The symmetries of McCullough-Miller space
We prove that if W is the free product of at least four groups of order 2, then the automorphism group of the McCullough-Miller space corresponding to W is isomorphic to group of outer automorphisms of W. We also prove that, for each integer n ≥ 3, the automorphism group of the hypertree complex of...
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| Zitieren: | The symmetries of McCullough-Miller space / A. Piggott // Algebra and Discrete Mathematics. — 2012. — Vol. 14, № 2. — С. 239–266. — Бібліогр.: 7 назв. — англ. |
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| citation_txt | The symmetries of McCullough-Miller space / A. Piggott // Algebra and Discrete Mathematics. — 2012. — Vol. 14, № 2. — С. 239–266. — Бібліогр.: 7 назв. — англ. |
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| description | We prove that if W is the free product of at least four groups of order 2, then the automorphism group of the McCullough-Miller space corresponding to W is isomorphic to group of outer automorphisms of W. We also prove that, for each integer n ≥ 3, the automorphism group of the hypertree complex of rank n is isomorphic to the symmetric group of rank n.
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h.Algebra and Discrete Mathematics RESEARCH ARTICLE
Volume 14 (2012). Number 2. pp. 239 – 266
c© Journal “Algebra and Discrete Mathematics”
The symmetries of McCullough-Miller space
Adam Piggott1
Communicated by R. I. Grigorchuk
Abstract. We prove that if W is the free product of at
least four groups of order 2, then the automorphism group of the
McCullough-Miller space corresponding to W is isomorphic to group
of outer automorphisms of W . We also prove that, for each integer
n ≥ 3, the automorphism group of the hypertree complex of rank n
is isomorphic to the symmetric group of rank n.
1. Introduction
A simplicial complex K is a geometric model for a group G if there
exists a homomorphism m : G → Aut(K), where Aut(K) denotes the
group of simplicial automorphisms of K—in other language, we would say
that K is equipped with a G-action. The smaller the kernel of m, the less
the model simplifies G; in the best case, m is injective, and the model
represents G precisely as a subgroup m(G) of Aut(K). The larger m(G)
in Aut(K), the greater the expectation that Aut(K) in its entirety, rather
than the subgroup m(G), can offer insights into G; in particular, it is
natural to believe that a model is better if m(G) is large (say, of finite
index) in Aut(K), and best if m(G) = Aut(K). Following [2], we say that
1Thanks to Murray Elder, and the University of Newcastle, Australia, for their
hospitality as this paper was written. Thanks to Andy Eisenberg, and the anonymous
referee, for carefully reading the paper, and suggesting improvements.
2010 MSC: 20E36; 05E18.
Key words and phrases: Autmorphisms of groups; group actions on simplicial
complexes; Coxeter groups; McCullough-Miller space; hypertrees.
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240 The symmetries of McCullough-Miller space
K is an accurate geometric model of G if there exists an isomorphism
m :G → Aut(K).
For each positive integer n, we write Wn for the universal Coxeter
group of rank n; that is, Wn is the free product of n groups of order two,
as presented 〈a1, . . . , an | a2
1, . . . , a2
n〉. We write Out(Wn) for the group of
outer automorphisms of Wn.
For each n ≥ 3, Out(Wn) is the outer automorphism group of the most
simple free product with n factors. The group Out(Wn) is related to, but
much more simple in structure than, Out(Fn−1) (see, for example, [3]),
where Fm denotes the free group of rank m. Even so, there are a number
of questions one may ask about a group which have been answered for
Out(Fn), but not for Out(Wn). In particular, one may wish to identify
an accurate geometric model for a group of interest. In [2], Bridson and
Vogtmann showed that if n ≥ 3, then the spine of the appropriate outer
space is an accurate geometric model for Out(Fn). In the present article
we prove that, provided n ≥ 4, a well-known geometric model of Out(Wn)
is in fact an accurate geometric model.
Given a group G and a fixed free-product decomposition of G, the
corresponding McCullough-Miller space K(G) is a contractible simplicial
complex which is a geometric model for the group of symmetric outer
automorphisms of G [6]—those outer automorphisms of G which map each
free factor in the fixed decomposition of G to a conjugate of a free-factor.
Further, it is known that the modeling homomorphism is injective. In
the present article we consider the case that G = Wn, equipped with the
canonical decomposition; we write Kn for the corresponding McCullough-
Miller space. In this case, each outer automorphism of Wn is a symmetric
outer automorphism, so Kn is a geometric model for Out(Wn). We show
that the modeling homomorphism is surjective, and hence prove the
following.
Theorem 1.1. For each integer n ≥ 4, the McCullough-Miller space Kn
is an accurate geometric model of Out(Wn).
Remark 1.2. The hypothesis n ≥ 4 is necessary because: Out(W3) is
finitely generated, and hence it is countably infinite; K3 is the barycentric
subdivision of the regular trivalent tree, and hence Aut(K3) is uncountably
infinite.
In general, a McCullough-Miller space is constructed by gluing together
copies of a finite complex HTn, called the hypertree complex of rank n,
which is the simplicial realization of a poset (HT n, ≤), called the hypertree
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A. Piggott 241
poset of rank n (see Remark 1.5 below). The poset (HT n, ≤) is the set
of hypertrees on n labeled vertices, partially-ordered by the operation
of folding. It is well-known, and easily seen, that HTn is a geometric
model for Σn, the symmetric group of rank n. As part of our proof of
Theorem 1.1, and for its independent interest, we prove the following.
Theorem 1.3. For each integer n ≥ 3, the hypertree complex HTn is an
accurate geometric model of the symmetric group Σn.
Remark 1.4. The hypothesis n ≥ 3 is necessary because there is only
one hypertree on 2 labeled vertices, so Aut(HT2) is the trivial group, and
it is not isomorphic to Σ2.
Remark 1.5. In McCullough and Miller’s original account [6] of the
construction now named for them, the hypertree complex is not explicitly
used. In its place is used a complex called the Whitehead complex, which
is the simplicial realization of a poset called the Whitehead poset. As
explained in [5], the Whitehead poset is isomorphic to the hypertree poset,
and thus the corresponding simplicial realizations are interchangeable in
the construction of McCullough-Miller space.
We now describe the structure of the paper, and proofs. In Section 2
we describe the hypertree poset, and a number of its subsets. In Section 3
we prove Theorem 1.3. Our argument proceeds by: identifying a subset of
HT n, the set of “star trees”, on which Σn acts as the full permutation
group; observing that the corresponding subset of vertices in the hypertree
complex is geometrically distinguishable, and hence must be fixed setwise
by an arbitrary simplicial automorphism; and showing that every other
vertex in the hypertree complex is uniquely identified either by its relative
proximities to vertices corresponding to star trees, or to vertices which
can be so identified.
In Section 4 we describe the construction of Kn and prove Theorem 1.1.
To do so we consider an arbitrary simplicial automorphism f of Kn. We
argue that: since Out(Wn) acts transitively on the copies of HTn from
which Kn is built, and Σn is the full automorphism group of HTn, there
exist elements σ, φ ∈ Out(Wn) such that the actions of σφ and f agree
pointwise on one of the copies of HTn. We then establish that overlapping
copies of HTn are sufficiently intertwined that if one copy is fixed pointwise
by a simplicial automorphism, overlapping copies are fixed pointwise too.
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242 The symmetries of McCullough-Miller space
2. The hypertree complex
In this section we introduce the hypertree complex. The interested
reader may find an alternative account of the hypertree complex in [5].
2.1. Hypertrees
A hypergraph Γ is an ordered pair (VΓ, EΓ) consisting of a set of
vertices VΓ, and a collection (often a set) EΓ of hyperedges, each of which
is a subset of VΓ containing at least two elements. When we want to
emphasize the vertex set of Γ, we say Γ is a hypergraph on VΓ. A graph
(without loops) is a hypergraph in which each hyperedge contains exactly
two vertices.
Hypergraphs Γ = (VΓ, EΓ) and Γ′ = (V ′
Γ, E′
Γ) are isomorphic as
unlabeled hypergraphs if there exists a bijection f : VΓ → V ′
Γ such that
for each subset S ⊂ VΓ, f(S) ∈ E′
Γ if and only if S ∈ EΓ; in this case
f is called a hypergraph isomorphism. Hypergraphs Γ = (VΓ, EΓ) and
Γ′ = (VΓ′ , EΓ′) are isomorphic as labeled hypergraphs if VΓ = VΓ′ , and the
identity map VΓ → VΓ is a hypergraph isomorphism Γ → Γ′. We shall
usually consider hypergraphs up to labeled-hypergraph isomorphism.
Let Γ be a hypergraph. Distinct vertices v, v′ ∈ VΓ are said to be
adjacent in Γ if {v, v′} ⊂ e for some hyperedge e ∈ EΓ. The valence in Γ
of a vertex v ∈ VΓ is the number of hyperedges containing v. A vertex
with valence one is called a leaf. The degree in Γ of a hyperedge e ∈ EΓ
is #e, the number of vertices it contains. In general, we shall write #S
for the cardinality of a set S.
Given a hypergraph Γ, and vertices v, v′ ∈ VΓ, a walk in Γ from v to
v′ is a diagram
v = v0
e1−→ v1
e2−→ . . .
ep
−→ vp = v′
with: p ≥ 0; each vk a vertex; each ek an edge; and vi−1 6= vi and
{vi−1, vi} ⊂ ei for each i = 1, . . . , p. Such a walk is said: to visit the
vertices v0, . . . , vp; to join the vertices v and v′; and to cross the hyperegdes
e1, . . . , ep. Such a walk is simple if the vertices v0, . . . , vp−1 are distinct,
and the edges e1, . . . , ep are distinct.
We say a hypergraph Θ is: connected if for each pair of vertices
v, v′ ∈ VΘ, there is at least one simple walk from v to v′; and a hypertree
if for each pair of vertices v, v′ ∈ VΘ, there is exactly one simple walk
from v to v′. It follows immediately that if Θ is a hypertree, then: Θ is
connected; the intersection of two or more distinct hyperedges contains
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A. Piggott 243
2
1
3
4
5
6
Γ
Figure 1. Γ is a hypergraph but not a hypertree, Θ is a hypertree.
at most one vertex; and the collection EΘ of edges is a set. A hypertree
Θ is a tree if it is a graph.
Remark 2.1. A hypergraph Γ may be represented as a labeled bipartite
graph B(Γ) as follows: each vertex in VΓ is a labeled vertex of B(Γ); each
hyperedge in EΓ is an unlabeled vertex of B(Γ); an unlabeled vertex u is
adjacent to a labeled vertex ℓ if ℓ ∈ u. Then Γ is a hypertree if and only
if B(Γ) is a tree. Thus there is a bijective correpondence between the
set of hypertrees on a set S, and the set of labeled bipartite trees with
labeled vertices in bijective correspondence with S. Two hypergraphs, one
of which is a hypertree, and the corresponding labeled bipartite graphs
are shown in Figure 1
2.2. The Hypertree Complex
For each positive integer n we write: [n] for the set {1, . . . , n}; and
HT n for the set of hypertrees on [n], considered up to labeled-hypergraph
isomorphism.
Remark 2.2. A general formula for #HT n, the number of hypertrees
on [n], was calculated in [4] and [7]. The sequence (#HT n) begins:
1, 1, 4, 29, 311, 4447, 79745, . . .
More terms of the sequence, and the general formula for #HT n, can be
found in the On-Line Encyclopedia of Integer Sequences [1, Sequence
A030019].
Example 2.3. The four hypertrees of HT 3 are depicted in Figure 2. The
29 hypertrees of HT 4 are depicted in Figure 4.
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244 The symmetries of McCullough-Miller space
2 1 3 1 2 3 1 3 2
1
3
2
Figure 2. (HT 3, ≤). Figure 3. HT3.
1 2 3 4
1 2 4 3
2 1 3 4
2 1 4 3
1 3 2 4
1 3 4 2
3 1 2 4
3 1 4 2
1 4 2 3
1 4 3 2
4 1 2 3
4 1 3 2
1
2
3
41
2
4
3
2
3
4
1 1
3
4
2
1
2
34
1
2
43
3
4
2 1
3
4
1 2
1
3
34
1
3
42
3
4
3 1
2
4
1 3
1
4
23
1
4
32
2
3
4 1
2
3
1 4
1
2
3
4
Figure 4. The 29 hypertrees in HT 4.
For the remainder of this section we fix an integer n ≥ 3.
There is a partial order ≤ on HT n, determined by an operation called
folding. Given hypertrees Θ, Θ′ ∈ HT n, we say Θ′ is obtained from Θ
by a single fold if there exist distinct hyperedges e, e′ ∈ EΘ such that
e ∩ e′ 6= ∅ and
EΘ′ = (EΘ \ {e, e′}) ∪ {e ∪ e′};
that is, EΘ′ is the result of replacing e and e′ by their union. The require-
ment that e ∩ e′ 6= ∅ ensures that, in such a case, Θ′ is also a hypertree
on [n]. For each pair Θ, Λ ∈ HT n, we write Θ ≤ Λ, and we say that Θ is
a result of folding Λ, if Θ may be obtained from Λ by a (possibly empty)
sequence of folds. Then (HT n, ≤) is a partially ordered set called the
hypertree poset of rank n.
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A. Piggott 245
Example 2.4. The Hasse diagram of (HT 3, ≤) is shown in Figure 2.
The simplicial complex HT3 is shown in Figure 3.
A hypertree Θ ∈ HT n must have at least one hyperedge, and may
have as many as n−1 hyperedges. For each h ∈ {0, . . . , n − 2}, we write
HT h
n for the set of hypertrees on [n] with h+1 hyperedges; a hypertree in
HT h
n is said to have height h. The unique hypertree of height 0 is denoted
Θ0
n. The hypertrees in HT n−2
n are precisely the trees on [n]; it follows
that there are nn−2 hypertrees in HT n−2
n .
Remark 2.5 (An alternative, but equivalent, definition of a hypertree).
Since the maximal elements in (HT n, ≤) are exactly the trees on [n],
and HT n is closed under folding, HT n is exactly the set of hypergraphs
obtained by folding trees on [n]. More generally, the set of hypertrees on
a vertex set V is the set of hypergraphs obtained by folding trees on V .
The hypertree complex of rank n, denoted HTn, is the simplicial realiza-
tion of (HT n, ≤). Recall that this means: there exists a bijection Vn from
HT n to the vertex set of HTn; for distinct hypertrees Θ1, . . . , Θk ∈ HT n,
the vertices Vn(Θ1), . . . , Vn(Θk) span a (k−1)-simplex if and only if there
exists a maximal chain in (HT n, ≤) which contains Θ1, . . . , Θk.
It is immediate that each single fold reduces the number of hyperedges
by one, and a hypertree can be folded provided it has more than one
hyperedge. It follows that each maximal chain in (HT n, ≤) contains
exactly n−1 hypertrees, the minimal element of which is Θ0
n, and the
maximal element of which is a tree. Thus the simplicial complex HTn has
dimension n−2.
We shall often consider the hypertree poset without its minimal ele-
ment. We write HT +
n := HT n \ {Θ0
n}, and we write HT+
n for simplicial
realization of (HT +
n , ≤). Equivalently, we may consider HT+
n to be the
subcomplex of HTn spanned by Vn(HT +
n ), or the link in HTn of Vn(Θ0
n).
We write V+
n (Θ) for the vertex in HT+
n corresponding to Θ.
Example 2.6. The complex HT+
4 is shown in Figure 5; this figure is
an adaptation of [6, Figure 8]. Some vertices are represented as stars,
some as filled circles and some as unfilled circles, for reasons described in
Section 3.
We shall make use of a metric on HT+
n which reflects the geometry of
the 1-skeleton of HT+
n .
Definition 2.7 (d+
n (·, ·)). For hypertrees Θ, Λ ∈ HT +
n , we write d+
n (Θ, Λ)
for the combinatorial length of the minimal length paths in the 1-skeleton
of HT+
n between the vertices V+
n (Θ) and V+
n (Λ).
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246 The symmetries of McCullough-Miller space
Figure 5. The endpoints of antipodal dashed edges should be identified
to create HT+
4 , the link in HT4 of Θ0
4.
2.3. Some subsets of HT n
In the arguments which follow, a number of subsets of HT n prove
important. We gather the definitions here for the convenience of the
reader. We have also included a table of notation at the end of the paper.
Definition 2.8 (Star trees). For each j ∈ [n], we write Ξj
n for the hyper-
tree on [n] with (n−1) hyperedges, each of which contains j; we say that
Ξj
n is the star tree of rank n and common vertex j. We write Sn for the
set of star trees of rank n.
It is immediate that the elements of Sn are isomorphic as unlabeled
hypergraphs; that is, any two elements of Sn differ by a permutation
of the vertices. It is also immediate that each star tree is a tree. The
hypothesis that n ≥ 3 ensures that there are n star trees in HT n.
Definition 2.9 (Line trees). A hypertree in which exactly two vertices
are leaves is called a line tree; we write Ln for the set of line trees.
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A. Piggott 247
It is immediate that the elements of Ln are isomorphic as unlabeled
hypergraphs. It is also immediate that each line tree is a tree, and that
each vertex in a line tree has valence one or valence two. There are n!/2
line trees in HT n.
Definition 2.10. For each h ∈ {1, . . . , n − 2}, we write Mh
n for those
hypertrees in HT h
n that contain a vertex of valence h+1, and a hyperedge
of degree n−h.
It is immediate that Mn−2
n = Sn, and, for each h, the elements of
Mh
n are isomorphic as unlabeled hypergraphs. Examples are shown in
Figure 6. The notation, a script M , was chosen because, amongst the
vertices in V+
n (HT h
n), the vertices in V+
n (Mh
n) prove to have maximal
valence in HT +
n , provided n ≥ 5 (Lemma 3.12 below).
2
5
n
3 1
4
2
4
n
3 1
2
3
n
1
2 n
3 1
4
Ξ n
1
Ωn
1, 2
Figure 6. From left to right, a hypertree in M1
n, M2
n, M3
n, and Mn−2
n ,
with n > 4.
