>k-homogeneous infinite graphs

0.41 MB·

by  Ove Ahlman

> k−HOMOGENEOUS INFINITE GRAPHS
OVE AHLMAN
6
1
0
2 Abstract. Inthisarticlewegiveanexplicitclassificationforthecount-

r ablyinfinitegraphsG whichare,forsomek,≥k-homogeneous. Itturns
a
out that a ≥k−homogeneous graph M is non-homogeneous if and only
M
if it is either not 1−homogeneous or not 2−homogeneous, both cases
which may be classified using ramsey theory.
8

]
O
1. introduction
C
. AgraphG iscalledk−homogeneousifforeachinducedsubgraphA ⊆ G
h
such that |A| = k and embedding f : A → G, f may be extended into
t
a an automorphism of G. If G is t−homogeneous for each t ≥ k (t ≤ k)
m
then G is called ≥k−homogeneous (≤k−homogeneous). A graph which
[
is both ≥k−homogeneous and ≤k−homogeneous is plainly called homoge-

2 neous. Lachlan and Woodrow [9] classified the countably infinite homoge-
v
neous graphs. Since then, the study of homogeneous structures have been
7
continued in many different ways. When it comes to countably infinite ho-
0
3 mogeneous structures Cherlin [3] classified all such digraphs and Lachlan
7 [10] classified all such tournaments. Even more kinds of infinite homo-
0
geneous structures have been classified, however few results about infinite
.
1
k−homogeneousstructureswhicharenothomogeneousseemtoexist. When
0
itcomestofinitestructuresGardiner[6]andindependentlyGolfandandKlin
6
1 [8]classifiedallhomogeneousfinitegraphs. Cameron[2]extendedthistothe
: k−homogeneouscontextandshowedthatany≤5−homogeneousgraphisho-
v
i mogeneous. Thus classifying the finite k−homogeneous graphs comes down
X
to the cases which are not t−homogeneous for some t ≤ 5. Among others,
r
a Chia and Kok [5] took on this task and characterized finite k−homogeneous
graphs with a given number of isolated vertices and nontrivial components.
In general though no known characterization of the finite k−homogeneous,
≥k−homogeneous or ≤k−homogeneous graphs exist. In the present article
howeverwedomakeprogressinthesubjectwhenitcomestoinfinitegraphs,
and provide a full classification of all ≥k−homogeneous infinite graphs.
For each t ∈ Z+ define the graph G as having universe G = Z×{1,...,t}
t t
and edges E = {{(a,i),(b,j)} : a (cid:54)= b}. If t ≥ 2 let H be the graph with
t,1
2010 Mathematics Subject Classification. 05C75, 05C63, 03C50.
Key words and phrases. >k-homogeneous, classification, countably infinite graph.
1
2 OVEAHLMAN
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Figure 2. The
Figure 1. The graph G ∪˙K
2 3
graph H
2,1
universe H = Z×{1,...,2t} and edges
t,1
E = {{(a,i),(b,j)} : i,j ≤ t or i,j > t, and a (cid:54)= b}.
t,1
Let H have the same universe as H but with edge set
t,2 t,1
E = E ∪{{(a,i),(b,j)} : i ≤ t,j > t and a = b}.
t,2 t,1
Lastly define the graph H as having universe Z × {1,2} and edge set
1,2
E = {{(a,i),(b,j)} : i = j or a = b but i (cid:54)= j}.
Theorem 1.1. Let k ∈ N. Then M is a ≥k−homogeneous graph if and
only if exactly one of the following hold.
(i) M is a homogeneous graph.
(ii) M is not 1−homogeneous and for some finite homogeneous graph H
and some t we have that M ∼= G ∪˙H or M ∼= (G ∪˙H)c.
t t
(iii) M is 1−homogeneous but not 2−homogeneous and for some t we
have that M ∼= H , M ∼= H , M ∼= Hc or M ∼= Hc .
t,1 t,2 t,1 t,2
Asthefinitehomogeneousgraphsareclassifiedaseitherthe3×3rookgraph,
the5−cycle, adisjointunionofcompletegraphsorthecomplementofoneof
the previous graphs, case (ii) is complete. We prove (ii) in Section 2 Lemma
2.1, (iii) in Section 3 Lemma 3.1 and lastly in Section 4 Lemma 4.1 we show
that any ≥k−homogeneous graph which does not fit into (ii) or (iii) has to
be homogeneous, in other words, (i) is proven.
A graph G is homogenizable if there exists a homogeneous structure M
with a finite amount of extra relational symbols in its signature compared
to G such the automorphism groups of M and G are the same and if we
remove all extra relations from M, we get G. For a more detailed definition
of homogenizable structures and explicit examples see [1, 11]. From the
proof of Theorem 1.1 we can draw the following two corollaries which relate
to being homogenizable.
>k−HOMOGENEOUS INFINITE GRAPHS 3
Corollary 1.2. If, for some k, M is a countably infinite ≥k−homogeneous
graph which is not 1−homogeneous then M is homogenizable by only adding
a single unary relation symbol.
Corollary 1.3. If, for some k, M is a countably infinite ≥k−homogeneous
graph which is 1−homogeneous but not 2−homogeneous then M is homog-
enizeable by adding only a single binary relation symbol and one of the fol-
lowing holds:
• k = 5 and M ∼= H or M ∼= Hc .
1,2 1,2
• k = 2n+1 > 3 and M ∼= H or M ∼= Hc .
n,1 n,1
• k = 4n+1 > 5 and M ∼= H or M ∼= Hc .
n,2 n,2
Note that an overlap between cases is to be expected. Corollary 1.2 does
not have an explicit classification of M depending on k, such as Corollary
1.3, since M ∼= G ∪˙H or M ∼= (G ∪˙H)c where both t and H affect for which
t t
k that M is ≥k−homogeneous. It is clear in Corollary 1.3 that we may
add a unary relation symbol to make the graph homogeneous. This however
does not follow the definition of being homogenizable since adding a unary
relation symbol changes the automorphism group. In the terminology of
Cherlin [4] we have proven that a ≥k−homogeneous graph has relational
complexity at most 2.
Notation and terminology. For each t ∈ Z+, K is the complete graph
t
on t vertices and K is the countably infinite complete graph. Whenever we
∞
talk about subgraphs H ⊆ G we mean induced subgraph in the sense that for
a,b ∈ H we have that aEGb if and only if aEHb. An embedding of graphs
f : G → H is an injective function such that for each a,b ∈ G we have aEGb
if and only if f(a)EHf(b). If we write K ⊆ G or K ⊆ G it means that for
t ∞
some subgraph H of G, H is isomorphic to K or K respectively. If G is
t ∞
a graph with a ,...,a ∈ G then K (a ,...,a ) is a complete subgraph of G
1 r t 1 r
containing t vertices, which includes a ,...,a . For t ∈ Z+, a t−orbit of G is
1 r
an orbit of t−tuples which arise when the automorphism group of G acts on
Gt. One of our main tools in the proofs is Ramsey’s famous theorem about
the existence of infinite complete or infinite independent subgraphs which
now a days is common practice and may be found in for instance [7].
Fact 1.4 (Ramsey’s Theorem). If G is an infinite graph then K ⊆ G or
∞
Kc ⊆ G.
∞
2. Graphs which are not 1−homogeneous
Lemma 2.1. For some k ∈ N, M is a ≥k−homogeneous graph which is not
1−homogeneous if and only if there exists t ∈ N and a finite homogeneous
graph H such that M ∼= G ∪˙H or M ∼= (G ∪˙H)c.
t t
Proof. Lemma2.8provesthatifMis≥k−homogeneousandK ⊆ Mthen
∞
M ∼= G ∪˙H for some t ∈ N and homogeneous finite graph H. It follows that,
t
using Ramsey’s theorem, if K (cid:54)⊆ M then Kc ⊆ M in which case it follows
∞ ∞
4 OVEAHLMAN
similarly that M ∼= (G ∪˙H)c.
t
In order to prove the second direction assume that M ∼= G ∪˙H and notice
t
that both G and H are homogeneous graphs and as subgraphs of M they
t
constitute distinct 1−orbits which are not connected to each other. Let
G(cid:48) ⊆ M be such that |G(cid:48)| = k = 3max(|H|,t). There are at least 2t vertices
in G(cid:48) which belong to G , thus there exists a vertex a ∈ G(cid:48) which is adjacent
t
to at least |H| vertices in M. Furthermore if b ∈ G then b is adjacent
t
either to a or an element in G(cid:48) which a is adjacent to. Thus any embedding
f : G(cid:48) → M has to map G(cid:48)∩G into G and G(cid:48)∩H into H. As H and G are
t t t
both homogeneous this means that f can be extended to an automorphism
of M, thus M is ≥k−homogeneous. (cid:3)
In the rest of this section we assume that M is ≥k−homogeneous but
not 1−homogeneous, thus M has more than one 1−orbit. Due to Ramsey’s
theorem K or Kc is embeddable in M. We will assume that K is em-
∞ ∞ ∞
beddable in M. The reader may notice that all the reasoning in this section
may be done in the same way if Kc would be embeddable by switching all
∞
references of edges and non-edges, thus producing a result for the comple-
ment.
Lemma 2.2. There are exactly two 1−orbits in M and one of them is finite.
Proof. Since M is ≥k−homogeneous, all elements which are in K ⊆ M
∞
have to be in the same 1−orbit, call it p. Assume that a,b ∈/ p and note that
a and b are adjacent to at most k−1 elements in K ⊆ M. Thus we may
3k
find G ⊆ K ⊆ M such that |G| = k and no element in G is adjacent to a or
3k
b. Let f : G∪{a} → G∪{b} be the embedding which maps G to G and a to
b. Since M is ≥k−homogeneous f may be extended to an automorphism,
thus a and b belong to the same orbit.
The orbit p has to be infinite since all elements in K ⊆ M belong to p.
∞
Assumethatthesecondorbit, callitq, isalsoinfinite. ByRamsey’stheorem
either K or Kc is embeddable in q, however K is impossible since M
∞ ∞ ∞
is ≥k−homogeneous and the orbits p and q are distinct. Each element in
Kc ⊆ q ⊆ Misadjacenttoatmostk−1elementsinK ⊆ p. Howeverthen
∞ k
there has to be a ∈ K ⊆ p and G ⊆ Kc ⊆ q such that |G| = k and a is not
k ∞
adjacenttoanyelementinG. Butthenanyembeddingf : G∪{a} → G∪{a}
which does not fixate a will be extendable to an automorphism, by the
≥k−homogeneity of M. Thus a ∈ q, which contradicts that a ∈ p. (cid:3)
We will keep notation from the previous Lemma and let p be the infinite
1−orbit and let q be the finite 1−orbit in M.
Lemma 2.3. If a ∈ p and b ∈ q then a is not adjacent to b.
Proof. If some element in p is adjacent to some element in q then for each
a ∈ p there exists some b ∈ q such that a is adjacent to b . As p is infinite
0 0 0 0
and q is finite, there has to exist b ∈ q such that b is adjacent to an infinite
amount of vertices in K ⊆ M. However, if G ⊆ K ⊆ M with |G| = k
∞ ∞
>k−HOMOGENEOUS INFINITE GRAPHS 5
such that all vertices in G are adjacent to b then there is an embedding
f : G∪{b} → G∪{b} which does not fixate b. Since M is ≥k−homogeneous
it is possible to extend f to an automorphism of M. Thus b ∈ p which is a
contradiction. (cid:3)
As Lemma 2.3 prove that p and q are not connected to each other, the
next lemma shows that each element in p is non-adjacent to at most k −1
elements in p.
Lemma 2.4. If a ∈ p then there are at most k+|q|−1 elements in M which
a is not adjacent to.
Proof. Assume a ∈ p is not adjacent to any elements in some G(cid:48) ⊆ M such
that |G(cid:48)| = k +|q| and let G = G(cid:48) ∩p. By Lemma 2.3 no element in G is
adjacent to any element in q. Assume b ∈ q. The function f : G ∪{a} →
G∪{b} mapping G to G and a to b is thus an embedding. Since |G| ≥ k it is
possible, by ≥k−homogeneity, to extend f into an automorphism. It follows
that a ∈ q which is a contradiction. (cid:3)
K (a) is any subgraph G of M which is isomorphic to K and contain a.
∞ ∞
Thus the following lemma proves that b is adjacent to all elements (except
a) in all K ⊆ M which contain a.
∞
Lemma 2.5. If a,b ∈ p are such that a is not adjacent to b then b is adjacent
to each element in K (a)−{a}.
∞
Proof. Assume that each element in p is non-adjacent to exactly t other
elements in p. Assume in search for a contradiction that d ∈ K (a)−{a} is
∞
not adjacent to b. Each pair of distinct elements from K (a) belong to the
∞
same 2−orbit. Thus for any distinct α,β ∈ K (a) there exists γ ∈ p such
∞
that α and β are both not adjacent to γ. This however implies either that a
would be non-adjacent to more than t different elements or that there would
exist some element c ∈ p which is non-adjacent to more than t elements
in K (a). Both of these conclusions lead to a contradiction since we have
∞
assumed each element in p to be non-adjacent to exactly t other elements in
p. (cid:3)
Lemma 2.6. If a,b,c ∈ p are such that a is not adjacent to b and a is not
adjacent c, then b is not adjacent to c.
Proof. Assume that b is adjacent to c. Lemma 2.5 implies that both b and c
are adjacent to each element in K (a)−{a}. Thus (K (a)∪{b,c})−{a}
∞ ∞
is a complete graph, where both b and c are non-adjacent to a. As b is not
adjacent to a and c ∈ K (b,c), Lemma 2.5 implies that c is adjacent to a
∞
which is a contradiction. (cid:3)
Lemma 2.7. If a,b ∈ p and a is adjacent to b then there exists G ⊆ M such
∼
that a,b ∈ G and G = K .
∞
6 OVEAHLMAN
Proof. If a is adjacent to an infinite amount of elements in some subgraph
∼
G ⊆ M such that G = K and b ∈ G, then the lemma holds. Assume, in
∞
search for a contradiction, that there exist elements c,d ∈ K (b) such that
∞
a is not adjacent to both c and d. Lemma 2.6 then implies that c is not
adjacent to d, which is a contradiction. (cid:3)
We now summarize all our knowledge about p and q into the following
lemma which proves the second part of this section’s main Lemma 2.1.
Lemma 2.8. For some t ∈ Z+, p ∼= G and M ∼= G ∪˙H for some homoge-
t t
neous finite graph H.
Proof. Assumethateachelementa ∈ pisnon-adjacenttotelementsb ,...,b
1 t
∈ p. By Lemma 2.6 these t elements always form a Kc. If c ∈ p is adjacent
t
to b , for some i ∈ {1,...,t}, then Lemma 2.5 and Lemma 2.7 implies that
i
∼
c is adjacent to all of b ,...,b . It is thus clear that p = G .
1 t t
IfwechooseG ⊆ Msuch that|G∩p| > k thenanyembeddingf : G → M
has to be possible to extend into an automorphism. Thus f has to preserve
the orbits. However, as q and p have no vertices which are adjacent to each
other any embedding g : q → q is possible to extend into an automorphism
of q, hence q is a finite homogeneous graph. We conclude that M ∼= G ∪˙H
t
for some finite homogeneous graph H. (cid:3)
3. Graphs which are 1−homogeneous but not 2−homogeneous
Lemma 3.1. A countably infinite graph M is ≥k−homogeneous, for some
k ∈ N, 1−homogeneous but not 2−homogeneous if and only if there exists t
such that M ∼= H , M ∼= H , M ∼= Hc or M ∼= Hc .
t,1 t,2 t,1 t,2
Proof. We prove that the suggested graphs are actually ≥k−homogeneous,
forsomek,andtheotherdirectionisdoneinLemma3.12. Wefirstnotethat
if M = H or M = H then there are two parts in M, each of which is
t,1 t,2
∼
isomorphic to G for some r. If M = H and G ⊆ M such that |G| ≥ 2t+1
r t,1
there has to be at least one edge between some elements a,b ∈ G. Thus if
f : G → M is an embedding, each vertex adjacent to a or b will then be
mapped to one of the G parts of M and each vertex not adjacent a nor
r
b, will be mapped the other G part. As no edges exist between the two
r
parts and each part is a homogeneous graph f may be extended into an
automorphism.
∼
If M = H and G ⊆ M with |G| ≥ 4t + 1 then there exist vertices
t,2
a,b,c ∈ G which are adjacent to each other. If f : G → M then a,b and c
needs to be mapped to the same part. An element d ∈ G is mapped to the
same part as a,b and c if and only if it is adjacent to two of these vertices.
As edges between parts are preserved by f and each part is homogeneous, f
may be extended into an automorphism.
If M ∼= Hc or M ∼= Hc then the reasoning is equivalent. (cid:3)
t,1 t,2
>k−HOMOGENEOUS INFINITE GRAPHS 7
In order to prove the second direction of Lemma 3.1, we will assume
throughout the rest of this section that M is ≥k−homogeneous, 1−homoge-
neous but not 2−homogeneous i.e. there are more than three 2−orbits but
only a single 1−orbit. Due to Ramsey’s theorem M has to contain either
K orKc . WewillassumethatMcontainsK andthereadermaynotice
∞ ∞ ∞
thatallthereasoninginthissectionmaybedoneinthesamewayforKc by
∞
switching all references to edges and non-edges. Since there is only a single
1−orbit, writing K (a) always makes sense for any vertex a ∈ M, while
∞
writing K (a,b) needs to be motivated in order to show existence.
∞
Lemma 3.2. There are at most two 2−orbits containing tuples of adjacent
elements in M and there are at most two 2−orbits containing tuples of dis-
tinct non-adjacent elements in M.
Proof. Assume (a,b ), (a,b ) and (a,b ) are three different 2−orbits such
1 2 3
that a is adjacent to b ,b and b . We may assume that a is the first co-
1 2 3
ordinate of all three parts without loss of generality, since we only have a
single 1−orbit in M. One of the 2−orbits may be assumed to be a part of
a K , say (a,b ). Since M is ≥k−homogeneous this property is unique for
∞ 1
the orbit of (a,b ). But then neither b nor b may be adjacent to more than
1 2 3
k−1 of the elements in K (a,b ). Thus we are able to find G ⊆ K (a,b )
3k 1 3k 1
such that a ∈ G, |G| ≥ k and nothing in G, except a, is adjacent to b or b .
2 3
The function f : G ∪{b } → G ∪{b } mapping G and b to G and b is an
1 3 2 3
embedding, and hence the ≥k−homogenity implies that (a,b ) and (a,b )
2 3
are of the same orbit, contradicting the assumption.
For the second part of the lemma, assume (c,d ), (c,d ) and (c,d ) are
1 2 3
different 2−orbits such that c is non-adjacent to all of d ,d and d . Since
1 2 3
the orbits are different the ≥k−homogenity implies that d is adjacent to
1
at least k of the vertices in K (c) if and only if d is adjacent to at most
2k 2
k−1 vertices in K (c). However, d has to be adjacent or non-adjacent to
2k 3
at least k vertices in K (c) and hence the orbit of (c,d ) can’t be distinct
2k 3
from the two other orbits by the ≥k−homogenity of M. (cid:3)
The previous lemma implies that we may assume there are at most five
2−orbits in M, out of which one is the orbit containing identical element
2−tuples (x,x). Call the 2−orbits where elements have an edge between
them, q and q and assume that q is the orbit of pairs of elements in K .
1 2 1 ∞
Itfollowsthat(a,b) ∈ q ifaisadjacenttolessthank−1elementsinK (b)
2 ∞
and a is adjacent to b.
Lemma 3.3. For each a ∈ M, there are only a finite, possibly zero, amount
of elements b ∈ M such that (a,b) ∈ q .
2
Proof. Let A ⊆ M be the subgraph containing all elements b such that
a
(a,b) ∈ q and assume in search for a contradiction that A is infinite. By
2 a
Ramsey’s theorem either K ⊆ A or Kc ⊆ A . If K ⊆ A then, since a is
k a k a k a
adjacent to each element in A , for any b ∈ K ⊆ A , (a,b) ∈ q which is a
a k a 1
contradiction against that (a,b) ∈ q .
2
8 OVEAHLMAN
OntheotherhandassumethatKc ⊆ A andb ∈ Kc. EachelementinA
k a k a
is adjacent to less than k−1 elements in K (a), thus there exists a vertex
∞
c ∈ K (a) such that none of the elements in Kc ⊆ A is adjacent to c. The
∞ k a
function f : Kc∪{a} → (Kc−{b})∪{a,c} mapping (Kc−{b})∪{a} back
k k k
to it self pointwise and b to c is then an embedding. Thus ≥k−homogenity
imply that (a,b) ∈ q , which is a contradiction. (cid:3)
1
Call the 2−orbits of tuples of elements which have no edge between them
p and p . Assume p is the orbit of pairs (a,b) such that b is adjacent to
1 2 1
at most k−1 of the elements in K (a). We note that p has to exist, since
∞ 1
elseq can’texist, whichwouldimplythatMhasatmostthree2−orbits. It
2
follows,usingthe≥k−homogenity,thateachelementbwhichisnon-adjacent
to at least k−1 elements in K (a) is such that (a,b) ∈ p . Thus the orbit
∞ 1
p contains all pairs (a,b) such that a and b are non-adjacent yet b is non-
2
adjacent to less than k −1 elements in K (a). It follows quickly from the
∞
definition, and the ≥k−homogenity, that the four orbits p ,p ,q and q are
1 2 1 2
symmetric in the sence that (a,b) ∈ r implies (b,a) ∈ r.
Lemma 3.4. Let a,b,c ∈ M. If (a,b) ∈ q ,(a,c) ∈ p and b is not adjacent
1 1
to c then (b,c) ∈ p .
1
Proof. Since (a,c) ∈ p , c is adjacent to at most k−1 elements in K (a,b),
1 ∞
thus (b,c) ∈ p . (cid:3)
1
We are now ready to prove that, similarly to q , the orbit p is finite if we
2 2
fixate one component.
Lemma 3.5. For each a ∈ M there are only a finite, possibly zero, amount
of elements b ∈ M such that (a,b) ∈ p .
2
Proof. Assume c is such that (a,c) ∈ p , let L be the set of all elements
1
b ∈ M such that (a,b) ∈ p and assume that L is infinite. By Ramsey’s
2
theorem, we either have K ⊆ L or Kc ⊆ L.
∞ ∞
If K ⊆ L then all these elements are non-adjacent to a but then the
∞
definition of p implies that (a,b(cid:48)) ∈ p for each b(cid:48) ∈ K ⊆ L. This is a
1 1 ∞
contradiction against (a,b(cid:48)) ∈ p .
2
Assume instead that Kc ⊆ L and let G ⊆ Kc ⊆ L be such that all
∞ ∞
elementsinG arenon-adjacenttoc. If|G| ≥ k−2thenanyinjectivefunction
f : G ∪{a} → G ∪{a,c} is an embedding, thus ≥k−homogenity imply that
(a,c) is in the same orbit as (a,d) for any d ∈ G. This is a contradiction,
since (a,d) ∈ p and (a,c) ∈ p , thus |G| ≤ k−3. Together with (a,b) ∈ p
2 1 2
thisimpliesthatthereisanelemente ∈ K (a)andH ⊆ Kc ⊆ L, suchthat
∞ ∞
|H| = 2k andbothcandeareadjacenttoallelementsinH. Assumewithout
loss of generality that b ∈ H. There are embeddings g : H∪{c} → H∪{a,e}
which map (b,c) to (a,e). This together with ≥k−homogenity imply that
(b,c) ∈ q . Lemma 3.4 together with (b,c) ∈ q , (a,c) ∈ p and (a,b) ∈ p
1 1 1 2
implies that (a,b) ∈ p which is a contradiction. (cid:3)
1
>k−HOMOGENEOUS INFINITE GRAPHS 9
The next lemma shows that the orbits q and p , in some sense, are closed
1 2
and together form a tight part of the graph M. This is a vital property
which will be used many times in order to handle q and p in the rest of
1 2
the section.
Lemma 3.6. Leta,b,c,d ∈ Msuchthatc (cid:54)= d,(a,b) ∈ q and(a,c),(a,d) ∈
1
p . Then (b,c) ∈ q and (c,d) ∈ p .
2 1 2
Proof. If b is not adjacent to c then, for every a(cid:48) ∈ K (a), since (a(cid:48),a) ∈
∞
q , there has to exist an element c(cid:48) such that (a,c(cid:48)),(a(cid:48),c(cid:48)) ∈ p . But
1 2
this contradicts Lemma 3.5, since each element c such (a,c ) ∈ p is non-
0 0 2
adjacent to at most k−1 elements in K (a). Thus we conclude that only
∞
a in K (a) can be non-adjacent to c, hence b is adjacent to c and more
∞
specifically (b,c) ∈ q .
1
Assume(c,d) ∈/ p andnotethat(a,c),(a,d) ∈ p impliesthatcanddare
2 2
non-adjacent to a finite amount of elements in K (a). Thus c is adjacent
∞
to an infinite amount of elements in K (b) and hence we conclude that
∞
(c,d) ∈ q . We previousply proved in this lemma though that (c,d) ∈ q
1 1
and (c,a) ∈ p implies that (d,a) ∈ q which is a contradiction against
2 1
(a,d) ∈ p . (cid:3)
2
Itismuchhardertogetagripoftheorbitsq andp . Thisisaconsequence
2 1
of that we assumed K ⊆ M and thus having neighbors which are also
∞
adjacent to some element is easy to handle. The rest of the section will be
dedicated to reasoning out how these orbits work in M.
Lemma 3.7. Foreacha,b ∈ Mif(a,b) ∈ q theneachelementc ∈ K (a)−
2 ∞
{a} will be such that (b,c) ∈ p .
1
Proof. Assume c is adjacent to b. For every element c ∈ K (a), (c ,a)
0 ∞ 0
is in the same 2−orbit as (c,a) thus there exists an element d such that
0
(a,d ),(c ,d ) ∈ q . This implies either that there is an infinite amount of
0 0 0 2
elements d(cid:48) such that (a,d(cid:48)) ∈ q or that there is an element d(cid:48)(cid:48) such that
2
(d(cid:48)(cid:48),c ) ∈ q for an infinite amount of elements c ∈ K (a). However both
0 2 0 ∞
of these conclusions are contradictions against Lemma 3.3. Thus c is not
adjacent to b. By the definition of q , there exists some G ⊆ K (a) such
2 ∞
that |G| ≥ k and each element in G is non-adjacent to b, thus (b,c) ∈ p . (cid:3)
1
In the upcoming two lemmas we will show that the orbit q and the orbit
2
p are very closely linked, and in fact most cases where q exist, also p has
2 2 2
to exist.
Lemma 3.8. If a,b ,b ∈ M, b (cid:54)= b and (a,b ),(a,b ) ∈ q then (b ,b ) ∈
1 2 1 2 1 2 2 1 2
p .
2
Proof. By Lemma 3.7 if d ∈ K (b ) then d can at most be adjacent to k−1
∞ 1
elements in K (a). If (b ,b ) ∈ q then b ∈ K (b ) which together with
∞ 1 2 1 1 ∞ 2
Lemma 3.7 implies that (a,b ) ∈ p which is a contradiction.
1 1
Assume instead that (b ,b ) ∈ q . Choose G ⊆ K (a) and d ∈ K (b )
1 2 2 ∞ ∞ 1
10 OVEAHLMAN
such that |G| = k and all elements in G are non-adjacent to b ,b and d.
1 2
The function f : G ∪{b ,d} → G ∪{b ,b } mapping G to G and (b ,d) to
1 1 2 1
(b ,b ) is then an embedding, thus ≥k−homogenity implies that (b ,b ) and
1 2 1 2
(b ,d) belong to the same orbit which is a contradiction as (b ,d) ∈ q . We
1 1 1
conclude that b must be nonadjacent to b .
1 2
Assume (b ,b ) ∈ p and let c ∈ K (a). It is then possible to find
1 2 1 ∞
G ⊆ K (b ) such that |G| = k and all elements in G are non-adjacent to a,c
∞ 1
and b . If f : G ∪{a,c} → G ∪{a,b } maps G to G and (a,c) to (a,b ) then
2 2 2
the ≥k−homogenity implies that (a,c) ∈ q , which is a contradiction. (cid:3)
2
Lemma 3.9. Let a,b,c ∈ M. If (a,b) ∈ q and (b,c) ∈ p then (a,c) ∈ q .
2 2 2
Proof. By Lemma 3.6 we know that (a,c) ∈/ q and (a,c) ∈/ p . Assume, in
1 2
search for a contradiction, that (a,c) ∈ p . Let d ∈ K (b)−{b} and note by
1 ∞
Lemma3.7that(a,d) ∈ p . Since(a,d)and(a,c)areinthesameorbit,there
1
has to exist an element e ∈ M, corresponding to what b is to (a,d), such
that(e,c) ∈ q and(e,a) ∈ q . Lemma3.8nowimpliesthat(e,b) ∈ p which
1 2 2
in turn together with Lemma 3.6 implies that (e,c) ∈ p . But (e,c) ∈ q ,
2 1
hence this is a contradiction and we can conclude that (a,c) ∈ q . (cid:3)
2
Lastly we figure out how the orbit p behaves. This is the hardest orbit
1
to handle, as it induces so little information about edges.
Lemma 3.10. Let a,b,c ∈ M. If (a,b) ∈ p and (a,c) ∈ p then (b,c) ∈ p .
1 2 1
Proof. Lemma 3.6 implies that (b,c) ∈/ q and (b,c) ∈/ p , and Lemma 3.9
1 2
implies that (b,c) ∈/ q , thus we conclude that (b,c) ∈ p . (cid:3)
2 1
Lemma 3.11. Let a,b,c ∈ M with b (cid:54)= c. If (a,b),(a,c) ∈ p then (b,c) ∈ q
1 1
or (b,c) ∈ p .
2
Proof. Assume (b,c) ∈ q and let G ⊆ K (a) be such that |G| = k and
2 ∞
both b and c are non-adjacent to each element in G. By the definition of p
1
and Lemma 3.7 there exists d ∈ K (b) such that d is non-adjacent to each
∞
element in G and to c. The function f : G ∪{b,d} → G ∪{b,c} mapping
G ∪ {b} pointwise to G ∪ {b} and d to c is then an embedding, which by
the ≥k−homogenity may be extended into an automorphism. This is a
contradiction since (b,c) ∈ q but (b,d) ∈ q .
2 1
Assume for the rest of this proof that (b,c) ∈ p . In order to reach a
1
contradiction in this case we will also need to make assumptions on which
of the orbits q and p exists. Assume that p exist, let d be such that
2 2 2
(a,d) ∈ p . Lemma 3.10 implies that (b,d) ∈ p , thus d is adjacent to at
2 1
most k − 1 elements in K (b). We may then find G ⊆ K (b) such that
∞ ∞
|G| = k and all elements in G are non-adjacent to a,c and d. Thus the
function f : G∪{a,d} → G∪{a,c} mapping G∪{a} to itself pointwise and d
to c is an embedding. Since M is ≥k−homogeneous f may be extended into
an automorphism which implies that (a,d) ∈ p which is a contradiction.
1
Assume q exists and that d is such that (a,d) ∈ q . If both b and c
2 2
are non-adjacent to less than k elements in K (d) then there is an infinite
∞