arXiv:1508.00622v2 [] 29 Jun 2016

SUBSPACE ARRANGEMENTS, BNS INVARIANTS, AND PURE
SYMMETRIC OUTER AUTOMORPHISMS OF RIGHT-ANGLED
ARTIN GROUPS
MATTHEW B. DAY AND RICHARD D. WADE
Abstract. We introduce a homology theory for subspace arrangements, and
use it to extract a new system of numerical invariants from the Bieri-NeumannStrebel invariant of a group. We use these to characterize when the set of basis
conjugating outer automorphisms (a.k.a. the pure symmetric outer automorphism group) of a right-angled Artin group is itself a right-angled Artin group.

1. Introduction
1.1. Motivation. Recall that a right-angled Artin group (RAAG) is a group AΓ
given by a finite presentation whose only relations are that some pairs of generators
commute (see Section 2.1 below). Outer automorphism groups of RAAGs form
a diverse and interesting family of groups. We are motivated by the following
question:
Question 1.1. When does the outer automorphism group Out(AΓ ) contain another
RAAG as subgroup of finite index? What combinatorial conditions on the defining
graph Γ characterize this?
We feel that this is an important test question in terms of the field’s understanding of these groups. Two important sequences of outer automorphism groups of
RAAGs, Out(Fn ) and GLn (Z), exhibit very different behavior when n = 2 compared to when n ≥ 3. In the case n = 2, both are virtually free, but if n ≥ 3 neither
is virtually a RAAG (see references below). The idea is to identify a ‘low rank’
or ‘low complexity’ type for the family of outer automorphism groups of RAAGs.
There are variants of this question where Out(AΓ ) is replaced by the automorphism group Aut(AΓ ), or where instead of asking about finite-index subgroups, we
ask more generally when Out(AΓ ) is commensurable to a RAAG.
As well as the above virtually free examples, there are examples of RAAGs whose
outer automorphism groups are finite [7, 9] or virtually free abelian [3]. There are
also some more interesting examples; for instance Out(F2 × F2 ) is commensurable
with F2 × F2 itself. On the other side of the spectrum:
• Out(AΓ ) may contain distorted cyclic subgroups (e.g. GL3 (Z)).
• Out(AΓ ) may contain a poison subgroup, forcing nonlinearity [10, 1].
• Out(AΓ ) may have an exponential Dehn function (e.g. Out(F3 ) [4]).
All of these phenomena rule out the possibility of finite index subgroups being
RAAGs. Despite these tools, a complete answer to Question 1.1 seems difficult. It
Date: 28 June, 2016.
1

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MATTHEW B. DAY AND RICHARD D. WADE

is often tricky to tell whether a group is a RAAG on the nose or not, let alone up
to finite index. For example, the group
G = ha, b, c, d, e|[a, b], [c, d], [ab, c], [cd, a]i
is a nonstandard presentation of the RAAG (F2 × F2 ) ∗ Z, but after adding the
innocent-looking relations [e, b] and [e, d], the group
G′ = ha, b, c, d, e|[a, b], [c, d], [ab, c], [cd, a], [e, b], [e, d]i
is not isomorphic to a RAAG (this can be shown using the methods in this paper).
This leads us to:
Question 1.2. Suppose G is a group given by a finite presentation whose only
relations are commutators (between words in the generators). Is there a procedure
to recognize if G is a RAAG?
This question is stated so generally that the answer is almost certainly ‘no’, but
for specific classes of groups the question is still interesting. To show such a group
is a RAAG, we need some kind of rewriting procedure for the presentation, and
to show it is not a RAAG, we usually need some kind of subtle invariant. One
such invariant is the BNS invariant; Koban and Piggott used the BNS invariant to
distinguish the non-RAAGs from a certain class of groups in a recent paper [14].
We discuss this below.
1.2. BNS invariants. The BNS invariant Σ of a finitely generated group G was
introduced in [2]. It is an open subset of the character sphere of G (i.e. the unit
sphere of Hom(G; R)) and it records the existence of certain kinds of actions on
R–trees. We review the BNS invariant in Section 4.1 below. There is a growing
collection of groups for which there is an explicit description of Σ. These examples
include:
• Fundamental groups of compact 3–manifolds [2, 17].
• Right-angled Artin groups [15].
• Pure braid groups [13].
• Pure symmetric automorphisms of right-angled Artin groups [14].
• Many hierarchies of groups over groups with trivial BNS invariants (see [6]
for a precise formulation).
In the above examples, the complement Σc is a union of linear subspheres of
the character sphere, so that the pre-image of Σc in Hom(G; R) determines a set
of subspaces VG of Hom(G; R). For an arbitrary pair (V, V) consisting of a vector
space V and a set of subspaces V of V , one can define a chain complex C∗ (V, V)
where C0 = V and each Cn is a formal direct sum of intersections of n subspaces in
V. We describe this chain complex in Section 3.1, although we would be interested
to know if it has appeared in the literature previously. This chain complex has
associated homology spaces H∗ (V, V). One can then study the homology
H∗ (VG ) = H∗ (Hom(G; R), VG )
given by the arrangement of maximal complementary subspaces VG in Hom(G; R).
In the above list of examples, VG is a finite set of subspaces of Hom(G; R), which
allows for H∗ (VG ) to be computed explicitly.
In general, one can still define VG to be the subspace arrangement consisting
of maximal subspaces V ⊂ Hom(G; R) such that the equivalence class [χ] of each

SUBSPACE ARRANGEMENTS, BNS INVARIANTS, AND PSO(AΓ )

3

nontrivial χ ∈ V is contained in Σc . This subspace arrangement only contains
every character in the complement of the BNS invariant when Σ is symmetric
in the character sphere (i.e. Σ = −Σ). Nevertheless, the Betti numbers for this
homology theory still provide a concrete set of numerical invariants for an arbitrary
group G.
The homology theory above is heavily influenced by a recent paper of Koban–
Piggott [14], who determine exactly when the pure symmetric automorphism group
of AΓ is itself a RAAG. This is directly related to our Question 1.1 because there are
many examples of RAAGs where the pure symmetric automorphisms form a finiteindex subgroup of Aut(AΓ ). This pure symmetric automorphism group PSA(AΓ ),
sometimes called the basis conjugating automorphism group is the subgroup consisting of automorphisms that take each element of a graphical basis of AΓ to a
conjugate of itself. The group PSA(AΓ ) has a standard generating set where each
a
generator πK
is given by a vertex a ∈ Γ and a component of K of Γ − st(a) (here
st(a) is the subgraph of Γ spanned by a and its adjacent vertices). The generator
a
πK
acts on each vertex of Γ by:
(
axa−1 if x ∈ K
a
πK
(x) =
x
otherwise
Toinet [18] gave a presentation of PSA(AΓ ) which was simplified by Koban and
Piggott to one that uses the above generators (see Theorem 2.5). This presentation
for PSA(AΓ ) is the standard presentation of a RAAG unless the graph Γ contains
a separating intersection of links, or SIL (often pronounced ‘sill’). A SIL occurs
when there is a common component K of both Γ − st(a) and Γ − st(b) for two
non-adjacent vertices a and b. This ‘no SIL’ RAAG presentation of PSA(AΓ ) first
appeared in work of Charney et al. [8].
In the converse direction, Koban and Piggott give an explicit description of the
BNS invariant Σ(PSA(AΓ )) and show that its complement is a set of rationally
defined linear subspheres of the character sphere. Furthermore, they find an invariant which allows them to prove that when the graph Γ contains a SIL, the
group PSA(AΓ ) is not a RAAG. The invariant they use coincides with the Euler
characteristic of H∗ (VG ). In our terminology, their results state:
Theorem A (Koban–Piggott, [14]). If G is a right-angled Artin group then the
Euler characteristic of H∗ (VG ) is equal to the rank of the center of G (in particular,
it is non-negative). If G = PSA(AΓ ) then either:
• the graph Γ does not contain a SIL and the Euler characteristic of H∗ (VG )
is zero; therefore G is a RAAG with trivial center; or
• the graph Γ contains a SIL and the Euler characteristic of H∗ (VG ) is strictly
negative; G is not a RAAG.
1.3. Results. In this paper, we study the image of PSA(AΓ ) in Out(AΓ ), which we
call the pure symmetric outer automorphism group of AΓ and denote by PSO(AΓ ).
We give a description of Σc (PSO(AΓ )) (see Proposition 4.15) as a finite set of
rationally defined subspheres of the character sphere and classify when PSO(AΓ ) is
itself a RAAG. Rather than being based on the existence of a SIL, this classification
depends on how SILs are arranged in Γ. Let us describe this in a precise way: when
b
a a
]
πL , πL
a and b have a separating intersection of links, relations of the form [πK
appear in the presentation of PSA(AΓ ), where b ∈ K and L is what we call a

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MATTHEW B. DAY AND RICHARD D. WADE

shared component of both Γ − st(a) and Γ − st(b). The following graph gives a
combinatorial description of when two components K and L of Γ − st(a) occur in
such a relation.
Definition 1.3. For each vertex a ∈ Γ the support graph ∆a has a vertex for each
component of Γ− st(a). There is an edge between two components K and L if there
exists a vertex b such that b ∈ K and L is a shared component of both Γ − st(a)
and Γ − st(b).
In particular, the graph Γ has no SIL if and only if each support graph ∆a is
discrete. The following theorem is the main result of our paper. The first part
describes H∗ (VG ) precisely when G is a RAAG, and the second part describes how
the support graphs determine when PSO(AΓ ) is isomorphic to a RAAG.
Theorem B. If G is a right-angled Artin group then:
(1) dim(H0 (VG )) is equal to the rank of the center of G.
(2) Hn (VG ) = 0 if n > 0.
If G = PSO(AΓ ) then either:
• Each support graph ∆a is a forest and G is isomorphic to a right-angled
Artin group.
• For some vertex a ∈ Γ the support graph ∆a contains a loop. Then H1 (VG )
is nontrivial and therefore G is not a right-angled Artin group.
Our methods give effective algorithms to determine whether PSO(AΓ ) is a RAAG
for a given Γ, and to identify which RAAG it is, if it is one. We encourage our
readers to try out several examples, but we only give two here.
Example 1.4. Let Γ be the edgeless graph on three vertices {a, b, c}, so AΓ is the free
group F3 . Koban–Piggott’s theorem shows that PSA(F3 ) is not a RAAG, because
a and b form a SIL. However, all three of the support graphs consist of a single
edge, so are trees. One can check that PSO(F3 ) is a free group generated by the
set {πba , πcb , πac }.
Example 1.5. Let Γ be the edgeless graph on four vertices {a, b, c, d}, so AΓ is the
free group F4 . Again, PSA(F4 ) is not a RAAG because there are SILs. All four of
the support graphs are triangles; for example the path labeled by {b}–{c}–{d}–{b}
is a loop in ∆a . We can use this loop to produce a nontrivial element of H1 (VG ),
which implies that G = PSO(F4 ) is also not a RAAG.
The paper is arranged as follows: Section 2 contains background material on
right-angled Artin groups and their symmetric automorphisms. Section 3 defines
the homology H∗ (V, V) associated to a subspace arrangement V in a vector space
V . It may be read independently from the rest of the paper. We describe some
simple examples and show that H∗ is functorial with respect to morphisms between
subspace arrangements. In Section 4 we apply this to BNS invariants of groups.
We first give the general definition of H∗ (VG ) before looking at the case when G is
equal to AΓ , PSA(AΓ ), or PSO(AΓ ) respectively. In particular we use Koban and
Piggott’s description of Σ(PSA(AΓ )) to give a description of Σ(PSO(AΓ )). Finally,
in Section 5 we give an explicit RAAG presentation for PSO(AΓ ) when each support
graph ∆a is a forest. The main contribution in this final section is a description of a
generating set for PSO(AΓ ) that serves as the standard basis for a graphical RAAG
presentation (if there are SILs then the original generating set will not work). This
uses the structure of the support graphs in an essential way.

SUBSPACE ARRANGEMENTS, BNS INVARIANTS, AND PSO(AΓ )

5

Acknowledgments. The authors would like to thank Dawid Kielak, Lance Miller,
Adam Piggott, Henry Schenck, Alex Suciu, and Uli Walther for helpful conversations. They would also like to thank an anonymous referee for comments that
improved the paper. Matthew Day was supported in part by NSF grant DMS1206981.
2. Pure symmetric automorphisms of RAAGs
2.1. Right-angled Artin groups. A finite graph Γ with vertex set V (Γ) and edge
set E(Γ) determines the right-angled Artin group AΓ with presentation:
E
D

AΓ = V (Γ) [v, w] : {v, w} ∈ E(Γ)

That is, the generators of AΓ are the vertices of Γ, and they commute if they are
connected by an edge in Γ. We call such a presentation a graphical presentation for
the RAAG. For v ∈ Γ, its link lk(v) is the set of vertices adjacent to v, and its star
st(v) is lk(v) ∪ {v}. For a word w, the support of supp(w) consists of each vertex
v such that v or v −1 appears in w. A word w is reduced if we cannot cancel any
inverse pairs of elements appearing in it: for any subword of the form v ǫ w′ v −ǫ , the
support of w′ is not contained in the star of v. The support of an element g ∈ AΓ is
the support of any reduced word representing g. This is independent of the reduced
representative. For any full subgraph Γ′ , the group AΓ′ naturally embeds in AΓ as
the subgroup generated by the vertices in Γ′ , so AΓ′ = hΓ′ i ⊂ AΓ .
For any vertex v ∈ Γ, its centralizer C(v) is the subgroup Ast(v) . This is an easy
special case of Servatius’s centralizer theorem [16]. The center Z(AΓ ) of AΓ is the
free abelian subgroup AΓ′ , where Γ′ is the span of the set of vertices adjacent to
every other vertex in the graph.
2.2. Symmetric automorphisms. A partially symmetric automorphism of AΓ is
an automorphism φ ∈ Aut(AΓ ) such that each vertex v ∈ AΓ is sent to a conjugate
gvg −1 under φ. The conjugating element g is allowed to vary with v. The set
PSA(AΓ ) forms a subgroup of Aut(AΓ ). We define PSO(AΓ ) to be the image of
PSA(AΓ ) in the outer automorphism group Out(AΓ ). If φ is an automorphism, we
use [φ] to denote the equivalence class represented by φ in Out(AΓ ). Each vertex
a
a ∈ Γ and component K of Γ − st(a) defines an automorphism πK
of PSA(AΓ ),
where:
(
axa−1 if x ∈ K
a
πK (x) =
x
otherwise
a
We refer to elements of the form πK
when K is a component of Γ − st(a) as
standard generators of PSA(AΓ ), and the set X of all such elements as the standard
generating set of PSA(AΓ ). If C = K1 ∪ K2 ∪ · · · ∪ Kn is a nontrivial union of
a
connected components of Γ − st(a), we may define πC
in the same way as above.
However, as
a
a
a
π a · · · πK
πC
= πK
n
1 K2

we leave these elements out of our generating set X. We will refer to all elements
a
as partial conjugations and reserve the term standard generator for
of the form πC
a
when K is a single connected component of Γ − st(a).
an element of the form πK
The element a is called the multiplier of the partial conjugation.

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MATTHEW B. DAY AND RICHARD D. WADE

2.3. Commutation in Out(AΓ ). The following lemma is a rephrasing of the classification of connected components given in [11]. This classification is used throughout the paper, so for completeness we give a brief proof.
Lemma 2.1. Let a and b be nonadjacent vertices of Γ. We can write the components of Γ − st(a) as A0 , . . . , Ak , C1 , . . . , Cl and the components of Γ − st(b) as
B0 , . . . , Bm , C1 , . . . , Cl where
• we have b ∈ A0 and a ∈ B0 , and
• A1 , . . . Ak ⊂ B0 and B1 , . . . , Bm ⊂ A0 .
We say that A0 and B0 are the dominating components, that A1 , . . . Ak and
B1 , . . . Bk are the subordinate components, and C1 , . . . Cl are the shared components
for the pair (a, b). We will sometimes use [b]a to denote the component of Γ − st(a)
containing b, i.e. [b]a is the dominating component of Γ − st(a) with respect to b.
Note that if we fix a the roles of the connected components of Γ − st(a) may change
as we vary b.
Proof of Lemma 2.1. Let K be a component of Γ − st(a) that is not the dominating component (so b ∈
/ K). It is enough to show that K is either a subordinate
component or a shared component, since by the symmetry between a and b this
will show that all components fall into the classification.
First we note that K ∩ st(b) = ∅. If this were not the case, there would be a
path of length one from b to an element of K, and since b is not adjacent to a (and
st(a) ∩ K = ∅) this would imply that b ∈ K, counter to our hypothesis. Since
K ∩ st(b) = ∅, K is a subset of a single component of Γ − st(b) (every path in K
avoids st(b), so every path in K is a path in Γ − st(b)).
We break into two cases: (1) there is an edge from K to an element of lk(a)−lk(b)
and (2) every edge from K to lk(a) connects to an element of lk(a) ∩ lk(b). In case
(1), there is a path from K to a avoiding st(b), by passing through an element of
lk(a)− lk(b). This means that there is a path from every element of K to a avoiding
st(b), so that K is a subset of the dominating component of Γ − st(b) with respect
to a, and therefore K is a subordinate component. In case (2), every path starting
in K and avoiding st(b) must also avoid st(a), since every edge from K to lk(a)
must connect to an element of st(b). This means that paths starting in K that
avoid st(b) cannot escape K; in other words the component of Γ − st(b) containing
K does not contain any elements outside of K and must equal K. So in case (2),
K is a shared component.

Guiterrez, Piggott and Ruane [11] give the following definition to describe when
there exist shared components for the pair (a, b):
Definition 2.2. We say that a pair (a, b) forms a separating intersection of links
or is an SIL-pair if a and b are nonadjacent and there is a connected component R
of Γ − (lk(a) ∩ lk(b)) with a, b 6∈ R.
Lemma 2.3 ([11], Lemma 4.5). A pair (a, b) is an SIL-pair if and only if the set
of shared components associated to (a, b) is nonempty.
Proof. From the above proof of Lemma 2.1, a component K is shared if and only
if K contains neither a nor b and every edge from K to lk(a) or lk(b) is an edge to
lk(a) ∩ lk(b). This means that K is a component of Γ − (lk(b) ∩ lk(a)) that does
not contain a or b.


SUBSPACE ARRANGEMENTS, BNS INVARIANTS, AND PSO(AΓ )

7

The above classification of components of Γ − st(a) and Γ − st(b) gives a quick
way of describing when generators of PSA(AΓ ) or PSO(AΓ ) commute. We will use
the commutator convention [g, h] = ghg −1 h−1 throughout.
Lemma 2.4. Let a and b be nonadjacent vertices in Γ. Then the commutator
a
b
[πK
, πL
] is nontrivial in Aut(AΓ ) if and only if one of the following conditions
hold:
• K and L are the dominating components for the pair (a, b).
• Either K or L is dominating and the remaining component is shared.
• We have K = L (they are identical shared components for the pair (a, b)).
The image of the commutator in Out(AΓ ) is nontrivial if and only if one of the
above cases holds and (a, b) is an SIL-pair.
Proof. The statement about Aut(AΓ ) is shown in Lemma 4.7 of [11]. We note
that the classification turns this statement into a straightforward exercise: in the
cases listed above, find a vertex that the commutator does not fix, and in the
other cases (some component is subordinate or the components are distinct shared
components), show that every vertex is fixed.
Now we show the statement about Out(AΓ ). First we suppose that we are not
in one of the listed cases, or (a, b) is not an SIL-pair. If we are not in one of
the three cases, then the commutator is trivial in Out(AΓ ) because it is trivial in
Aut(AΓ ). If (a, b) is not an SIL-pair, then there are no shared components and the
only interesting case is where K and L are both dominating. In this case, let K ∗ be
the union of the remaining (subordinate) components A1 , . . . Ak of Γ − st(a). The
a
a
a
b
product πK
∗ πK is an inner automorphism. As [πA , πL ] = 1 for all i, the elements
i
b
a
a
a −1
a
b
πL and πK ∗ commute. As [πK ] = [πK ∗ ] in Out(AΓ ), it follows that [πK
] and [πL
]
commute.
We are left to show that if (a, b) is an SIL-pair and the components K and L
satisfy one of the above cases, then the commutator in question is also nontrivial in Out(AΓ ). Suppose that K and L are the dominating components. Then
a
b
[πK
, πL
](a) = [a, b]a[a, b]−1 , and for any vertex x in a shared component we have
a
b
a
b
[πK , πL ](x) = x. Suppose [πK
, πL
] is an inner automorphism, conjugating all elements by some g ∈ AΓ . Then gxg −1 = x, so g is in the centralizer of x and the
support of g is a subset of st(x). It follows that the support of gag −1 is a subset of
st(x) ∪ {a}. This is a contradiction as b is not in st(x) and
a
b
supp(gag −1 ) = supp([πK
, πL
](a)) = {a, b}.
a
b
Hence [πK
, πL
] is nontrivial in Out(AΓ ). A similar argument applies in the remaining two cases.

a
b
In particular, one sees that πL
and πK
commute in PSO(AΓ ) unless (a, b) is an
SIL-pair. Lemma 2.4 makes it easy to identify the standard generators in the center
of PSO(AΓ ); we leave this as an exercise to the reader.

2.4. Presentations for PSA(AΓ ) and PSO(AΓ ). Toinet [18] gave a presentation
of PSA(AΓ ), and Koban–Piggott adapted Toinet’s presentation as follows:
Theorem 2.5 (Toinet ([18], Theorem 3.1), Koban–Piggott ([14], Theorem 3.3)).
The group PSA(AΓ ) has a finite presentation consisting of the standard generating
set and relations of the form:
a
b
(R1) [πK
, πL
] = 1 when [a, b] = 1.

8

MATTHEW B. DAY AND RICHARD D. WADE

a
b
(R2) [πK
, πL
] = 1 when K ∩ L = ∅, b 6∈ K and a 6∈ L.
a
b
(R3) [πK , πL
] = 1 when {a} ∪ K ⊂ L or {b} ∪ L ⊂ K.
b
a a
(R4) [πK
πL , πL
] = 1 when b ∈ K and a 6∈ L.

Note that the case [a, b] = 1 includes when a = b. In the language of Lemma 2.1,
the relation (R2) corresponds to distinct non-dominating components, and the relation (R3) corresponds to when one component is dominating and the remaining
component is subordinate for the pair (a, b). The repetition of L in (R4) is not
a misprint—the only time such a relation is not implied by (R1)–(R3) is when K
is a dominating component and L is a shared component for a and b (in particular, (a, b) forms an SIL). It follows that if Γ contains no SILs then PSA(AΓ ) is
isomorphic to a right-angled Artin group (this was originally shown by Charney–
Ruane–Stambaugh–Vijayan in [8]). We therefore call relations of the form (R4)
SIL relations.
As PSO(AΓ ) is obtained from PSA(AΓ ) by taking the quotient by the normal
subgroup consisting of inner automorphisms, this implies:
Corollary 2.6. The group PSO(AΓ ) is finitely presented, with a presentation given
by the image of the standard generating set in Out(AΓ ) and relations of the form:
(R1)
(R2)
(R3)
(R4)
(R5)

a
b
[πK
, πL
] = 1 when [a, b] = 1
a
b
[πK , πL ] = 1 when K ∩ L = ∅, b 6∈ K and a 6∈ L
a
b
, πL
] = 1 when {a} ∪ K ⊂ L or {b} ∪ L ⊂ K.
[πK
a a
b
[π
π
,
] = 1 when b ∈ K and a 6∈ L.
QK L πL
a
π
K∈Ia K = 1 where the product is taken over the set Ia of connected components of Γ − st(a).

2.5. The support graph. The support graph ∆a gives a combinatorial description
of how the roles of the components of Γ − st(a) for the pair (a, b) change as we vary
b in Γ. We repeat the definition from the introduction:
Definition 2.7. For each vertex a ∈ Γ the support graph ∆a has a vertex for each
component of Γ − st(a). There is an edge between two components K and L if
there exists a vertex b such that K is the dominating component with respect to b
(equivalently b ∈ K) and L is a shared component of both Γ − st(a) and Γ − st(b).
In other words, each edge in ∆a is a dominating-shared pair : a pair of components
of the form {[b]a , L}, where L is a shared component for the pair (a, b). Furthermore:
Lemma 2.8 (Star Lemma). Let (a, b) be an SIL-pair. There is a unique connected
component C of ∆a containing the dominating component [b]a and all shared components of Γ − st(a) for the pair (a, b). These vertices consist of a subset of the star
of [b]a . If ∆a is a forest, then every shared component L is adjacent to [b]a and a
(possibly empty) set of subordinate components for the pair (a, b).
Proof. From the definition of ∆a each shared component L for the pair (a, b) is
connected by an edge to [b]a and makes up a subset of the star of [b]a in ∆a . Hence
[b]a and the shared components for (a, b) lie in the same connected component of
∆a . If two shared components L and L′ are adjacent then there exists a loop in
∆a through L, L′ and [b]a . This cannot happen if ∆a is a forest.


SUBSPACE ARRANGEMENTS, BNS INVARIANTS, AND PSO(AΓ )

9

The support graphs let us define a large set of central elements in PSO(AΓ ). For
a
C a component of ∆a , let ζC
denote the partial conjugation
Y
a
a
ζC
=
πK
K∈C

a
Proposition 2.9. Let C be a component of ∆a . The element ζC
is central in
PSO(AΓ ).

a
b
Proof. This follows from the fact that pairs of standard generators πK
, πL
commute
in Out(AΓ ) unless (a, b) is an SIL-pair and K and L fall into one of the three cases
b
from Lemma 2.4. Fix a standard generator πL
of PSO(AΓ ). If (a, b) does not form
a
b
b
an SIL-pair, then [πK , πL ] = 1 for all components K of Γ − st(a), so πL
commutes
a
with ζC . The same assertion also holds if L is subordinate for the pair (a, b). We
may therefore assume that (a, b) is an SIL-pair and L is either a dominating or
shared component of Γ − st(b). The Star Lemma tells us that either every vertex
of C is a subordinate component for (a, b), or C contains all of the dominating and
a
shared components. In the first case, ζC
is a product of elements which commute
b
a
b
with πL , so ζC commutes with πL . Otherwise, as we are in Out(AΓ ):
Y
a
a −1
=
ζC
(πK
) .
K6∈C

b
Each K in this product is subordinate, so πL
commutes with every term and also
a
a
commutes with ζC . Hence ζC commutes with every generator and is central in
PSO(AΓ ).

a
Note that ζC
is inner and trivial in PSO(AΓ ) if and only if ∆a is connected.

Remark 2.10. When PSO(AΓ ) is a RAAG, our graphical presentation of PSO(AΓ )
a
will prove that elements of the form ζC
form a free (abelian) generating set of the
center of PSO(AΓ ). It would be interesting to know whether the center is still
free abelian, and whether these elements form a generating set, in the case that
PSO(AΓ ) is not RAAG.
3. Subspace arrangements in vector spaces
3.1. A chain complex for subspace arrangements. We fix a field K and work
with vector spaces over K. A subspace arrangement is a pair (V, V) where V is a
vector space and V = (Vj )j∈J is a collection of subspaces. We may define a chain
complex C∗ (V, V) as follows. We define Ck to be trivial for k < 0 and we define C0
to be the vector space V . For k ≥ 1 we define Ck by a vector space presentation.
Ck is the vector space over K spanned by tuples (V1 , . . . , Vk , v) such that:
• V1 , . . . , Vk ∈ V,
• v ∈ V1 ∩ . . . ∩ Vk ,
subject to the relations that:
•
•
•
•

λ · (V1 , . . . , Vk , v) = (V1 , . . . , Vk , λv) for all λ ∈ K
(V1 , . . . , Vk , v) + (V1 , . . . , Vk , w) = (V1 , . . . , Vk , v + w)
for any permutation σ, we have (V1 , . . . , Vk , v) = sign(σ)(Vσ(1) , . . . , Vσ(k) , v);
if Vi = Vj for some i, j, then (V1 , . . . , Vk , v) = 0 (this is implied by the
above bullet point unless the field K is of characteristic 2).

10

MATTHEW B. DAY AND RICHARD D. WADE

The boundary map ∂k : Ck → Ck−1 is defined by:
∂k ((V1 , . . . , Vk , v)) =

k
X

(−1)i−1 (V1 , . . . , V̂i , . . . , Vk , v),

i=1

where (V1 , . . . , V̂i , . . . , Vk , v) is the element of Ck−1 given by deleting the ith entry
from the tuple. For k = 1, the boundary map is defined by:
X
∂1 ((vj )j∈J ) =
vj .
j∈J

This makes sense because C1 = ⊕j∈J Vj is simply the direct sum of the subspaces
from V.

Remark 3.1. One can allow repetitions of subspaces in V. In this case one must be
careful to view the symmetrization given by bullet points (3) and (4) by treating
vector spaces as equivalent if Vi = Vj as indexed elements of V rather than just as
subspaces of V . Adding a redundant subspace does not change the homology (see
Proposition 3.7). We allow redundancy because it will simplify a later argument.
Proposition 3.2. With the boundary maps ∂k , the vector spaces Ck (V, V) form a
well defined chain complex.
Proof. This is a straightforward exercise and we omit the details. The most interesting part of the proof is the fact that ∂k−1 ◦ ∂k = 0. As often happens
with chain complex boundary maps, this is a result of the sign convention: for
v̄ = (V1 , . . . , Vk , v) ∈ Ck , the sum that we get by expanding ∂k−1 ◦ ∂k (v̄) contains
each (V1 , . . . , V̂i , . . . , V̂j , . . . , Vk , v) twice, with opposite signs. This is because Vj is
in the jth position of v̄, but in the (j − 1)st position of (V1 , . . . , V̂i , . . . , Vk , v). 
Definition 3.3. For any subspace arrangement (V, V) we define H∗ (V, V) to be the
homology of the chain complex C∗ (V, V).
As the image of ∂1 is equal to the span of V, we have a description of H0 as:
∼ V /span(V).
H0 (V, V) =
For finite collections of subspaces, there is a more explicit description of the
chain complex, which we give in the next section.
3.2. Finite subspace arrangements. Suppose that V = {V1 , . . . , Vn } is a finite
collection of subspaces of a vector space V indexed by the set I = {1, . . . , n} with
the natural ordering. We let J = {j1 , . . . , jk } vary over all subsets of I of size k
with j1 < j2 < · · · < jk and define
VJ = Vj1 ∩ . . . ∩ Vjk .
The ordering of I removes the need to symmetrize with respect to permuting terms
in tuples, and gives a simpler description of each Ck as the direct sum:
M
VJ .
Ck =
J⊂I,|J|=k

For any k and any J ⊂ I with |J| = k, let Ji be the set obtained from J by removing
the ith term, so that
VJi = Vj1 ∩ . . . ∩ V̂ji ∩ . . . ∩ Vjk .

SUBSPACE ARRANGEMENTS, BNS INVARIANTS, AND PSO(AΓ )

11

Let ∂Ji : VJ → VJi be the inclusion map, let pJ : Ck → VJ be the natural projection
onto VJ and let ιJi : VJi → Ck−1 be the inclusion map of VJi as a factor of Ck−1 .
The boundary map defined in Section 3.1 may be rewritten as:
∂i (v) =

k
XX
J

(−1)i−1 ιJi ◦ ∂Ji ◦ pJ (v),

i=1

where the left hand sum ranges over all J ⊂ I with |J| = k. We define ∂1
the same way as before. These maps are reasonably easy to write explicitly in
examples. For instance, ∂2 : C2 → C1 maps a vector v ∈ Vi ∩ Vj to the tuple
(0, 0, . . . , v, 0, . . . , −v, 0 . . . , 0) in C1 = V1 ⊕ V2 · · · ⊕ Vn , where the nonzero terms
occur in the ith and jth positions.
Example 3.4. Let V be K2 with basis x = (1, 0) and y = (0, 1). Let V1 be the x-axis,
let V2 be the y-axis, let V3 be the subspace given by the diagonal line spanned by
x + y, and let V = {V1 , V2 , V3 }. Then C0 = K2 , the space C1 = hxi ⊕ hyi ⊕ hx + yi
is 3-dimensional, and each Ck for k ≥ 2 is trivial as no pair of distinct subspaces
intersects nontrivially. As these subspaces span K2 , the map ∂1 : C1 → C0 is
surjective, and H0 (V, V) = 0. The space of 1-cycles is 1-dimensional, and is spanned
by the cycle (x, y, −x − y). Since there are no nontrivial 1-boundaries, this means
that H1 (V, V) is 1-dimensional, and all other homology vector spaces are trivial.
Example 3.5. Let V be K3 with basis x, y, z. Let V be the collection of subspaces
defined by
V1 = hy, zi

V2 = hx + y, zi

V3 = hx, y + zi

V4 = hx, yi

There are 6 intersections Vi ∩ Vj with i < j given by:
V1 ∩ V2 = hzi

V1 ∩ V3 = hy + zi

V1 ∩ V4 = hyi

V2 ∩ V3 = hx + y + zi

V2 ∩ V4 = hx + yi

V3 ∩ V4 = hxi

The above calculation implies that each intersection of distinct triples in V is trivial,
so that the chain complex is of the form
∂

∂

1
2
V →0
C1 −→
0 → C2 −→

with dim V = 3, dim C1 = 8 and dim C2 = 6. The map ∂1 is surjective, so that
dim(ker ∂1 ) = 5. One can check that ∂2 surjects onto ker ∂1 , so that dim(ker ∂2 ) = 1.
It follows that H2 (V, V) is 1-dimensional and the homology is trivial everywhere
else.
3.3. Functoriality. Suppose that (V, V) and (W, W) are subspace arrangements
in two vector spaces V and W over the same field K. A morphism of subspace
arrangements f : (V, V) → (W, W) is a linear map f : V → W such that for each
V ′ ∈ V, its image f (V ′ ) is contained in some element of W. In other words, for any
morphism there exists a map α : V → W such that f (V ′ ) ⊂ α(V ′ ) for all V ′ ∈ V.
Note that if v ∈ V1 ∩ . . . ∩ Vk then f (v) ∈ α(V1 ) ∩ . . . ∩ α(Vk ). Hence every choice
of α as above gives a map
αC : C∗ (V, V) → C∗ (W, W)
of chain complexes induced by the linear extension of the map:
αC ((V1 , . . . , Vk , v)) = (α(V1 ), . . . , α(Vk ), f (v)).

12

MATTHEW B. DAY AND RICHARD D. WADE

On C0 we define αC from C0 (V, V) = V to C0 (W, W) = W to be the linear map f .
It is easy to check that αC is a chain map, so we have an induced map on homology
α∗ : H∗ (V, V) → H∗ (W, W).
′

Given V ∈ V, the subspace f (V ′ ) may be contained in more than one element of
W, which means that the map α need not be unique. However, the next proposition
shows that the induced map on homology depends only on f .
Proposition 3.6. Let f : (V, V) → (W, W) be a morphism of subspace arrangements. Let α, β : V → W be maps such that f (V ′ ) ⊂ α(V ′ ), β(V ′ ) for all V ′ ∈ V.
Then α∗ = β∗ .
Proof. We will construct an explicit chain homotopy between the maps αC and βC .
We use an easy modification of the prism operators used to show that homotopic
maps between two topological spaces induce the same map on homology. We define
a degree-one map
P : C∗ (V, V) → C∗+1 (W, W)
that is trivial on C0 , and for k ≥ 1 is defined on generators of Ck by:
P ((V1 , . . . , Vk , v)) =

k
X

(−1)i+1 (α(V1 ), . . . , α(Vi ), β(Vi ), . . . , β(Vk ), f (v))

i=1

and extend this map linearly. Following the proof in Hatcher’s book [12, Theorem 2.10], one can check that
∂P − P ∂ = βC − αC .
Hence P is a chain homotopy between αC and βC and α∗ = β∗ .



We have shown that any morphism of subspace arrangements induces a welldefined map on homology. A further application of the above proposition allows us
to show that the homology H∗ (V, V) only depends on the maximal subspaces in V.
Proposition 3.7. Suppose that V is a subspace arrangement in V such that each
element of V is contained in a maximal element of V (this is true if V is finite
dimensional). Let V ′ be the family of maximal elements of V. Then H∗ (V, V) ∼
=
H∗ (V, V ′ ).
Proof. Let α : V ′ → V be the natural injection of V ′ into V. We may also choose
a map β : V → V ′ by picking a maximal subspace β(W ) containing each element
W ∈ V. Let αC and βC be the induced maps on chain complexes with respect to
the identity map from f : V → V to itself. Note that β ◦α is the identity map on V ′ ,
and it follows that βC ◦ αC is the identity map on C∗ (V, V ′ ). Hence β∗ ◦ α∗ induces
the identity map on H∗ (V, V ′ ). In the other direction, αC ◦ βC is the map from
C∗ (V, V) to itself induced by the map αβ : V → V. We may apply Proposition 3.6
to the identity morphism f = idV : (V, V) → (V, V). Here we take α′ : V → V to be
the identity map on the family V and take map β ′ = αβ : V → V; the Proposition
implies α′∗ = β∗′ . As α′∗ is the identity map, so is β∗′ = α∗ ◦ β∗ . It follows that α∗
and β∗ are isomorphisms.

Corollary 3.8. If V ∈ V then H∗ (V, V) is trivial.
Proof. This follows from the above as H∗ (V, {V }) is trivial and isomorphic to
H∗ (V, V).


SUBSPACE ARRANGEMENTS, BNS INVARIANTS, AND PSO(AΓ )

13

Remark 3.9. It is possible to characterize H∗ (V, V) as a derived functor. We do not
use this in this paper, but we outline it in this remark.
Consider a category C of subspace arrangements with a fixed index set J, whose
morphisms are linear maps that send the jth subspace into the jth subspace for each
j ∈ J (this is much more restrictive than the definition we use above). This category
C is an additive category, but not an abelian category because epimorphisms and
monomorphisms are not necessarily normal. We consider the category D of cubical
diagrams of vector spaces; this is the functor category from the opposite category
of the category of subsets of J (with inclusions) to the category of vector spaces
over K. It turns out that C embeds in D by sending an arrangement to the diagram
of inclusions of intersections of subspaces in the arrangement.
It follows from standard arguments that D is an abelian category, and it is possible to show that every object is a quotient of a projective object. The H0 functor
we define above corresponds to a functor H0 from D to vector spaces. Specifically,
if an object (V, f ) of D is given by {VS }S⊂J and {fS,T : VS → VT }T ⊂S⊂J , then
H0 ((V, f )) = V∅ /span({f{j},∅ (V{j} )}j∈J ).
This functor turns out to be right-exact. Our homology theory functors are then
the left-derived functors of the functor H0 .
3.4. Inclusion-exclusion. Our next statement has a connection to the inclusionexclusion principle, which we explain in the following remark.
Remark 3.10. Recall that the inclusion-exclusion principle allows us to count a finite
union of sets {Sj }j∈J by taking an alternating sum of the counts of the intersections
of these sets:


[

j∈J

Sj =

|J|
X

(−1)k+1 

k=1

X

\

I⊂J,|I|=k j∈I

Sj  .

One might hypothesize an analogous statement for vector spaces, asserting that
the dimension of a span of vector subspaces {Vj }j∈J is an alternating sum of the
dimensions of the intersections:


|J|
\ 
X
X
?
dim
Vj  .
dim span({Vj }j∈J ) =
(−1)k+1 
k=1

I⊂J,|I|=k

j∈I

This is famously false, although it holds in many simple examples. It fails in
different ways in Examples 3.4 and 3.5, by overcounting in the first one and undercounting in the second one.
Suppose (V, V) is an arrangement where V is finite-dimensional and the subspaces
in V span V (in other words, H0 (V, V) = 0). For such an arrangement, the validity
of the “inclusion-exclusion principle for vector spaces” is equivalent to the vanishing
of the Euler characteristic of H∗ (V, V). We do not use this fact, but we leave it as
an exercise for the interested reader.
We do require one result that is related to inclusion-exclusion. We are interested
in the case where all subspaces in our collection V are generated by subsets of a
fixed basis for V . (We will see below that this is true for BNS invariants of RAAGs.)
In this case, if dim(V ) is finite, then inclusion-exclusion clearly holds. This means

14

MATTHEW B. DAY AND RICHARD D. WADE

that the alternating sum in the remark above is dim(span(V)), and that the Euler
characteristic of H∗ (V, V) is
dim V − dim span(V) = dim H0 (V, V).
In fact, more is true: in this special case, the homology is trivial, except possibly
for H0 (V, V). Our proposition refines a lemma of Koban–Piggott [14], which uses
an inclusion-exclusion sum involving the BNS invariant of a RAAG to count the
number of non-central vertices in the defining graph. We state and prove our proposition assuming that V is finite-dimensional, although this can be easily extended
to the general case.
Proposition 3.11. Let V be a vector space with basis S = {s1 , . . . , sn } and let V
be a collection of subspaces of V such that each V ′ ∈ V is spanned by a subset of
S. Then Hn (V, V) = 0 for all n ≥ 1.
Proof. We use induction on the dimension of V . When dim(V ) = 1, either all spaces
in the collection V are trivial, or V ∈ V. The result then follows from Corollary 3.8.
Now suppose the result holds for all such arrangements in vector spaces of dimension
n − 1. Let V and V = {Vi }i∈I be as in the statement of the theorem with basis
S = {s1 , . . . , sn }. Let P be the subspace spanned by S ′ = {s1 , . . . , sn−1 } and let
p : V → P be the projection given by
p

n
X
i=1

Let



λi si =

n−1
X

λi si .

i=1

P = {Pi = p(Vi ) : i ∈ I}
be the projected subspace arrangement in P . Let Q = hsn i and let
Q = {Qi = Q ∩ Vi : i ∈ I}
be the induced subspace arrangement in Q. Note that for both (P, P) and (Q, Q)
we allow for repetitions of subspaces as described in Remark 3.1.
Let αC : C∗ (V, V) → C∗ (P, P) be the induced map on chain complexes coming
the from projection p : V → P and the map α : V → P given by α(Vi ) = Pi . The
element (V1 , . . . , Vk , v) is mapped to (P1 , . . . , Pk , p(v)) under αC . If w ∈ P1 ∩· · ·∩Pk ,
there exists v ∈ V1 ∩ · · · ∩ Vk with p(v) = w. It follows that αC is surjective.
The kernel chain complex of αC is spanned in Ck (V, V) by elements of the form
(V1 , . . . , Vk , λsn ), and in C0 (V, V) = V the kernel is the subspace Q = hsn i. This
kernel chain complex is naturally isomorphic to C∗ (Q, Q). We then have a short
exact sequence of chain complexes
α

C
C∗ (P, P) → 0
0 → C∗ (Q, Q) → C∗ (V, V) −→

which induces the long exact sequence in homology
· · · → Hk (Q, Q) → Hk (V, V) → Hk (P, P) → · · ·
As each vector space in in V is spanned by a subset of S, each element of P is
spanned by a subset of S ′ . Hence both (P, P) and (Q, Q) are subspace arrangements
where each subspace is spanned by a fixed subset of some basis. For k ≥ 1, the
space Hk (P, P) is trivial by the inductive hypothesis and Hk (Q, Q) is trivial by the
dimension 1 case. This implies that Hk (V, V) is trivial for k ≥ 1 also.


SUBSPACE ARRANGEMENTS, BNS INVARIANTS, AND PSO(AΓ )

15

4. BNS invariants and subspace arrangements
4.1. BNS invariants. The Bieri–Neumann–Strebel invariant is a subset Σ of the
character sphere of a finitely generated group G. The character sphere S of G is
the set
(Hom(G; R) \ {0})/ ∼
where characters are identified if they lie in the same ray in Hom(G; R): χ1 ∼ χ2
if and only if there is λ > 0 with χ1 = λχ2 . The original definition of the BNS
invariant from [2] states that [χ] ∈ S is in Σ if and only if [G, G] is finitely generated
over a finitely generated submonoid of χ−1 ([0, ∞)). Bieri–Neumann–Strebel also
give a convenient characterization in terms of a generating set in Proposition 2.3
of [2]: [χ] ∈ S is in Σ if and only if the preimage under χ of the closed half-line
[0, ∞) in the Cayley graph of G is connected. We do not use the original definition
or the equivalent one from the original paper; instead we prefer another equivalent
definition due to Brown that we state below.
Remark 4.1. Sometimes Σ is viewed as the first invariant in a collection Σ = Σ1 ⊃
Σ2 ⊃ Σ3 ⊃ · · · (see [2]). We will not be considering these higher invariants in this
paper.
Recall that an R-tree is a geodesic metric space in which a unique arc connects
any two points. An action of G on an R-tree T is abelian if there exists a character
χ such that |χ(g)| = kgkT for all g ∈ G, where kgkT is the translation length of
g as an isometry of T . We say that T realizes χ. Note that for each χ there is a
natural abelian action of G on a line realizing χ. Any abelian action realizing a
nontrivial character fixes one or two points in the boundary ∂T of T . When there
is a unique fixed point in ∂T we say that the action is exceptional.
Let T be an exceptional action realizing a character χ with fixed end e ∈ ∂T .
Let (gn ) be a sequence of elements of G such that for some (equivalently, any) point
x ∈ T the orbit gn · x converges to e. The sequence (χ(gn )) converges to either +∞
or −∞. We say that the invariant end is at +∞ in the former case, and −∞ in the
latter. This is independent of any choices made above. Swapping χ with −χ will
then swap the location of the invariant end. The following definition of Σ is due to
Brown [5], who showed that it is equivalent to the original definition from [2].
Definition 4.2. An element [χ] ∈ S is in Σ if there exists no exceptional action of
G on an R-tree T realizing χ with the invariant end at −∞.
Note that Brown’s definition allows one to consider Σ even in the case that G is
not finitely generated.
Rather than considering the BNS invariant as a subset of the character sphere,
for most of the paper we will consider the preimage of Σ in Hom(G; R). Let
p : (Hom(G; R) − {0}) → S
be the quotient map to the character sphere. We say that χ ∈ Hom(G; R) lies in
the complement of the BNS invariant if χ 6∈ p−1 (Σ). The complement of the BNS
invariant may then be viewed as a subspace arrangement in Hom(G; R).
Definition 4.3. Let G be group. We define VG to be the set of maximal subspaces
in Hom(G; R) contained in the complement of the BNS invariant. We define H∗ (VG )
to be the subspace arrangement homology H∗ (V, VG ), where our ambient space V
is always Hom(G; R).

16

MATTHEW B. DAY AND RICHARD D. WADE

More generally, we can consider the collection of all subspaces of Hom(G; R) in
p−1 (Σc ) ∪ {0}; Proposition 3.7 shows that this gives the same homology spaces as
the collection of maximal subspaces VG .
Remark 4.4. Recall that the BNS invariant Σ of a group G is symmetric if Σ = −Σ,
meaning that it is invariant under the antipodal map. In this case, each character
χ with [χ] ∈ Σc determines an entire line in p−1 (Σc ) ∪ {0}. As χ is contained
in some subspace of p−1 (Σc ) ∪ {0}, it is also contained in a maximal one. Hence
p−1 (Σc ) ∪ {0} is exactly the union of the elements of VG . Conversely, if Σ is not
symmetric then ∪VG is a proper subset of p−1 (Σc ) ∪ {0}. Even if Σ is symmetric
and Hom(G; R) is finite dimensional, as far as we know it is still possible for VG to
be an infinite family.
+
Remark 4.5. One can instead take the larger family VG
spanned by characters χ
which are realized by some exceptional action on an R-tree (in other words, either χ
or −χ lies in p−1 (Σc )). One can view VG as the arrangement obtained by removing
+
characters corresponding to Σ ∪ −Σ from V , whereas for VG
one only removes
characters corresponding to elements of Σ ∩ −Σ. When Σ is non-symmetric there
+
are examples where H∗ (VG , V ) and H∗ (VG
, V ) are different (such examples can be
found in [2, 5]).

