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Intrinsic Chirality of Multipartite Graphs

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Abstract

We classify which complete multipartite graphs are intrinsically chiral.
J Math Chem (2013) 51:1853–1863
DOI 10.1007/s10910-013-0187-y
ORIGINAL PAPER
Intrinsic chirality of multipartite graphs
Erica Flapan ·Will Fletcher
Received: 20 March 2013 / Accepted: 29 April 2013 / Published online: 7 May 2013
© Springer Science+Business Media New York 2013
Abstract We classify which complete multipartite graphs are intrinsically chiral.
Keywords Chiral ·Achiral ·Intrinsic chirality ·Spatial graphs ·Multipartite graphs
Mathematics Subject Classification (1991) 57M25 ·57M15 ·92E10 ·05C10
1 Introduction
The chemical properties of molecules and interactions between molecules often
depend on whether the molecules have mirror image symmetry. A molecule is said to
be chiral if it cannot transform into its mirror image, otherwise it is said to be achi-
ral. Chemical chirality is determined experimentally, but predicting chirality based on
molecular formulae and bond connectivity is important, especially in the process of
designing new pharmaceuticals. Physical models can be used to determine whether
rigid molecules will have mirror image symmetry, but large molecules pose greater
difficulty, especially if they can attain numerous conformations by rotating around
multiple bonds. One method for determining the chirality of such complex molecules
is to model them as topological graphs embedded in 3-dimensional space, where ver-
tices and edges correspond to atoms and bonds. In topology, a structure is chiral if there
is no ambient isotopy taking it to its mirror image, otherwise it is achiral. If a structure
The first author was supported in part by NSF Grant DMS-0905087.
E. Flapan (B
)
Department of Mathematics, Pomona College, Claremont, CA 91711, USA
e-mail: eflapan@pomona.edu
W. Fletcher
Department of Physics, University of Cambridge, Cambridge CB30DS, UK
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is topologically achiral, then there exists an orientation reversing homeomorphism
under which the structure is setwise invariant, and if it is topologically chiral then no
such homeomorphism exists. Since molecular motions do not alter bond connectivity,
such motions are ambient isotopies. Thus a molecule whose corresponding embedded
graph is topologically chiral must itself be chemically chiral.
A graph is called intrinsically chiral if all possible embeddings of the graph in 3-
dimensional space are chiral. If the graph corresponding to a molecule is intrinsically
chiral, it follows that the molecule and all its stereoisomers (which have the same bond
connectivity) are chemically chiral. Intrinsic chirality has been demonstrated for sev-
eral families of graphs, including complete graphs of the form K4n+3(n1)[8] and
Möbius ladders with an odd number of rungs (at least three) [1]. A number of mole-
cules have also been shown to be intrinsically chiral, including the molecular Möbius
ladder [2], the Simmons-Paquette molecule [3], triple-layered naphalenophane [5], fer-
rocenophenone [6], and two different fullerenes with caps [10]. Liang and Mislow [9]
classify molecules according to a hierarchy of types of chirality and provide examples
of each type, including 21 additional examples of intrinsically chiral molecules.
While it is not hard to check that every complete bipartite graph is achirally embed-
dable in 3-dimensional space, the task of determining whether or not a complete mul-
tipartite graph has an achiral embedding is more complex. In this paper, we provide
such a characterization. Note that our results imply that the intrinsically linked graph
K3,3,1is achirally embeddable (like its cousin K6), while the intrinsically knotted
graph K3,3,1,1is intrinsically chiral (like its cousin K7).
2 Intrinsic chirality of multipartite graphs
We work in the 3-sphere S3=R3∪{}, however the reader should note that a graph
is intrinsically chiral in S3if and only if it is intrinsically chiral in R3. To construct
achiral embeddings, we will make use of the following Lemma.
Lemma 2.1 [7]Let G be a finite group of homeomorphisms of S3and let γbe a graph
whose vertices are embedded in S3as a set V such that G induces a faithful action
on γ. Let Y denote the union of the fixed point sets of all of the nontrivial elements of
G. Suppose that adjacent pairs of vertices in V satisfy the following hypotheses:
(a) If an adjacent pair is pointwise fixed by nontrivial elements h,gG , then fix(h)=
fix(g)
(b) No adjacent pair is interchanged by an element of G
(c) Any adjacent pair that is pointwise fixed by a nontrivial g G bounds an arc in
fix(g)whose interior is disjoint from V (Yfix(g))
(d) Every adjacent pair is contained in a single component of S3Y
Then there is an embedding of the edges of γsuch that the resulting embedding of γ
is setwise invariant under G.
Define the join G1+G2of graphs G1and G2as the graph G1G2together with
additional edges joining every vertex in G1to every vertex in G2. Then for any natural
number nand any graph G, we define nG as the join of ncopies of the graph G.Fora
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complete multipartite graph Km1,m2,...,mn, we will refer to the sets mias partite sets,
and we define the size of each partite set as the number of vertices it contains.
The next three theorems present various forms of complete multipartite graphs that
have an achiral embedding. