B ∩G ⊆ Iso(G ). Thus B ∩G = ∅. Now
C C C
B = B \G = B[V ],
C γ
γ∈B\Iso(G)
[
and in particular, s(B) ⊆ [V ].
γ∈B γ
Fix x ∈ s(B). We will construct a compact open bisection M such that x ∈ s(M ),
x x
S
r(M ) ⊆ U, and and M−1BM = ∅. Choose γ with x ∈ [V ] and choose η ∈ V Gx.
x x x γ γ
First suppose that r(η) 6= x. Since G is ample and Hausdorff, there exist compact
open neighbourhoods U of x and U of r(η) such that U ∩ U = ∅. We have
x r(η) x r(η)
η ∈ s−1(U ) ∩ r−1(U ). Thus by intersecting an open compact bisection containing η
x r(η)
with the closed set s−1(U )∩r−1(U ) we obtaina compact open bisection M containing
x r(η) x
η such that r(M ) ∩ s(M ) = ∅. Since r(η) ∈ V , we can replace M with V M to
x x γ x γ x
obtain r(M ) ⊆ V ⊆ U (the replacement makes the range smaller, so we still have
x γ
r(M )∩s(M ) = ∅). We then have
x x
M−1BM = M−1V BV M = ∅.
x x x γ γ x
Now suppose that r(η) = x. Then x = r(η) ∈ V and so M := BV satisfies x ∈ s(M ),
γ x γ x
r(M ) ⊆ r(B) ⊆ U, and r(M )∩s(M ) = ∅. Furthermore,
x x x
M−1BM = V B−1BBV = V s(B)BV = V BV = ∅.
x x γ γ γ γ γ γ
We have s(B) ⊆ s(M ), and since s(B) is compact, there exist M ,...,M ∈
x∈s(B) x 1 n
n
{M : x ∈ s(B)} such that s(B) ⊂ s(M ). For 1 ≤ i ≤ n let
x S i=1 i
N :=SM \ M s(M ) .
i i i j
j<i
(cid:0)[ (cid:1)
These N satisfy (a)–(c) because the M do. Since each s(N ) = s(M )\ s(M ), the
i i i i j<i j
s(N ) are mutually disjoint and s(N ) = s(M ) = s(B), giving (d). (cid:3)
i i i i i S
Lemma 4.3. Let G be an ampleSHausdorff Sgroupoid which is strongly effective, and let R
be a commutative ring with identity. Let I be an ideal in A (G). If f ∈ I, then f| ∈ I.
R G(0)
Proof. Fix f ∈ I. Since G(0) is closed and open, we can write f = r 1 +
V∈F0 V V
r 1 , where F and F are finite collections of mutually disjoint compact open
C∈F1 C C 0 1 P
bisections in G(0) and G \ G(0) respectively, and the r and r are all nonzero in R. It
U V
P
suffices to show that r 1 ∈ I for all V ∈ F . Fix U ∈ F . We have
V V 0 0
1 f1 = r 1 + r 1 ∈ I,
U U U U B UBU
BX∈F1
and {UBU : B ∈ F } is a set of mutually disjoint compact open bisections contained in
1
G\G(0). We will show that r 1 ∈ I; we will do this by induction.
U U
Let n ≥ 0. Our inductive hypothesis is: if r1 + r 1 ∈ I where U ⊆ G(0) is
U B∈F B B
compact openandF is aset ofnmutually disjoint compact openbisections inUGU\G(0),
P
then r1 ∈ I. When n = 0 the induction hypothesis holds trivially.
U
Now let g ∈ I be of the form
g = r1 + r 1
U B B
B∈H
X
10 LISA ORLOFFCLARK,CAIN EDIE-MICHELL,ASTRID ANHUEF, ANDAIDANSIMS
where U ⊆ G(0) is compact open and H is a collection of n+1 mutually disjoint compact
open bisections in UGU \G(0).
Fix B ∈ H. We will first show that a1 ∈ I. Since G is strongly effective, we apply
0 s(B0)
Lemma 4.2 to U and B to obtain compact open bisections {N ,...,N } satisfying prop-
0 1 m
erties (a)–(d) of the lemma. For 0 ≤ i ≤ m, we have s(N ) = N−1UN and N−1B N = ∅
i i i i 0 i
by properties (a) and (b) of Lemma 4.2, respectively. Hence
h := 1 g1 = r1 + r 1 ∈ I.
i Ni−1 Ni s(Ni) B Ni−1BNi
B∈HX,B6=B0
m
By property (d) of Lemma 4.2 we have r1 = r1 , and thus
i=1 s(Ni) s(B0)
m Pm
(4.1) h = r1 + r 1 ∈ I.
i s(B0) B Ni−1BNi
Xi=1 Xi=1 B∈HX,B6=B0
For i 6= j, by property (c) of Lemma 4.2 we have
s(N−1BN )∩s(N−1BN ) ⊆ s(N )∩s(N ) = ∅,
i i j j i j
r(N−1BN )∩r(N−1BN ) ⊆ s(N )∩s(N ) = ∅.
i i j j i j
Hence N−1BN ∩N−1BN = ∅. For B ∈ H with B 6= B set
i i j j 0
n
D := N−1BN .
B i i
i=1
G
Then each D is a compact open bisection contained in UGU because the source and
B
range of the N are contained in U by properties (a) and (d) of Lemma 4.2. Hence (4.1)
i
is
(4.2) r1 + r 1 ∈ I.
s(B0) B DB
B∈HX\{B0}
To apply the inductive hypothesis, we must verify that each D ∩G(0) = ∅. Fix γ ∈ D .
B B
Then γ ∈ N−1BN for some i. If r(γ) 6= s(γ), then γ ∈/ G(0). So suppose that r(γ) = s(γ).
i i
Since N is a bisection, there is a unique element α ∈ N such that s(α) = s(γ) and
i i
γ = α−1βα where β is the unique element of B with s(β) = r(α). Since B ∩ G(0) = ∅,
β ∈/ G(0), and so γ ∈/ G(0). Thus D ∩G(0) = ∅. Now the inductive hypothesis applies to
B
(4.2), giving r1 ∈ I.
s(B0)
Since our choice of B was arbitrary, we obtain r1 ∈ I for every B ∈ H. We may
0 s(B)
also assume the collection {s(B)} is disjoint (by disjointification). So
B∈H
V := s(B) ⊆ U
B∈H
[
satisfies
r1 = r1 ∈ I.
V s(B)
B∈H
X
Since s(B) ∩ U \ V = ∅ for B ∈ H we have 1 g1 = r1 ∈ I. Thus r1 =
U\V U\V U\V U
r1 +r1 ∈ I as well. (cid:3)
V U\V