4.1.1. Maps between groups. When f : G → H is a surjective homomorphism, an
exceptional abelian action of H on a tree induces an exceptional abelian action
of G. This does not change the location of the invariant end with respect to the
characters χ : H → R and f ∗ (χ) : G → R. Hence we have the following well-known
fact:
Proposition 4.6. Let f : G → H be a surjective map and
f ∗ : Hom(H; R) → Hom(G; R)
the induced map on character spaces. If χ ∈ Hom(H; R) is in the complement of the
BNS invariant of H, then f ∗ (χ) = χ ◦ f is in the complement of the BNS invariant
of G.
It follows that if f : G → H is surjective, then f induces a morphism of subspace
arrangements
f ∗ : (Hom(H; R), VH ) → (Hom(G; R), VG ).
This in turn gives a map (f ∗ )∗ : H∗ (VH ) → H∗ (VG ) on homology as described in
Section 3.3, although we will not need this in the work that follows.
To summarize, we have defined
G 7→ H∗ (VG ),
a contravariant functor from the category of groups with surjective homomorphisms
to the category of graded vector spaces over R. Such a thing superficially resembles
a cohomology theory of groups. It would be interesting to characterize this invariant
in terms of cohomology.
4.2. Right-angled Artin groups. Suppose that G is a right-angled Artin group
AΓ . For a vertex a of Γ, let χa : AΓ → R be the character defined on generators by
(
1 if v = a
χa (v) =
0 if v 6= a

SUBSPACE ARRANGEMENTS, BNS INVARIANTS, AND PSO(AΓ )

17

The abelianization of AΓ is a free abelian group generated by the images of the
vertices in H1 (AΓ ; Z) and the characters χa define a basis of Hom(G; R).
For any character χ ∈ Hom(AΓ ; R), we define the support supp(χ) to be the
full subgraph of Γ spanned by the vertices v such that χ(v) 6= 0. The support is
dominating if every vertex in Γ is either contained in, or adjacent to, a vertex in
supp(χ).
Theorem 4.7 (Meier–VanWyk, [15]). Let χ ∈ Hom(AΓ ; R) − {0}. Then [χ] ∈
Σ(AΓ ) if and only if supp(χ) is connected and dominating.
Proposition 4.8. The set VG of maximal subspaces in the complement of the BNS
invariant is the set of vector spaces of the form
VS = hχa : a ∈ Si
for each maximal subset S of vertices in Γ spanning a disconnected subgraph of Γ.
Proof. If Γ′ is a subgraph of Γ which is not dominating, then there is a vertex v ′
which is not adjacent to Γ′ , so that Γ′ ∪ v ′ spans a disconnected subgraph of Γ.
Hence every subgraph which is either disconnected or non-dominating is contained
in a maximal disconnected subgraph of Γ. Combining this with Theorem 4.7, the
support of every character that lies in the complement of the BNS invariant of AΓ
is contained in a maximal disconnected subgraph of Γ, hence lies in VS for some S
as above.

Corollary 4.9. Let G = AΓ be a right-angled Artin group. Then:
(
rank(Z(AΓ )) if n = 0
dim(Hn (VG )) =
0
if n > 0
Proof. Each element of VG is spanned by a subset of our basis for Hom(AΓ ; R). For
n ≥ 1, each Hn (VG ) is trivial by Proposition 3.11. We are then left to find
H0 (VG ) = Hom(G; R)/span(VG ).
A vertex a ∈ Γ lies in a disconnected full subgraph of Γ if and only if st(a) if not
equal to the whole of Γ. In other words, χa is contained in some element of VG
unless a is central in AΓ . It follows that dim(H0 (VG )) is equal to the rank of the
center of AΓ .

In particular, the Euler characteristic of H∗ (VG ) is equal to the rank of the center
of AΓ and is non-negative (cf. Theorem 4.13).
4.3. Pure symmetric automorphisms. Now suppose that G = PSA(AΓ ) and
let X be the standard generating set of G. For a ∈ Γ and K ∈ ∆b , we let χaK be
the character defined on generators by
(
a
b
1 if πK
= πL
b
χaK (πL
)=
0 otherwise
It follows from Toinet’s presentation that the abelianization of PSA(AΓ ) is a free
abelian group, and the standard generators map bijectively to a free generating
set. This means that each χaK is a well-defined element of Hom(PSA(AΓ ); R) and
the elements χx form a basis of Hom(PSA(AΓ ); R). As before, we may define the
support supp(χ) of a character χ to be the subset of the standard generating set X
a
consisting of all generators such that χ(πK
) 6= 0. Koban and Piggott characterize

18

MATTHEW B. DAY AND RICHARD D. WADE

elements of Σ(PSA(AΓ )) according to their support in a similar fashion to Meier–
VanWyk. They first define the following nice subsets of the standard generating
set:
Definition 4.10. A subset S ⊂ X is a p-set if
• for each vertex a of Γ, there is at most one partial conjugation in S with
multiplier a, and
a
• S has a nontrivial partition S = S1 ∪ S2 such that for every πK
∈ S1
b
and πL ∈ S2 , we have a ∈ L and b ∈ K (L and K are the dominating
components for the pair (a, b)).
A subset S ⊂ X is a δ-p-set if
• for each vertex a of Γ, there are exactly two or zero partial conjugations in
S with multiplier a, and
a
• S has a nontrivial partition S = S1 ∪ S2 such that for every πK
∈ S1 and
b
πL in S2 , we have a ∈ L or b ∈ K or K = L (so L is the dominating
component [a]b or K is the dominating component [b]a or K and L are the
same shared component).
The p-sets here give exceptional characters similar to those occurring in RAAGs,
whereas the δ-p-sets only appear when Γ has an SIL-pair. The complement of the
BNS invariant of PSA(AΓ ) can be characterized as follows:
Theorem 4.11 (Koban–Piggott, [14]). Let χ : PSA(AΓ ) → R be nonzero character.
Then χ is in the complement of the BNS invariant if and only if
• χ is nontrivial on some inner automorphism and the support of χ is a subset
of a p-set, or
• χ is trivial on every inner automorphism and the support of χ is a subset
of a δ-p-set.
In the second case of the above theorem, as χ is trivial on every inner automora
a
a
a
phism, it follows that χ(πK
) = −χ(πL
) for each pair of elements πK
, πL
with the
same multiplier in its associated δ-p-set. This gives enough information to describe
VG .
Proposition 4.12. Let G = PSA(AΓ ) and let VG be the set of maximal subspaces
in the complement of the BNS invariant. For each maximal p-set S ⊂ X there is a
subspace VS ∈ VG given by
a
VS = h{χaK : πK
∈ S}i

and for each maximal δ-p-set S ⊂ X there is a subspace VS ∈ VG of the form
a
a
VS = h{χaK − χaL : πK
, πL
∈ S}i

Furthermore, each element of VG is one of these two types.
Koban and Piggott used this description to take an alternating sum of dimensions
of intersections of spheres in Σc (G). Intersections of spheres in Σc (G) correspond
to intersection of subspaces in VG . Using our terminology, we rephrase their result
as follows:
Theorem 4.13 (Koban–Pigott [14]). Let G = PSA(AΓ ). If Γ contains no separating intersection of links then the Euler characteristic of H∗ (VG ) is zero. Otherwise,
the Euler characteristic of H∗ (VG ) is strictly negative.

SUBSPACE ARRANGEMENTS, BNS INVARIANTS, AND PSO(AΓ )

19

4.4. Pure symmetric outer automorphisms. We now turn our attention to
PSO(AΓ ). Let
f : PSA(AΓ ) → PSO(AΓ )
be the quotient map, and let
f ∗ : Hom(PSO(AΓ ); R) → Hom(PSA(AΓ ); R)
be the dual map on characters given by f ∗ (χ) = χ ◦ f . As f is surjective, the map
f ∗ is injective, with image given by the characters χ ∈ Hom(PSA(AΓ ); R) that are
trivial on the inner automorphisms. In other words, if ∆a is the support graph for
some vertex a ∈ Γ, we have
X
a
χ(πK
) = 0.
K∈∆a

We identify Hom(PSO(AΓ ); R) with this subspace of Hom(PSA(AΓ ); R). This allows us to talk about the support of a character on PSO(AΓ ); it is the support of
of the character on PSA(AΓ ) we get by composing with the projection f .
To proceed, we need the following well-known fact, which is stated in [14]. We
do not give a proof here.
Lemma 4.14 ([14]). If χ : G → R is a nontrivial character on a group G that
factors through a surjective map G → A ∗ B, where A ∗ B is a nontrivial free
product, then [χ] is in the complement of the BNS invariant of G.
Proposition 4.15. Let AΓ be a RAAG and let χ : PSO(AΓ ) → R be nonzero. The
class [χ] is not in the BNS invariant if and only if the support of χ is a subset of
a δ-p-set.
Proof. By Proposition 4.6, if χ is in the complement of the BNS invariant of
PSO(AΓ ), then f ∗ (χ) is in the complement of the BNS invariant of PSA(AΓ ).
Since f ∗ (χ) is in the image of f ∗ , it is trivial on every inner automorphism. Then
by Theorem 4.11 it has a support which is a subset of a δ-p-set.
Conversely, given any character χ whose support is a δ-p-set, we need to show
that [χ] is not in Σ. Following [14], we find a surjection φ : PSO(AΓ ) → A1 ∗ A2 to
a nontrivial free product which χ factors through. By Lemma 4.14, it will follow
that [χ] ∈ Σc . Let S be the δ-p-set which is the support of χ, and let S1 ∪ S2
be a partition of S given in Definition 4.10. Each multiplier a that appears in
a
a
S has two elements πK
and πK
, both of which lie on one side of the partition
1
2
a
a
S1 ∪ S2 . Furthermore, χ(πK1 ) = −χ(πK
). Let A1 be the free abelian group on the
2
multipliers that appear in S1 and A2 the free abelian group on the set of multipliers
a
that appear in S2 . We map PSO(AΓ ) to A1 ∗ A2 by sending [πK
] to a, sending
1
a
[πK2 ] to −a, and every other generator with multiplier a to the trivial element. If
b is a multiplier that occurs on the other side of the partition with corresponding
b
b
a
b
elements πL
and πL
, then each commutator [πK
, πL
] is nontrivial in PSO(AΓ ).
1
2
i
j
Furthermore, one can check that the map to A1 ∗ A2 respects all relations in the
presentation of PSO(AΓ ) and is therefore well-defined. Hence [χ] ∈ Σc .

Corollary 4.16. Let G = PSO(AΓ ) and let VG be the set of maximal subspaces
in the complement of the BNS invariant. The family VG consists exactly of the
subspaces of the form
a
a
VS = h{χaK − χaL : πK
, πL
∈ S}i

for each maximal δ-p-set S.

20

MATTHEW B. DAY AND RICHARD D. WADE

Our next goal is to show that H1 (VG ) is nontrivial for G = PSO(AΓ ) under certain conditions. To do this, we build a cycle and show that it represents a nontrivial
homology class. As is often the case with homology theories, it is convenient to do
this by pairing our cycle with a cocycle.
We do not give a full treatment of a cohomology theory of subspace arrangements
here. However, we make the following definition: for a subspace arrangement (V, V)
over K, we define
C ∗ (V, V) = Hom(C∗ (V, V), K),
and for f ∈ C n (V, V), define df = f ◦ ∂. This is a cochain complex and we define
cocycles, coboundaries and cohomology as usual.
This means that a 1-cochain f in C 1 (V, V) is determined by a family {fW }W ∈V
of linear functionals
on each subspace; each fW is the restriction of f to the W L
summand of
V = C1 (V, V). Such a collection of functionals determines a cocycle
if, for any two subspaces W1 and W2 , the linear maps fW1 and fW2 agree on W1 ∩W2
(this is easily seen to be equivalent to f ◦ ∂ = 0). The cocycle f represents the
trivial cohomology class if and only if there exists a linear functional f˜: V → R
such that each fW is the restriction of f˜ to W (this is the same as saying that
f = f˜ ◦ ∂).
L
Suppose c = (cW )W ∈V is a 1-chain in C1 (V, V) =
V. If the 1-cocycle
f is exP
pressed as a family of functionals {fW }W ∈V , then f (c) is the sum W ∈V fW (cW ).
As usual, the evaluation of 1-cocycles on 1-cycles descends to a well defined evaluation of cohomology classes on homology classes. In particular, if c is a 1-boundary,
then f (c) = 0 for any 1-cocycle f . So if f (c) 6= 0 for some cocycle, then c represents
a nontrivial homology class.
Proposition 4.17. Suppose AΓ is a RAAG such that for some vertex a ∈ Γ,
the support graph ∆a contains a loop. Let G = PSO(AΓ ) and let VG be the excluded subspace configuration for the BNS invariant of G in V = Hom(G; R). Then
H1 (VG ) 6= 0.
Proof. Let (K1 , . . . , Kn ) be a loop in ∆a involving n ≥ 3 distinct vertices. By the
definition of ∆a , for each i (from 1 to n and counting modulo n), there is an element
bi such that either (1) bi ∈ Ki , and Ki+1 is a shared component of bi and a; or
(2) bi ∈ Ki+1 , and Ki is a shared component of bi and a. This implies that either
bi
bi
bi
bi
a
a
a
a
{πK
, πK
, πK
} ∪ {π[a]
} or {πK
, πK
, πK
} ∪ {π[a]
} is a δ-p-set. Each of these
i
i+1
i
i+1
i+1
i
sets is contained in a maximal δ-p-set. So for i = 1, . . . , n, let Si be a maximal
a
a
δ-p-set with πK
, πK
∈ Si , and for i = n + 1, . . . , m, let {Si }i label the remaining
i
i+1
maximal δ-p-sets in any order. Let Vi be the span of Si for i = 1, . . . , m; then
Corollary 4.16
Lm says that VG = {V1 , V2 , . . . , Vm }. We build the following element of
C1 (VG ) = i=1 Vi :
x = (χaK1 − χaK2 , χaK2 − χaK3 , . . . , χaKn−1 − χaKn , χaKn − χaK1 , 0, . . . , 0) ∈ C1 (VG ).

This x a cycle for H1 (VG ), since the sum of its components is zero.
To show that x represents a nontrivial homology class, we build a cocycle. Define
a set T ⊂ {1, . . . , m} by

a
a
} ⊂ Si .
, πK
T = i ∈ {1, . . . , m} : {πK
2
1

We define functionals fVi : Vi → R for i = 1, . . . , m as follows: if i ∈ T , then
a
fVi (χ) = χ(πK
) for χ ∈ Vi ; if i ∈
/ T , then fVi = 0.
1

SUBSPACE ARRANGEMENTS, BNS INVARIANTS, AND PSO(AΓ )

21

To show that these functionals patch together to form a cocycle f , we need to
show that they agree on the intersections of their domains. Let 1 ≤ i < j ≤ m.
If both i and j are in T , or both i and j are not in T , then clearly fVi |Vi ∩Vj =
/ T . Let χ ∈ Vi ∩ Vj . By Proposition 4.6
fVj |Vi ∩Vj . So suppose that i ∈ T and j ∈
and Theorem 4.11, supp(χ) contains exactly zero or two standard generators with
multiplier a. Since χ ∈ Vj , we know supp(χ) ⊂ Sj , so supp(χ) does not contain
a
a
both πK
and πK
. But χ ∈ Vi and i ∈ T , so if supp(χ) ⊂ Si contains a standard
1
2
a
a
a
a
generator with multiplier a it contains both πK
and πK
with χ(πK
) = −χ(πK
).
1
2
1
2
Therefore supp(χ) does not contain any generators with multiplier a. This means
a
/ T is identical,
that fVi (χ) = χ(πK
) = 0 = fVj (χ). The case where j ∈ T and i ∈
1
so we have that the {fVi }i agree on all pairwise intersections of spaces from VG .
This means that these functionals patch together to form a cocycle f in C 1 (VG ).
Now it is enough to show that f (x) 6= 0. By our numbering of S1 , . . . , Sm , we
know that T ∩ {1, . . . , n} = {1}, and the (n + 1)st through mth components of x
a
are 0. So f (x) = (χaK1 − χaK2 )(πK
) = 1. Hence H1 (VG ) 6= 0.

1
5. Finding a graphical RAAG presentation for PSO(AΓ ).
We now give a right-angled Artin presentation for PSO(AΓ ) when all support
graphs are forests. We will be working with outer automorphism classes of elements
throughout, however for ease of reading we suppress the bracket notation and write
a
a
elements as πK
∈ PSO(AΓ ) rather than [πK
].
5.1. An alternative generating set for PSO(AΓ ). Throughout this section we
suppose that each support graph ∆a is a forest with ka + 1 maximal subtrees
(connected components) C0a , . . . , Ckaa . Since the vertices of ∆a represent connected
components of Γ − st(a), and ∆a has its own connected components, we usually
refer to the connected components of ∆a as maximal subtrees to avoid confusing
repetition of the term “component”. We pick a basepoint xai in each tree Cia . We
say that xa0 is the preferred basepoint of the forest ∆a .
We need two kinds of generators for our generating set for PSO(AΓ ). We use
a set of partial conjugations that are not necessarily standard generators We have
already introduced the first kind. Suppose C = Cia is a maximal subtree of ∆a .
a
Recall that ζC
denotes be the product
Y
a
a
ζC
=
πK
K∈C

over all elements K of the vertex set of C (each K is a connected component of
Γ − st(a)). These elements are central in PSO(AΓ ) by Proposition 2.9.
We also introduce an element ηea associated to each edge in ∆a .

Definition 5.1. Let e be an edge in a maximal subtree C of ∆a with basepoint
x ∈ C. The interior of the edge e separates C into two pieces. Let L be the
component of C − e which does not contain the basepoint x. We define
Y
a
ηea =
πK
K∈L

The choice of basepoint gives a uniform way of choosing a component of C − e,
however this choice does not matter too much, at least in terms of commuting
elements in PSO(AΓ ):

22

MATTHEW B. DAY AND RICHARD D. WADE

Lemma 5.2. Let L′ be the component of C − e which contains the basepoint of C,
and let
Y
a
(ηea )′ =
πK
.
K∈L′

Then ηea (ηea )′ is central in PSO(AΓ ). In particular an element commutes with ηea if
and only if it commutes with (ηea )′ .
Proof. We simply observe that, as V (L) ∪ V (L′ ) = V (C):
a
ηea (ηea )′ = ζC
,

which is central, by Proposition 2.9. Since (ηea )′ is the product of ηea with a central
element, we see that anything that commutes with ηea also commutes with (ηea )′ .
By symmetry, they have exactly the same centralizers.

Although elements of the form ηea are not central in PSO(AΓ ), there is quite a
strong requirement for a commutator of the form [ηea , ηfb ] to be nonzero.
Proposition 5.3. Let e and f be edges of ∆a and ∆b respectively. Then ηea and
ηfb commute unless:
• (a, b) is an SIL pair and
• the edges e and f are of the form {[b]a , L} and {[a]b , L}, where L is a shared
component of (a, b).
Proof. Suppose that [ηea , ηfb ] 6= 1. If (a, b) do not form an SIL-pair then ηea and
a
b
and πL
commute. We may
ηfb commute as all standard generators of the form πK
therefore assume that (a, b) is an SIL-pair. Let C be the maximal subtree of ∆a
containing the dominating component [b]a , and let D be the maximal subtree of
∆b containing [a]b . The elements ηea and ηfb will commute unless e ∈ C and f ∈ D,
as otherwise one of the products ηea or ηfb will consist of standard generators only
corresponding to subordinate components for (a, b). Let C ′ ⊂ C be the star of
[b]a in ∆a ; so C ′ contains [b]a together with all the shared components of (a, b).
Suppose for contradiction that e is not an edge of C ′ ; then one component L or L′
of ∆a − e is disjoint from C ′ and contains only vertices of subordinate components.
Hence ηfb commutes with either ηea or (ηea )′ , so by Lemma 5.2, it commutes with ηea .
This contradicts our hypothesis, so e must be an edge of C ′ . The same argument
applies with the location of f in D. It follows that both e and f are of the form
{[b]a , L} and {[a]b , L′ } respectively, where L and L′ are shared components for
(a, b). Lemma 5.2 allows us to assume that the component of C − e (respectively
D − f ) which does not contain the basepoint is the one containing L (respectively
L′ ), so that
Y
Y
a
a
b
b
ηea = πL
πK
and ηfb = πL
πK
′
′
a
πK

K
b
πK
′)

K′

where each
(respectively
in the product is subordinate for the pair (a, b).
As partial conjugations along distinct shared components commute, it follows that

L = L′ when [ηea , ηfb ] 6= 1.