The set M1
n is particularly important as, provided n ≥ 5, the vertices
in V+
n (M1
n) prove to be the vertices in HT+
n of maximal valence. It
is convenient to define notation for the elements of M1
n. For each pair
j, k ∈ [n] with j 6= k, we write Ωj,k
n for the hypertree on [n] with hyperedges
{j, k} and [n] \ {k}. It follows that M1
n = {Ωj,k
n | j, k ∈ [n], j 6= k}.
3. Automorphisms of HTn
In this section we prove Theorem 1.3.
Fix an integer n ≥ 3. Recall that we write Σn for the symmetric
group of rank n, which we identify with the group of bijections [n] → [n].
For a bijection σ ∈ Σn, and a subset S ⊂ [n], we write σ(S) for the set
{σ(s) | s ∈ S}. For a bijection σ ∈ Σn, and a hypertree Θ ∈ HT n, we
write σ(Θ) for the hypertree on [n] such that Eσ(Θ) = {σ(e) | e ∈ EΘ};
that is, Eσ(Θ) is obtained by replacing each hyperedge e ∈ EΘ with σ(e).
Evidently, the map Θ 7→ σ(Θ) preserves the partial order ≤, and hence
determines an automorphism of (HT n, ≤); thus we have a homomorphism
Σn → Aut(HT n, ≤). Since Aut(HT n, ≤) embeds in Aut(HTn), we also
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248 The symmetries of McCullough-Miller space
have a homomorphism ιn : Σn → Aut(HTn). Theorem 1.3 is proved if
we show that ιn is bijective. We can simplify this task a little using the
following lemma. The lemma follows immediately from the observation
that for each hypertree except Θ0
n, there is another hypertree with the
same number of hyperedges.
Lemma 3.1. For each integer n ≥ 3, Vn(Θ0
n) is the unique vertex in HTn
of maximal valence.
It follows from the lemma that for each simplicial automorphism
f ∈ Aut(HTn), f fixes Vn(Θ0
n), and restricts to a simplicial automor-
phism f+ ∈ Aut(HT+
n ). The restriction f 7→ f+ is an isomorphism
Aut(HTn) → Aut(HT+
n ). Pre-composing this isomorphism with ιn gives a
homomorphism ι+
n :Σn → Aut(HT+
n ). To prove Theorem 1.3 it suffices to
show that ι+
n is bijective.
That ι+
n is injective follows immediately from an analysis of how Σn
acts on Sn = {Ξ1
n, . . . , Ξn
n}, the set of star trees.
Lemma 3.2. For each integer n ≥ 3, the homomorphism ι+
n : Σn →
Aut(HT+
n ) is injective.
Proof. Since n ≥ 3, there are n distinct star trees, and Σn acts on Sn by
permuting superscripts. It follows that Σn acts as the full permutation
group on V+
n (Sn), and distinct elements of Σn act distinctly on V+
n (Sn).
Thus distinct elements of Σn act distinctly on HT+
n .
Notation 3.3. Empowered by the lemma, we shall not distinguish be-
tween an element of Σn and the corresponding automorphisms of HTn,
HT+
n and (HT n, ≤).
It remains to show that ι+
n is surjective. The cases n = 3 and n = 4
are easily dispatched by inspection of HT+
3 and HT+
4 respectively.
Lemma 3.4. The homomorphism ι+
3 is surjective.
Proof. The complex HT+
3 consists of three disjoint vertices. The ver-
tices are the elements of V+
3 (S3). Evidently, each automorphism of HT+
3
permutes these vertices. The result follows.
Lemma 3.5. The homomorphism ι+
4 is surjective.
Proof. Let f+ ∈ Aut(HT+
4 ). It suffices to show that there exists σ ∈ Σ4
such that σ−1f+ fixes pointwise V+
4 (HT +
4 ), because then σ−1f+ is the
identity automorphism of HT+
4 , and f+ = σ ∈ Σ4.
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A. Piggott 249
Consider the complex HT+
4 , as shown in Figure 5. The vertices in
V+
4 (S4) are shown as stars, the vertices in V+
4 (L4) are shown as filled
circles, and the vertices in V+
4 (HT 1
4) are shown as unfilled circles. Note
that HT +
4 = S4 ∪ L4 ∪ HT 1
4. Inspection of Figure 5 shows that, for each
Θ ∈ HT +
4 , Θ ∈ S4 if and only V+
4 (Θ) has valence three, and V+
4 (Θ) is
not adjacent to a vertex of valence two. Since the elements of V+
4 (S4) can
be identified geometrically, f+ fixes setwise V+
4 (S4). Since Σ4 acts as the
full permutation group on V+
4 (S4), there exists σ ∈ Σ4 such that σ−1f+
fixes pointwise V+
4 (S4).
Inspection also shows that, for each Θ ∈ HT +
4 , Θ ∈ HT 1
4 if and only if
V+
4 (Θ) has valence three, and V+
4 (Θ) is adjacent to a vertex of valence two.
Since the elements of V+
4 (HT 1
4) can be identified geometrically, σ−1f+
fixes setwise V+
4 (HT 1
4), and hence also fixes setwise V+
4 (L4).
Further inspection shows that distinct elements of V+
4 (HT 1
4) (that is,
distinct unfilled circles) can be distinguished by their relative proximities in
HT+
4 to the elements of V+
4 (S4); that is, if Θ and Λ are distinct elements
of HT 1
4, then there exists Υ ∈ S4 such that d+
4 (Θ, Υ) 6= d+
4 (Λ, Υ). It
follows that σ−1f+ fixes pointwise V+
4 (HT 1
4).
Finally, further inspection shows that distinct elements in V+
4 (L4) can
be distinguished by their relative proximities in HT+
4 to the elements of
V+
4 (HT 1
4). It follows that σ−1f+ fixes pointwise V+
4 (L4).
We have that σ−1f+ fixes pointwise V+
4 (HT +
4 ), as required.
To prove that ι+
n is surjective for n ≥ 5, we adapt the argument for
n = 4, replacing each use of inspection by a general argument. Although
our general argument still works with the subsets Ln, Sn and HT 1
n, it must
recognize that these sets no longer combine to give the entirety of HT+
n .
In particular, it takes more effort to show that an arbitrary simplicial
automorphism of HT+
n fixes setwise V+
n (HT 1
n).
To ensure the structure of our general argument is easily visible, we
describe it assuming a series of technical claims, to be proved immediately
after. For each n ≥ 5, we claim the following:
(A) The vertices in V+
n (M1
n) are exactly the vertices in HT+
n of maximal
valence.
(B) The vertices in V+
n (Sn) are exactly the vertices in HT+
n which are
adjacent to n−1 vertices in V+
n (M1
n).
(C) The vertices in V+
n (Ln) are exactly the vertices in HT+
n which,
although not adjacent to any vertex in V+
n (Sn), are distance exactly
two from n−2 vertices in V+
n (Sn).
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250 The symmetries of McCullough-Miller space
(D) The vertices in V+
n (HT 1
n) are exactly the vertices in HT+
n which are
adjacent to some vertex in V+
n (Sn), and adjacent to some vertex in
V+
n (Ln).
(E) For all Θ, ∆ ∈ M1
n, Θ = ∆ if and only if d+
n (Θ, Υ) = d+
n (∆, Υ) for
all Υ ∈ Sn.
(F) For all Θ, ∆ ∈ HT 1
n, Θ = ∆ if and only if d+
n (Θ, Υ) = d+
n (∆, Υ) for
all Υ ∈ M1
n.
(G) For all Θ, ∆ ∈ HT +
n , Θ = ∆ if and only if d+
n (Θ, Υ) = d+
n (∆, Υ) for
all Υ ∈ HT 1
n.
Proposition 3.6. For each integer n ≥ 5, ι+
n is surjective.
Proof, assuming claims (A) through (G). Considered in order, Claims (A)
through (D) combine to give that an arbitrary automorphism of HT+
n
fixes setwise V+
n (M1
n), V+
n (Sn), V+
n (Ln) and V+
n (HT 1
n).
Let f+ ∈ Aut(HT+
n ). As in the case that n = 4, it suffices to show
that there exists σ ∈ Σn such that σ−1f+ fixes pointwise V+
n (HT +
n ).
Since f+ fixes setwise V+
n (Sn), and Σn acts as the full permutation
group of V+
n (Sn), there exists σ ∈ Σn such that σ−1f+ fixes pointwise
V+
n (Sn). Since σ−1f+ fixes pointwise V+
n (Sn), and fixes setwise V+
n (M1
n),
(E) implies that σ−1f+ fixes pointwise V+
n (M1
n). Since σ−1f+ fixes point-
wise V+
n (M1
n), and fixes setwise V+
n (HT 1
n), (F) implies that σ−1f+ fixes
pointwise V+
n (HT 1
n). Finally, since σ−1f+ fixes pointwise V+
n (HT 1
n) and
fixes setwise V+
n (HT +
n ), (G) implies that σ−1f+ fixes pointwise V+
n (HT +
n ),
as required.
Remark 3.7. Claim (A) fails in the case n = 4 because, as is evident in
Figure 5, the vertices in V+
4 (HT 1
4) (shown as unfilled circles) are not the
only vertices of maximal valence.
3.1. Proving Claims (A) through (G)
Throughout this subsection we assume n ≥ 5.
Claim (A) requires that we understand the valences of vertices in
HT+
n . For each Θ ∈ HT +
n : we write A+
n (Θ) for the set of hypertrees in
HT +
n , distinct from Θ, which fold to Θ; and we write B+
n (Θ) for the set
of hypertrees in HT +
n , distinct from Θ, which can be obtained from Θ by
folding. Thus the valence in HT+
n of V+
n (Θ) is #A+
n (Θ) + #B+
n (Θ).
A convenient formula for #A+
n (Θ) follows from the observation that
the operation of folding has a natural inverse. Let Θ, Λ ∈ HT+
n , let e ∈ EΘ,
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A. Piggott 251
and let ∆ be a hypertree on e (that is, ∆ is a hypertree with vertex set
e). We say Λ is obtained from Θ by unfolding e to ∆, or just unfolding e,
if EΛ = (EΘ \ {e}) ∪ E∆; that is, if EΛ is the result of replacing e ∈ EΘ
by the elements of E∆. If, in the above, ∆ has only one hyperedge, then
Θ = Λ and we say the unfolding is trivial; otherwise the unfolding is
nontrivial, and Λ has more hyperedges than Θ. Evidently, Λ is obtained
from Θ by unfolding if and only if Θ is obtained from Λ by folding.
Lemma 3.8 (c.f Lemma 2.5(1) [5]). For each Θ ∈ HT+
n ,
#A+
n (Θ) = −1 +
∏
e∈EΘ
#(HT #e).
Proof. Let Θ ∈ HT+
n . For each hyperedge e ∈ EΘ, there are #(HT #e)
distinct hypertrees on e, and hence #(HT #e) distinct hypertrees, including
Θ itself, which may be obtained from Θ by unfolding e. The result
follows.
Let h ∈ {1, . . . , n−2}. If Θ ∈ HT h
n, then a hyperedge in Θ has degree
at most n−h. Our next result records that those hypertrees in HT h
n that
contain a hyperedge of degree n−h are precisely the hypertrees in HT h
n
for which #A+
n (Θ) is maximal. We note that if Θ ∈ HT h
n has a hyperedge
of degree n−h, then the other hyperedges in Θ have degree exactly two.
Lemma 3.9. Let h ∈ {1, . . . , n−2}, let Θ ∈ HT h
n, and let Λ ∈ HT h
n be
such that Λ has a hyperedge of degree n−h. If Θ also has a hyperedge of
degree n−h, then #A+
n (Θ) = #A+
n (Λ); otherwise #A+
n (Θ) < #A+
n (Λ).
The lemma follows inductively from Lemma 3.8, and the following
result.
Lemma 3.10. For all integers p, q ≥ 3, #HT p . #HT q < #HT p+q−2.
Proof. Let Λ be the hypertree on [p+q−1] with hyperedges {1, . . . , p−1}∪
{p+q −1} and {p, p+1, . . . , p+q −1}. By Lemma 3.8 we have #A+
n (Λ) =
−1 + #HT p . #HT q; recall that #A+
n (Θ0
p+q−2) = −1 + #HT p+q−2. Thus
to prove the lemma it suffices to exhibit an injective, but not surjective,
map A+
n (Λ) → A+
n (Θ0
p+q−2).
Let Υ be an arbitrary hypertree in A+
n (Λ). Note that if e ∈ EΥ,
and p + q − 1 ∈ e, then either e \ {p + q − 1} ⊂ {1, . . . , p − 1} or
e\{p+q −1} ⊂ {p, . . . , p+q −2}. Let a ∈ {1, . . . , p−1} be maximal such
that a is adjacent in Υ to p + q − 1, and let ea ∈ EΥ be the hyperedge
such that {a, p + q − 1} ∈ ea; let b ∈ {p, . . . , p + q − 2} be minimal such
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252 The symmetries of McCullough-Miller space
that b is adjacent in Υ to p + q − 1, and let eb ∈ EΥ be the hyperedge
such that {b, p + q − 1} ∈ eb. We construct a hypertree Υ′ ∈ HT p+q−2 as
follows: (ea ∪ eb) \ {p + q − 1} is a hyperedge in Υ′; for each hyperedge
e ∈ EΥ with p + q − 1 ∈ e and e \ {p + q − 1} ⊂ {1, . . . , p − 1} and e 6= ea,
we replace e by {b} ∪ e \ {p + q − 1}; for each hyperedge e ∈ EΥ with
p + q − 1 ∈ e and e \ {p + q − 1} ⊂ {p, . . . , p + q − 2} and e 6= eb, we
replace e by {a} ∪ e \ {p + q − 1}.
To show that the map Υ 7→ Υ′ is injective, we now explain how, given
Υ′, we can reconstruct Υ. Suppose Υ′ is as described above. We can identify
a because it is maximal amongst the elements of {1, . . . , p − 1} which are
adjacent in Υ′ to at least one element of {p, . . . , p + q − 2}; similarly, we
can identify b. Hence we can identify the hyperedge (ea ∪ eb) \ {p + q − 1},
and we can recover ea and eb. Having done this, it is now evident that we
can recover the other hyperedges of Υ.
In this paragraph we show that the map Υ 7→ Υ′ is not surjective.
It follows from the construction that any Υ′ must have the property
that, amongst the elements of {1, . . . , p − 1} which are adjacent to an
element of {p, . . . , p + q − 2}, only the maximal element may be adjacent
to more than one element of {p, . . . , p + q − 2}. Hence the hypertree with
hyperedges
{1, p + q − 2}, {1} ∪ {p, . . . , p + q − 3}, {2, p + q − 2}, {2, 3, . . . , p − 1}
is contained in A+
n (Θ0
p+q−2), but it is not the image of any hypertree
Υ ∈ A+
n (Λ) (it is here we use the hypothesis that p, q ≥ 3).
Next we look to understand #B+
n (Θ) for an arbitrary hypertree Θ ∈
HT +
n . We begin by observing that each hypertree Λ ∈ B+
n (Θ) corresponds
to an equivalence relation on EΘ, but, in general, only certain equivalence
relations on EΘ correspond to hypertrees in B+
n (Θ). Given Θ ∈ HT n and
Λ ∈ B+
n (Θ), we define a relation ∼Λ on EΘ as follows: if e, e′ ∈ EΘ, then
e ∼Λ e′ if there exists u ∈ EΛ such that e ∪ e′ ⊂ u; that is, hyperedges in
Θ are related if and only if they are eventually folded in Λ. We leave the
reader to verify that ∼Λ is an equivalence relation. The restriction, that
two hyperedges must intersect nontrivially if they are to be merged in a
single fold, implies that each equivalence class must satisfy the following
condition:
(∗) if e ∈ EΘ and i, j ∈ ∪e∼e′e′, then the unique simple walk in Λ from
i to j visits only vertices in ∪e∼e′e′.
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It is easily verified that an equivalence relation ∼ on EΘ corresponds to a
hypertree Θ∼ if and only if the equivalence relation satisfies (∗); in which
case, the hypertree Θ∼ is in B+
n (Θ) unless there is only one equivalence
class of hyperedeges (because then Θ∼ = Θ0
n), or all equivalence classes
are singletons (because then Θ∼ = Θ).
If Θ ∈ HT h
n has a vertex j of valence h+1, then j is contained in
every hyperedge of Θ, and every vertex but j is a leaf (otherwise Θ would
fail to be a hypertree). We now show that the hypertrees in HT h
n that
contain a vertex of valence h+1 are precisely the hypertrees in HT h
n for
which #B+
n (Θ) is maximal.
Lemma 3.11. Let h ∈ {1, . . . , n−2}, let Θ ∈ HT h
n, and let Λ ∈ HT h
n be
such that Λ has a vertex of valence h+1. If Θ also has a vertex of valence
h+1, then #B+
n (Λ) = #B+
n (Θ); otherwise, #B+
n (Θ) < #B+
n (Λ).
Proof. Let Υ ∈ HT h
n. If some vertex j has valence h+1, then j is contained
in every hyperedge of Υ, and every partition of EΥ satisfies (∗); otherwise,
there exist partitions of EΥ which do not satisfy (∗). The result follows.
Lemmas 3.9 and 3.11 combine with the definition of Mh
n (Defini-
tion 2.10) to give the following.
Lemma 3.12. Let h ∈ {1, . . . , n−2}. If Θ ∈ HT h
n \ Mh
n, and Λ ∈ Mh
n,
then the valence in HT+
n of V+
n (Λ) strictly exceeds that of V+
n (Θ).
We now compare the valences of vertices in Mh
n for different values of
h, and so establish Claim (A). Recall that for distinct integers j, k ∈ [n],
we write Ωj,k
n for the hypertree with hyperedges {j, k} and [n] \ {k}; and
we write M1
n := {Ωj,k
n | j, k ∈ [n], j 6= k}. We begin by observing that
there is a simple way to characterize those vertices in V+
n (HT n) which
are adjacent to V+
n (Ωj,k
n ).
Definition 3.13 ((j, k)-tag). Given Θ ∈ HT +
n , and distinct integers
j, k ∈ [n] with j 6= k, we say Θ has a (j, k)-tag if {j, k} is a hyperedge in
Θ, and no other hyperedge in Θ contains k.
Lemma 3.14. For each pair j, k ∈ [n] with j 6= k, and for each hypertree
Λ ∈ HT +
n , V+
n (Λ) and V+
n (Ωj,k
n ) are adjacent in HT+
n if and only if Λ has
at least three hyperedges, and Λ has a (j, k)-tag.
Proof. Since Ωj,k
n is minimal in (HT +
n , ≤), V+
n (Λ) and V+
n (Ωj,k
n ) are adja-
cent in HT+
n if and only if Ωj,k
n ∈ B+
n (Λ).
If Λ has only two hyperedges, then B+
n (Λ) = ∅.
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254 The symmetries of McCullough-Miller space
If Λ has at least three hyperedges and Λ has a (j, k)-tag, then we
obtain Ωj,k
n by folding, in some allowable order, all of the hyperedges in Λ
except {j, k}; hence Ωj,k
n ∈ B+
n (Λ).
Suppose that Λ does not have a (j, k)-tag. Then there exists some
vertex ℓ 6= j such that k and ℓ are adjacent in Λ. It follows from the
definition of folding that k and ℓ are adjacent in each element of B+
n (Λ);
thus Ωj,k
n 6∈ B+
n (Λ).
Proposition 3.15. [Claim (A)] The vertices in V+
n (M1
n) are exactly the
vertices in HT+
n of maximal valence.
Proof. Let Ω ∈ M1
n, and let Θ be a hypertree in Mh
n for some h ∈
{2, . . . , n}. In light of Lemma 3.12, it suffices to show that the valence
in HT+
n of V+
n (Ω) exceeds that of V+
n (Θ). To do so we will exhibit an
injective, but not surjective, map p :A+
n (Θ) ∪ B+
n (Θ) → A+
n (Ω).
Since the hypertrees in M1
n are isomorphic as unlabeled hypertrees,
the corresponding vertices have the same valences in HT+
n , and we may
assume Ω = Ω1,2
n . Similarly, we may assume
EΘ = {{1, 2}, {1, 3}, . . . , {1, h+1}, {1, h+2, h+3, . . . , n}},
as shown in Figure 7.
h+2
2
n
1
Θ
3
h+1
3
2
n
1
Ω
Figure 7. Θ and Ω as in the proof of Proposition 3.15.
We define p(Ω) = Θ, and p(Υ) = Υ for each Υ ∈
(
A+
n (Θ) ∪ B+
n (Θ)
)
∩
A+
n (Ω). Since Θ has a (1, 2)-tag, Θ ∈ A+
n (Ω); it follows that A+
n (Θ) is
a subset of
(
A+
n (Θ) ∪ B+
n (Θ)
)
∩ A+
n (Ω). Thus it remains only to define
p(Υ) for Υ ∈ B+
n (Θ) \ (A+
n (Ω) ∪ {Ω}). Consider a hypertree Υ ∈ B+
n (Θ) \
(A+
n (Ω) ∪ {Ω}). Since Υ ∈ B+
n (Θ), 1 is the only non-leaf vertex; since
Υ 6∈ A+
n (Ω) ∪ {Ω}, the unique hyperedge e ∈ EΥ such that e contains 2
is such that e 6= {1, 2}. Let j be minimal such that j ≥ 3 and j ∈ e. We
let EΥ′ be the set of hyperedges obtained from EΥ by removing 2 from e;
swapping each occurrence of 1 with j, and vice-versa; and then adding
the hyperedge {1, 2}. An example of this process is shown in Figure 8.
Since Υ′ has at least three hyperedges, and it has a (1, 2)-tag, Υ′ ∈ A+
n (Ω).
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A. Piggott 255
Since j is a non-leaf in Υ′, Υ′ 6∈ B+
n (Θ). We define p(Υ) = Υ′. Since j
and 1 are the only non-leaves in Υ′, we can recognize j in Υ′, and hence
we can recover Υ from Υ′. It follows that p is injective.
1 2 4
5 ...n
3
1
2
4
3
14
3
41
3
We remove 2
from e
Then we swap 1 and
the next least element
left in e
Finally, we restore 2 to
the picture, this time in
a (1,2)-tag
Υ is in B(Θ), but
not A(Ω)U{Ω}
The hyperedge e
Υ Υ'5 ...n 5 ...n 5 ...n
Figure 8. An example computing Υ′ for Υ ∈ B+
n (Θ) \ (A+
n (Ω) ∪
{Ω, Θ0
n}), as in the proof of Proposition 3.15.
It remains only to show that p is not surjective. To do so, consider
∆ ∈ HT n such that
E∆ = {{1, 2}, {1, 3}, {3, 4}, {4, 5, . . . , n}}
(recall that, by hypothesis, n ≥ 5). Since ∆ has a (1, 2)-tag, ∆ ∈ A+
n (Ω);
since 1 is not the only non-leaf in ∆, ∆ 6∈ B+
n (Θ); evidently, ∆ 6= Θ;
since ∆ has 3 non-leaves, and any Υ′ constructed as above has only two
non-leaves, ∆ 6= Υ′. Since ∆ ∈ A+
n (Ω), and ∆ is not in the image of p, p
is not surjective.
We now turn our attention to claims (B) through (G).
Lemma 3.16 (Claim (B)). The vertices in V+
n (Sn) are exactly the vertices
in HT+
n which are adjacent to n−1 vertices in V+
n (M1
n).
Proof. The lemma follows immediately from Lemma 3.14, and the obser-
vation that the star trees are exactly the hypertrees with n−1 distinct
tags.
Lemma 3.17. Let j ∈ [n] and let Θ ∈ HT +
n . Then
1) d+
n (Θ, Ξj
n) ≤ 1 if and only if j is the only vertex in Θ that is not a
leaf;
2) d+
n (Θ, Ξj
n) = 2 if and only if j is not a leaf in Θ, and there is at
least one other vertex in Θ which is not a leaf.
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256 The symmetries of McCullough-Miller space
Proof. Since Ξj
n is maximal in (HT +
n , ≤), d+
n (Θ, Ξj
n) = 1 if and only if
Θ ∈ B+
n (Ξj
n). Evidently, j is the only non-leaf vertex in Ξj
n, folding cannot
make a leaf into a non-leaf, and no element of HT +
n has n-leaves. Thus
each element of B+
n (Ξj
n) has exactly n−1 leaves. Property (1) follows.
Property (2) follows immediately from the observations that:
d+
n (Θ, Ξj
n) = 2 if and only if Θ 6∈ B+
n (Ξj
n)∪{Ξj
n}, but B+
n (Θ)∩B+
n (Ξj
n) 6= ∅;
if j is a non-leaf in a hypertree Λ, then we can arrange that j is the only
non-leaf by repeatedly folding two hyperedges which both contain some
i ∈ [n] \ {j}.
Corollary 3.18 (Claim (C)). The vertices in V+
n (Ln) are exactly the
vertices in HT+
n which, although not adjacent to any vertex in V+
n (Sn),
are distance exactly two from n−2 of the n vertices in V+
n (Sn).
Proof. Line trees and star trees have the same height, and so cannot be
adjacent in HT +
n . By definition, the lines trees are exactly the hypertrees
with exactly two leaves. Equivalently, the line trees are exactly the hy-
pertrees in which there are n−2 vertices which are not leaves. The result
now follows from Lemma 3.17.
Lemma 3.19 (Claim (D)). The vertices in V+
n (HT 1
n) are exactly the
vertices in HT+
n which are adjacent to some vertex in V+
n (Sn), and adjacent
to some vertex in V+
n (Ln).
Proof. Suppose V+
n (Θ) is adjacent to both V+
n (Ξ) and V+
n (Λ) for some
star-tree Ξ, and some line-tree Λ. Then, since Ξ and Λ are maximal in
(HT +
n , ≤), Θ ≤ Ξ and Θ ≤ Λ. Since Θ ≤ Ξ, Θ has only one non-leaf
vertex. Since distinct hyperedges in a hypertree can contain at most one
common vertex, a single fold can turn at most one non-leaf into a leaf.
Since Λ has n−2 non-leaves, it will take at least n−3 single folds to obtain
from Λ a hypertree that has only one non-leaf. Since Λ ∈ HT n−2
n , and Θ
is obtained by at least n−3 single folds, Θ ∈ HT 1
n.
Conversely, suppose that Θ ∈ HT 1
n. Then EΘ = {A ∪ {j}, B ∪ {j}}
for some j ∈ [n] and some non-trivial partition {A, B} of [n] \ {j}. Since
j is the only non-leaf of Θ, Θ ∈ B+
n (Ξj
n). Let a ∈ A and let ΛA be
a line tree on A in which a is a leaf; let b ∈ B and let ΛB be a line
tree on B in which b is a leaf. Let Λ be the hypertree on [n] such that
EΛ = EΛA
∪ EΛB
∪ {{a, j}, {b, j})}. Then Λ is a line tree on [n] and
Θ ∈ B+
n (Λ).
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j [n]\{i,j}i
k-1
k+1 k
Ξ n
k
Ωn
1, 2
1
n
j [n]\{i,j, k}i k j [n]\{i,j, k}i k
unfolds to unfolds tofolds to
Figure 9. A 3-step unfolding and folding sequence.
The author thanks Andy Eisenberg for observing an error in an earlier
proof of the following lemma, and suggesting the proof which appears
below.
Lemma 3.20 (Claim (E)). For all Ω, Ω′ ∈ M1
n, Ω = Ω′ if and only if
d+
n (Ω, Ξ) = d+
n (Ω′, Ξ) for all Ξ ∈ Sn.
Proof. Let i, j, k be distinct elements of [n]. It is clear that d+
n (Ωi,j
n , Ξi
n) =
1.
We claim that d+
n (Ωi,j
n , Ξk
n) = 3. The unfolding and folding sequence in
Figure 9 shows that d+
n (Ωi,j
n , Ξk
n) ≤ 3. Since k is a leaf in Ωi,j
n , Lemma 3.17
implies that d+
n (Ωi,j
n , Ξk
n) > 3.
Finally, we claim that d+
n (Ωi,j
n , Ξj
n) > 3. Because it is minimal in
(HT +
n , ≤), we cannot fold Ωi,j
n and stay in HT +
n . Suppose that Λ′ is
obtained by unfolding Ωi,j
n . Then Λ′ has a (i, j)-tag. Since j is a leaf in
Λ′, Lemma 3.17 implies that d+
n (Λ′, Ξj
n) > 2. Hence d+
n (Ωi,j
n , Ξj
n) > 3 as
claimed. The result follows.
Lemma 3.21 (Claim (F)). For all Θ, ∆ ∈ HT 1
n, Θ = ∆ if and only if
d+
n (Θ, Υ) = d+
n (∆, Υ) for all Υ ∈ M1
n.
Proof. Let Θ ∈ HT 1
n \ M1
n. Then EΘ = {A ∪ {j}, B ∪ {j}} for some
j ∈ [n] and some disjoint nonempty subsets A, B ⊂ [n] \ {j} such that
A ∪ B = [n] \ {j}, and #A, #B ≥ 3.
Let Υ ∈ M1
n. Then Υ = Ωk,ℓ
n for some k, ℓ ∈ [n]. Since Θ and Ωk,ℓ are
distinct elements of the same height, d+
n (Θ, Ωk,ℓ) ≥ 2.
If k and ℓ are adjacent in Θ, and ℓ is a leaf, then Θ can be unfolded
to a hypertree with a (k, ℓ)-tag, so d+
n (Θ, Ωk,ℓ
n ) = 2. If k and ℓ are not
adjacent in Θ, or ℓ is not a leaf, then Θ cannot be unfolded to a hypertree
with a (k, ℓ)-tag, so d+
n (Θ, Ωk,ℓ
n ) ≥ 3.
It follows that the function Ξ 7→ d+
n (Θ, Ξ), for Ξ ∈ Sn, can be used to
identify which pairs of vertices in Θ are adjacent (and which vertex is a
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258 The symmetries of McCullough-Miller space
non-leaf). Since the hyperedges in Θ are precisely the maximal subsets of
[n] with the property that elements are pairwise adjacent in Θ, the result
follows.
Lemma 3.22 (Claim (G)). For all Θ, ∆ ∈ HT +
n , Θ = ∆ if and only if
d+
n (Θ, Υ) = d+
n (∆, Υ) for all Υ ∈ HT 1
n.
Proof. Let Λ ∈ HT 1
n. Then EΛ = {A∪{j}, B∪{j}} for some vertex j, and
some disjoint non-empty subsets A, B ⊂ [n]\{j} such that A∪B = [n]\{j}.
For all Θ ∈ HT +
n , Θ ∈ A+
n (Λ) if and only if, for all a ∈ A and b ∈ B,
a and b are not adjacent in Θ. Since there is an element of HT 1
n for
each choice j ∈ [n], and subsequent choices of A and B, the function
Υ 7→ d+
n (Θ, Υ), for Υ ∈ HT 1
n, contains sufficient information to establish
exactly which vertices are not adjacent in Θ, and hence exactly which
vertices are adjacent in Θ. It follows from the definition of a hypertree
that the hyperedges in Θ are precisely the maximal subsets of [n] with
the property that elements are pairwise adjacent in Θ. Thus we can use
the knowledge of which vertices are adjacent in Θ to reconstruct EΘ. The
result follows.
4. The symmetries of McCullough-Miller space
4.1. Automorphisms of Wn
Fix a positive integer n. There are exactly n conjugacy classes of
involutions in Wn, each represented by a generator. Each permutation
of the generators induces an automorphism of Wn which permutes these
conjugacy classes; we write Σn for the group of these automorphisms. It
follows that Aut(Wn) acts transitively on the set of conjugacy classes of
involutions; we write Aut0(Wn) for the kernel of this action. It is easily
verified that Aut(Wn) = Aut0(Wn)⋊Σn. It follows that, writing Out0(Wn)
for quotient Aut0(Wn)/ Inn(Wn), we have Out(Wn) ∼= Out0(Wn) ⋊ Σn.
Thus for each α ∈ Out(Wn), there exist unique automorphisms φ ∈
Out0(Wn) and σ ∈ Σn such that α = φσ. For n ≥ 3, Out0(Wn) is an
infinite group.
Definition 4.1 (Partial conjugation). For an integer i ∈ [n], and a proper
subset D ⊂ [n] \ {i}, we write xiD for the outer automorphism of Wn
determined by the map:
aj 7→
{
aiajai if j ∈ D,
aj if j ∈ [n] \ D;
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A. Piggott 259
we say that xiD is the partial conjugation with acting letter i and do-
main D.
If xiD is a partial conjugation, then xiD is an involution (if D were
not a proper subset of [n] \ {i}, then xiD would be the identity outer
automorphism). If i ∈ [n] and D, D′ are disjoint proper subsets of [n]\{i}
such that D ∪ D′ = [n] \ {i}, then xiD = xiD′ . We adopt the convention
that whenever we write xiD, it is assumed that either i = 1 and 2 6∈ D,
or i 6= 1 and 1 6∈ D.
The partial conjugations generate Out0(Wn) (see, for example [3]). The
following definition and lemma, due to McCullough and Miller, together es-
tablish a relationship between the hypertree poset and the automorphisms
of Wn.
Definition 4.2 (Carried by). Given a partial conjugation xiD, and a
hypertree Θ ∈ HT n, we say that xiD is carried by Θ if: for all d ∈ D
and for all j ∈ [n] \ D, the simple walk in Θ from j to d visits i. Given
an automorphism α ∈ Out0(Wn) and a hypertree Θ ∈ HT n, we say that
α is carried by Θ if α = xipDp . . . xi1D1
for some partial conjugations
xipDp , . . . , xi1D1
, each of which is carried by Θ.
Remark 4.3. It follows that xiD is carried by Θ if and only if D is a
union of connected components of Θ \ {i}.
Lemma 4.4. Let xi1D1
, . . . , xipDp be partial conjugations and let Θ ∈
HT n. If Θ carries the product xipDp . . . xi1D1
, then the partial conjugations
xi1D1
, . . . , xipDp pairwise commute.
Proof. Suppose that Θ carries xijDj
for each j ∈ {1, . . . , k}. Let p, q ∈
{1, . . . , k}. It follows from [6, p.14, second paragraph] that xipDp commutes
with xiqDq if ip 6= iq. Because the factors in the free product decomposition
of Wn are abelian, xipDq commutes with xiqDq if ip = iq (this is not
necessarily true in the more general setting considered by McCullough
and Miller).
Given a hypertree Θ ∈ HT n, and an integer j ∈ [n], it is easy to count
the partial conjugations xjD carried by Θ: there is one for each collection
of connected components of Θ \ {j}, provided the collection excludes the
connected component containing the least vertex of [n] \ {j}. The next
lemma follows.
Lemma 4.5. For each h ∈ {0, . . . , n−2}, and each hypertree Θ ∈ HT h
n, Θ
carries exactly 2h automorphisms, including the identity automorphism.
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260 The symmetries of McCullough-Miller space
4.2. McCullough-Miller space Kn
We define a relation ∼ on Out0(Wn)×HT n as follows: (α, Θ) ∼ (β, Λ)
if and only if Θ = Λ, and α−1β is carried by Θ. It is easily verified that
∼ is an equivalence relation. We write [α, Θ] for the ∼-equivalence class
of (α, Θ), and we write Kn for the set of ∼-equivalence classes.
We define a partial order ≤ on Kn as follows: [α, Θ] ≤ [β, Λ] if and
only if α−1β is carried by Λ, and Λ folds to Θ. Equivalently, [α, Θ] ≤ [β, Λ]
if and only if Λ folds to Θ and [β, Λ] = [α, Λ].
McCullough-Miller space Kn is the simplicial realization of (Kn, ≤).
We write Vn(α, Θ) for the vertex in Kn corresponding to [α, Θ].
Remark 4.6. Since Θ0
n carries only the identity in Out0(Wn), equivalence
classes of the form [α, Θ0
n] are singletons, and [α, Θ0
n] ≤ [β, Λ] if and only
of β = α. Thus Kn consists of copies of HTn, one copy for each element of
Out0(Wn), glued appropriately. Vertices of the form Vn(α, Θ0
n) are called
nuclear vertices.
4.3. A map Out(Wn) → Aut(Kn)
The set Out0(Wn) × HT n is naturally equipped with a left Out(Wn)-
action: for all φ ∈ Out0(Wn), σ ∈ Σn, and (α, Θ) ∈ Out0(Wn) × HT n,
we define φσ.(α, Θ) = (φσασ−1, σΘ). It is easily verified that this action
preserves the equivalence relation ∼, and that the induced action on Kn
preserves the partial order ≤. Thus we have a homomorphism Out(Wn) →
Aut(Kn, ≤), which induces a homomorphism χn :Out(Wn) → Aut(Kn).
Lemma 4.7. For each integer n ≥ 3, the homomorphism χn :Out(Wn) →
Aut(Kn) is injective.
Proof. Recall that each element of Out(Wn) has the form φσ for some
φ ∈ Out0(Wn) and some σ ∈ Σn. If φ is non-trivial and ι denotes the
identity in Out(Wn), then
χn(φσ)Vn(ι, Θ0
n) = Vn(φσισ−1, σΘ0
n) = Vn(φ, Θ0
n) 6= Vn(ι, Θ0
n).
If φ is trivial, but σ is not, then σΘ 6= Θ for some Θ ∈ HT n, hence
χn(φσ)Vn(ι, Θ) = χn(σ)Vn(ι, Θ) =
= Vn(σισ−1, σΘ) = Vn(ι, σΘ) 6= Vn(ι, Θ).
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A. Piggott 261
To prove Theorem 1.1 it suffices to show that χn is surjective, which is
achieved in the next proposition. To ensure the structure of the argument
is most clear, we describe it assuming the following technical claims, to
be proved immediately after. We claim the following:
(H) For each integer n ≥ 3, the nuclear vertices are exactly the vertices
of maximal valence in Kn.
(I) Let n ≥ 3, let xiD ∈ Out0(Wn) be a partial conjugation, and let
β ∈ Out0(Wn) be an automorphism such that, for each Θ ∈ HT n,
Θ carries β if and only if Θ carries xiD. Then β = xiD.
(J) Let n ≥ 4, let g ∈ Aut(Kn), let α ∈ Out0(Wn), and let xiD be a
non-trivial partial conjugation. If g fixes Vn(αxiD, Θ0), and fixes
pointwise the star of Vn(α, Θ0), then g fixes pointwise the star of
Vn(αxiD, Θ0).
Remark 4.8. Claim (J) fails in the case that n = 3 because, in that
case, Kn is the barycentric subdivision of the regular trivalent tree, and
an automorphism of the tree may fix pointwise the star of a valence-three
vertex v without fixing pointwise the star of those valence-three vertex
distance two from v.
Proof that χn is surjective, for n ≥ 4, assuming Claims (H),(I) and (J).
Consider an arbitrary simplicial automorphism f ∈ Aut(Kn). It follows
from Claim (H) that f maps nuclear vertices to nuclear vertices; that
is, f maps Vn(ι, Θ0
n) to Vn(α, Θ0
n) for some α ∈ Out0(Wn). It follows
that χn(α−1)f fixes Vn(ι, Θ0
n), and fixes setwise the star of Vn(ι, Θ0
n). By
Theorem 1.3, there exists σ ∈ Σn such that χn(α−1)fVn(ι, Θ) = Vn(ι, σΘ)
for all Θ ∈ HT n; hence χn(σ−1α−1)f fixes pointwise the star of Vn(ι, Θ0
n).
Now suppose that χn(σ−1α−1)f fixes pointwise the star of Vn(α, Θ0
n),
for some α ∈ Out0(Wn). Let xiD be a partial conjugation. By Claims (H)
and (I), amongst the vertices in Kn of maximal valence, Vn(αxiD, Θ0
n) is
distinguished by the set of vertices in the star of Vn(α, Θ0
n) to which it
is adjacent. It follows that χn(σ−1α−1)f fixes Vn(αxiD, Θ0
n). Claim (J)
then gives that χn(σ−1α−1)f fixes pointwise the star of Vn(xiDα, Θ0
n).
By induction we have that χn(σ−1α−1)f fixes pointwise the star of
Vn(α, Θ0
n) whenever α can be written as a product of partial conjugations.
Since the partial conjugations generate Out0(Wn), χn(σ−1α−1)f fixes
pointwise the star of every nuclear vertex, and hence the entire space Kn.
The result follows.
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262 The symmetries of McCullough-Miller space
It remains only to prove Claims (H), (I) and (J). Before addressing
Claim (H), it is convenient to define some notation. For each hypertree
Θ ∈ HT n, we write A(Θ) for the set of hypertrees in HT n, distinct from
Θ, which fold to Θ; and B(Θ) for the set of hypertrees in HT n, distinct
from Θ, which can be obtained from Θ by folding.
Proposition 4.9 (Claim (H)). For each integer n ≥ 3, the nuclear vertices
are exactly the vertices of maximal valence in Kn.
Proof of Proposition 4.9. Consider first the case that n = 3. Nuclear
vertices have valence three, and since each hypertree θ ∈ HT n carries
exactly one partial conjugation, and folds only to the nuclear hypertree,
each non-nuclear vertex in K3 is adjacent to two nuclear vertices, and no
non-nuclear vertices. Thus the result holds.
We therefore assume that n ≥ 4. It is clear that each nuclear vertex
has valence #A(Θ0
n).
Consider an arbitrary element [α, Θ] ∈ Kn, with Θ ∈ HT h
n for some
h ∈ {1, . . . , n − 2}. The definitions immediately give that, for any other
element [β, Λ] ∈ Kn:
1) [α, Θ] < [β, Λ] if and only if β−1α is carried by Θ, and Θ < Λ (from
which it follows that [α, Λ] = [β, Λ]);
2) [β, Λ] < [α, Θ] if and only if β−1α is carried by Θ, and Λ < Θ (which
does not necessarily imply that [β, Λ] = [α, Λ]).
It follows that the valence in Kn of Vn(α, Θ) is at most
#A(Θ) + #(automorphisms carried by Θ) . #B(Θ).
By Proposition 3.15, #A(Θ) ≤ #A(Ω1,2
n ). By Lemma 4.5, there are 2h
automorphisms carried by Θ. It follows from Lemma 3.11 that #B(Θ) ≤
#B(Ξ1
n) = −1 + Bn−1, where Bn−1 is the number of partitions of [n − 1]
(Bn−1 is the (n−1)-th Bell number). Thus we have that the valence in
Kn of Vn(α, Θ) is at most
#A(Ω1,2
n ) + 2n−2(−1 + Bn−1),
and the proposition is proved if we show that
#A(Ω1,2
n ) + 2n−2(−1 + Bn−1) < #A(Θ0
n).
We make the following choices, in order:
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A. Piggott 263
L 1
L 2
L 3
1 6 3 2
5 41
789102
1 6 3 254 78910
L 4 112
11
Θ
Figure 10. The construction of Θ, as in the proof of Proposition 4.9.
• we choose a partition P = {P1, . . .Pp} of {2, . . . , n} subject only to
the restriction that p > 1 (there are (−1 + Bn−1) such partitions);
• we then choose a function m :{1, . . . , p} → {1, 2} subject only to the
restriction that m(j) = 1 if 2 ∈ Pj (there are 2p−1 such functions);
• for each j ∈ {1, . . . , p}, we choose a composition Cj := (c1
j , . . . , cq
j)
of #Pj (so c1
j , . . . , cq
j are positive integers which sum to #Pj ; there
are 2−1+#Pj such compositions).
In making these choices, we have chosen one combination of data from a
possible
(−1 + Bn−1)2p−1
p
∏
j=1
2−1+#Pj = (−1 + Bn−1)2n−2
combinations.
Now for each j ∈ {1, . . . , p}, we construct a set Λj of hyperedges as
follows: if Pj = {s1, . . . , sq} with s1 > · · · > sq, then we define
Λj :=
{
{m(j), s1, . . . , sc1
j
}, {sc1
j
, . . . , sc1
j
+c2
j
}, . . . , {s
−c
q
j
+#Pj
, . . . , sq}
}
.
Finally, we define Θ to be the hypertree on [n] such that EΘ =
p
∪
j=1
Λj .
An example construction is shown in Figure 10, using the data: n = 11;
p = 4; P1 = {2, 3, 6}, m(1) = 1, C1 = (1, 2); P2 = {4, 5}, m(2) = 1,
C1 = (2); P3 = {7, 8, 9, 10}, m(3) = 2, C3 = (3, 1); P4 = {11}, m(4) = 2,
C4 = (1).
Let H denote the set of hypertrees constructed in the manner described
above. Given Θ ∈ H: the corresponding partition P, and the function m,
can be recovered from Θ by considering the connected components of
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264 The symmetries of McCullough-Miller space
Θ\{1} and Θ\{2}; the compositions Cj can be recovered from considering
the subhypertrees of Θ corresponding to each partition set. It follows
that distinct choices of input data determine distinct hypertrees. The
requirement that p > 1 ensures that Θ 6= Θ0
n. Thus we have exhibited
(−1 + Bn−1)2n−2 distinct hypertrees in H ⊂ HT n.
We wish to identify at least 1 + #A(Ω1,2
n ) more hypertrees in HT n.
First we consider the elements in A(Ω1,2
n ). Recall that these are precisely
the hypertrees with a (1, 2)-tag and at least three hyperdges. Because
some elements of H have (1, 2)-tags, this falls short of the extra hypertrees
required by 1 + #A(Ω1,2
n ) ∩ H.
Suppose Θ ∈ A(Ω1,2
n ) ∩ H. Since Θ ∈ H: the valence of j in Θ is at
most two for each j ∈ {3, . . . , n}. Since Θ ∈ A(Ω1,2
n ), it has a (1, 2) tag.
The only way this can happen is if {2} is a partition set, and m(j) = 1
for each j ∈ {2, . . . , p}. It follows that 1 has valence in Θ at least three,
and no other vertex in Θ has valence exceeding two. We write Θ′ for the
hypertree obtained from Θ by swapping the vertices 1 and 3. Then Θ′ is
not contained in H (because 3 has valence in Θ′ at least three), and it
does not have a (1, 2)-tag (because 1 and 2 are not adjacent in Θ′).
Evidently, distinct choices of Θ ∈ H ∩ A(Ω1,2
n ) give distinct hypertrees
Θ′. Hence the set
H ∪ A(Ω1,2
n ) ∪ {Θ′ | Θ ∈ H ∩ A(Ω1,2
n )}
contains exactly
(−1 + Bn−1)2n−2 + #A(Ω1,2
n )
hypertrees.
It remains only to find one more hypertree in HT n. The star tree Ξ4
n
suffices because: it does not have a (1, 2)-tag, and hence is not contained
in A(Ω1,2
n ); the valence in Ξ4
n of 4 exceeds two, and hence Ξ4
n cannot be
an element of H ∪ {Θ′ | Θ ∈ H ∩ A(Ω1,2
n )}.
Lemma 4.10 (Claim (I)). Let n ≥ 3, let xiD ∈ Out0(Wn) be a partial
conjugation, and let β ∈ Out0(Wn) be an automorphism such that, for
each Θ ∈ HT n, Θ carries β if and only if Θ carries xiD. Then β = xiD.
Proof. There exists at least one hypertree Θ ∈ HT n which carries xiD
.
It follows, by Definition 4.2 and hypothesis, β = xipDp . . . xi1D1
for some
partial conjugations xi1D1
, . . . , xipDp , each of which is carried by Θ. Thus
it suffices to show that if xjF is a partial conjugation and xjF 6= xiD, then
there exists a hypertree Λ ∈ HT n such that Λ carries xiD but Λ does
not carry xjF . Equivalently, it suffices to show that if xjF is a partial
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A. Piggott 265
conjugation and xjF 6= xiD, then there exists a hypertree Λ ∈ HT n such
that D is a union of connected components of Λ \ {i}, but F is not a
union of connected components of Λ \ {j}. We leave the reader to verify
this statement.
Lemma 4.11 (Claim (J)). Let n ≥ 4, let g ∈ Aut(Kn), let α ∈ Out0(Wn),
and let xiD be a non-trivial partial conjugation. If g fixes Vn(αxiD, Θ0),
and fixes pointwise the star of Vn(α, Θ0), then g fixes pointwise the star
of Vn(αxiD, Θ0).
Proof. Suppose g fixes Vn(αxiD, Θ0), and fixes pointwise the star of
Vn(α, Θ0).
Recall that a line tree is a hypertree which has exactly two leaves. It is
immediate from the definitions that xiD is carried by exactly (#D)!(n −
#D − 1)! line trees; let X denote this set of line trees. It follows that the
star of Vn(α, Θ0) shares at least (#D)!(n−#D −1)! vertices with the star
of Vn(xiDα, Θ0). Since n ≥ 4 and 1 ≤ #D ≤ n−2, (#D)!(n−#D−1)! ≥ 2.
Since g fixes Vn(xiDα, Θ0), it fixes setwise the star of Vn(xiDα, Θ0). Hence
g fixes setwise the set of vertices common to the stars of Vn(xiDα, Θ0) and
Vn(xiDα, Θ0), and this set contains at least two vertices corresponding to
line trees.
Now each line tree which carries xiD is fixed by exactly one nontrivial
element of Σn, and no two line trees are fixed by the same nontrivial
element of Σn. It follows that the pointwise stabilizer in Aut(HTn) of X
is the trivial subgroup of Σn. By Theorem 1.3, g acts as an element of
Σn on the star of Vn(xiDα, Θ0). But since g is contained in the pointwise
stabiliser of the vertices shared with the star of Vn(α, Θ0), g acts as the
identity on the star of Vn(α, Θ0). That is, g fixes pointwise the star of
Vn(αxiD, Θ0), as required.
References
[1] The On-Line Encyclopedia of Integer Sequences. Published electronically at
http://oeis.org, 2011.
[2] M. R. Bridson and K. Vogtmann. The symmetries of outer space. Duke Math. J.,
106(2):391–409, 2001.
[3] N. D. Gilbert. Presentations of the automorphism group of a free product. Proc.
London Math. Soc. (3), 54(1):115–140, 1987.
[4] L. Kalikow. Enumeration of parking functions, allowable permutation pairs, and
labeled trees. PhD thesis, Brandeis University, 1999.
[5] J. McCammond and J. Meier. The hypertree poset and the l
2-Betti numbers of
the motion group of the trivial link. Math. Ann., 328(4):633–652, 2004.
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266 The symmetries of McCullough-Miller space
[6] D. McCullough and A. Miller. Symmetric automorphisms of free products. Mem.
Amer. Math. Soc., 122(582), 1996.
[7] D. Warme. Spanning trees in hypergraphs with applications to Steiner trees. PhD
thesis, University of Virginia, 1998.
Appendix. Table of notation
[n] the set {1, . . . , n}
Θ0
n the hypertree on [n] with exactly one hyperedge
Ξj
n the hypertree on [n] with exactly n−1 hyperedges, each of
which contains j
Ωj,k
n the hypertree on [n] with exactly two hyperedges, {j, k} and
[n] \ {j}
HT n the set of hypertrees on [n]
HT +
n the set of hypertrees on [n] that have at least two hyperedges
HT h
n the set of hypertrees on [n] that have exactly h+1 hyperedges
Sn the set {Ξj
n | j ∈ [n]}; elements of Sn are called star trees
Ln the set of hypertrees on [n] that have exactly two leaves;
elements of Ln are called line trees
Mh
n the set of hypertrees on [n] which have exactly h+1 hyperedges,
a vertex of valence h+1, and a hyperedge of degree n−h
(note: M1
n = {Ωj,k
n | j, k ∈ [n], j 6= k} and Mn−2
n = Sn =
{Ξj
n | j ∈ [n]})
A+
n (Θ) the set of hypertrees on [n], distinct from Θ, which fold to Θ
B+
n (Θ) the set of hypertrees on [n], distinct from Θ, which can be
obtained by folding Θ
HTn the simplicial realization of (HT n, ≤), called the hypertree
complex of rank n
HT+
n the simplicial realization of (HT +
n , ≤)
Table 1. Notation relating to hypertrees
Contact information
A. Piggott Department of Mathematics, Bucknell Univer-
sity, Lewisburg PA 17837
E-Mail: adam.piggott@bucknell.edu
Received by the editors: 19.12.2011
and in final form 16.03.2012.
|
| id | nasplib_isofts_kiev_ua-123456789-152242 |
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| issn | 1726-3255 |
| language | English |
| last_indexed | 2025-12-07T15:12:56Z |
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| publisher | Інститут прикладної математики і механіки НАН України |
| record_format | dspace |
| spelling | Piggott, A. 2019-06-09T06:10:02Z 2019-06-09T06:10:02Z 2012 The symmetries of McCullough-Miller space / A. Piggott // Algebra and Discrete Mathematics. — 2012. — Vol. 14, № 2. — С. 239–266. — Бібліогр.: 7 назв. — англ. 1726-3255 2010 MSC:20E36; 05E18. https://nasplib.isofts.kiev.ua/handle/123456789/152242 We prove that if W is the free product of at least four groups of order 2, then the automorphism group of the McCullough-Miller space corresponding to W is isomorphic to group of outer automorphisms of W. We also prove that, for each integer n ≥ 3, the automorphism group of the hypertree complex of rank n is isomorphic to the symmetric group of rank n. Thanks to Murray Elder, and the University of Newcastle, Australia, for their hospitality as this paper was written. Thanks to Andy Eisenberg, and the anonymous referee, for carefully reading the paper, and suggesting improvements. en Інститут прикладної математики і механіки НАН України Algebra and Discrete Mathematics The symmetries of McCullough-Miller space Article published earlier |
| spellingShingle | The symmetries of McCullough-Miller space Piggott, A. |
| title | The symmetries of McCullough-Miller space |
| title_full | The symmetries of McCullough-Miller space |
| title_fullStr | The symmetries of McCullough-Miller space |
| title_full_unstemmed | The symmetries of McCullough-Miller space |
| title_short | The symmetries of McCullough-Miller space |
| title_sort | symmetries of mccullough-miller space |
| url | https://nasplib.isofts.kiev.ua/handle/123456789/152242 |
| work_keys_str_mv | AT piggotta thesymmetriesofmcculloughmillerspace AT piggotta symmetriesofmcculloughmillerspace |