5.2. The right-angled Artin presentation. We are now in a position to give
an explicit right-angled Artin presentation for the group PSO(AΓ ).
Definition 5.4. Let AΘ be the right-angled Artin group with defining graph Θ
given by vertices of the form:

SUBSPACE ARRANGEMENTS, BNS INVARIANTS, AND PSO(AΓ )

23

• vea for each vertex a ∈ Γ and each edge e in ∆a .
a
• vC
for each vertex a ∈ Γ and each maximal subtree C of ∆a not equal to
the tree C0a containing the preferred basepoint.
The graph Θ is given the following edges:
a
• There is an edge between each vertex vC
and every other vertex in Θ.
a
b
• There is an edge between ve and vf unless (a, b) forms an SIL-pair and
e = {[b]a , L} and f = {[a]b , L} for some shared component L of (a, b).
Note that the definition of AΘ depends on the location of the preferred basepoints but is independent of the remaining basepoints. Propositions 2.9 and 5.3
immediately imply the following:
a
a
Proposition 5.5. The map on generators given by φ(vC
) = ζC
and φ(vea ) = ηea
induces a homomorphism φ : AΘ → PSO(AΓ ).

5.3. Constructing an inverse map. To show that PSO(AΓ ) ∼
= AΘ we will construct an inverse map ψ : PSO(AΓ ) → AΘ . The first step is to write each standard
a
a
generator πK
as a product of elements of the form ζC
and ηea .
a
Lemma 5.6. Let πK
be a standard generator of PSO(AΓ ). Let C0 , . . . , Ck be the
set of maximal trees in the support graph ∆a , and let e0 , . . . , en be the edges of the
support graph adjacent to K.
(1) If K is not the basepoint of its subtree in ∆a and e0 is the edge adjacent to
K in the direction of the basepoint then
a
πK
= ηea0 (ηea1 )−1 · · · (ηean )−1 .

(2) If K is the basepoint of some tree C 6= C0 then
a
a
πK
= ζC
(ηea0 )−1 (ηea1 )−1 · · · (ηean )−1 .

(3) If K is the preferred basepoint of ∆a , then:
a −1
a
a −1
) · · · (ζC
) · (ηea0 )−1 (ηea1 )−1 · · · (ηean )−1 .
πK
= (ζC
1
k

Proof. We explain the proof of item (1). Let C be the maximal subtree of ∆a
containing K and let Li be the component of C − ei disjoint from the basepoint.
The vertex set of L0 is the disjoint union of {K} with the vertex sets of the Li for
i ≥ 1. Equation (1) then follows from the definition of ηeai . A similar calculation
applies to cases (2) and (3).

Corollary 5.7. The homomorphism φ : AΘ → PSO(AΓ ) is surjective.
Proof. In Lemma 5.6, we wrote each element of the standard generating set as a
product of elements in the image of φ.

Lemma 5.6 gives an obvious candidate for an inverse map.
Definition 5.8. With e0 , . . . , en and C0 , . . . , Ck as in Lemma 5.6, let:

a −1
a a −1

if K is not a basepoint of ∆a
ve0 ve1 · · · ven
a
−1
−1
a
a
a
ψ(πK ) = vC ve0 · · · ven
if K is a basepoint but not preferred

 a −1
a −1
a −1
a −1
vC1 · · · vC
if
K is the preferred basepoint of ∆a
·
·
·
v
·
v
en
e0
k

This defines a map ψ : PSO(AΓ ) → AΘ .

24

MATTHEW B. DAY AND RICHARD D. WADE

We owe the reader a proof that this map, as defined on generators, extends to
a well defined homomorphism. The following lemma reduces the number of cases
which we need to run through:
a
Lemma 5.9. Let πK
be a standard generator of PSO(AΓ ) and let e0 , . . . , en be the
edges in ∆a adjacent to K. If K is not a basepoint of its tree in ∆a we assume
that e0 is the edge in the direction of the basepoint. There exists a central element
g ∈ AΘ such that
a
ψ(πK
) = g · (vea0 )ǫ (vea1 )−1 · · · (vean )−1 ,
where ǫ ∈ {1, −1}. If K is a basepoint in ∆a then ǫ = 1, otherwise ǫ = −1.
a
Proof. This follows from the definition of ψ and the fact that each element vC
is
central in AΘ .

a
b
Lemma 5.10. Suppose that [ψ(πK
), ψ(πL
)] 6= 1. Then (a, b) forms an SIL-pair
and either:
• K and L are both dominating for the pair (a, b), or
• K is dominating for the pair (a, b) and L is shared, or
• L is dominating for the pair (a, b) and K is shared, or
• K = L is a shared component for the pair (a, b).

Proof. Let
′

b
a
) = g ′ .(vfb0 )ǫ (vfb1 )−1 · · · (vfbm )−1
ψ(πK
) = g.(vea0 )ǫ (vea1 )−1 · · · (vean )−1 and ψ(πL
a
b
be the decompositions of ψ(πK
) and ψ(πL
) respectively given by Lemma 5.9. If
these two elements do not commute in AΘ , then as g and g ′ are central, the elements
′

(vea0 )ǫ (vea1 )−1 · · · (vean )−1 and (vfb0 )ǫ (vfb1 )−1 · · · (vfbm )−1
also do not commute in AΘ . In particular there exist i and j such that veai and vfbj
do not commute in AΘ . From the definition of AΘ , this implies that (a, b) is an
SIL-pair and ei = {[b]a , L′ }, fj = {[a]b , L′ } for some shared component L′ of (a, b).
As K is an endpoint of ei and L is an endpoint of fj , one of the four cases listed
above must hold.

Proposition 5.11. The map ψ : PSO(AΓ ) → AΘ as defined on generators extends
to a well defined homomorphism.
Proof. We need to check the relations (R1)–(R5) in Corollary 2.6 are sent to the
identity under the induced map from the free group on the standard generators of
a
b
PSO(AΓ ) to AΘ . The relations in (R1)–(R3) are commutators [πK
, πL
] corresponding to the following situations:
• The commutator [a, b] = 1, so in particular (a, b) is not an SIL-pair.
• The components K and L are disjoint and non-dominating for the pair
(a, b).
• One component is dominating and the other is subordinate for the pair
(a, b).
a
b
By Lemma 5.10, in all three situations we have [ψ(πK
), ψ(πL
)] = 1.
b
a a
The relations in (R4) are of the form [πK πL , πL ], where (a, b) is an SIL-pair, K
is dominating, and L is shared for the pair (a, b). Let e be the edge e = {K, L} in
the support graph ∆a . Let x be the basepoint of the component of ∆a containing
K and L. If x is closer to K (including x = K) then vea occurs with exponent −1

SUBSPACE ARRANGEMENTS, BNS INVARIANTS, AND PSO(AΓ )

25

a
b
in the decomposition of ψ(πK
) and exponent +1 in the decomposition of ψ(πL
).
a
Similarly, if x is closer to L (including x = L) then ve occurs with exponent −1 in
a
a
) and with exponent +1 in the decomposition of ψ(πK
).
the decomposition of ψ(πL
In either case, this term cancels out in any reduced word representing the product
a
a
ψ(πK
)ψ(πL
). More precisely, one can show that there exists a central element g in
AΘ such that:
Y
a
a
ψ(πK
)ψ(πL
) = g (veai )−1 ,
ei

where this product is taken over all edges ei 6= e in ∆a adjacent to either K or L.
b
In contrast, the image of πL
in AΘ is of the form
b
ψ(πL
) = g ′ .(vfb0 )ǫ (vfb1 )−1 · · · (vfbm )−1 ,

where g ′ is central and f0 , . . . , fm are the edges adjacent to L in ∆b . As L is
shared, the Star Lemma tells us that one of these edges fi = {L, [a]b } has the
dominating component [a]b as its other vertex, and the remaining edges are of the
form {L, L′ }, where L′ is a subordinate component of (a, b) (this uses the fact that
a
a
∆a is a forest). As vea does not occur in our decomposition of ψ(πK
)ψ(πL
), it
follows that [veai , vfbj ] = 1 in AΘ for all ei and fi in the above two decompositions.
a
a
b
Hence ψ(πK
)ψ(πL
) and ψ(πL
) commute in AΘQ
.
a
. We want to show that
Finally, each relation in (R5) is of the form K∈∆a πK
Y
a
ψ(πK
) = 1.
K∈∆a

a
ψ(πK
),

From the definition of
one can check that each element vea occurs with
exponents +1 and −1 exactly once each in the above product (corresponding to the
images of the generators given by the endpoints of the edge e under ψ). Similarly, if
a
C0 , . . . , Ck are the components of ∆a and i ≥ 1 then vC
also occurs with exponents
i
+1 and −1 exactly once (the exponent +1 appears in the image of the element given
by the basepoint of Ci , and the exponent −1 appears in the image of the generator
corresponding to our
Q preferred abasepoint). As all the above elements commute in
) = 1.

AΘ , it follows that K∈∆a ψ(πK
Theorem 5.12. The group PSO(AΓ ) is isomorphic to AΘ .

Proof. It only remains to show that φ and ψ are mutual inverses. The fact that
a
a
a
φ(ψ(πK
)) = πK
follows directly from the definitions and Lemma 5.6. If vC
is a
generator of AΘ then
Y
a
a
ψ(φ(vC
)) =
ψ(πK
)
K∈C

a
The element vC
occurs exactly once with exponent +1 under the image of the
standard generator πxa corresponding to the basepoint of C. If e ∈ C, the generator
vea occurs twice in the above product (corresponding to the two endpoints of e),
once with exponent +1 and once with exponent −1. As all these elements commute,
a
a
ψ(φ(vC
)) = vC
. Similarly
Y
a
ψ(φ(vea )) =
ψ(πK
),
K∈L

where L is the component of C − e which does not contain the basepoint. The
element vea occurs once in this product with exponent +1 in the image of the
standard generator given by the one endpoint of e which is contained in L. Every

26

MATTHEW B. DAY AND RICHARD D. WADE

edge e′ ∈ L then occurs twice, once with exponent +1 and once with exponent −1.
These appear in the images of the generators corresponding to the endpoints of e′ ,
both of which lie in L. It follows that ψ(φ(vea )) = vea . Hence the compositions φ ◦ ψ
and ψ ◦ φ are the identity maps on PSO(AΓ ) and AΘ respectively, so that both
maps are isomorphisms.

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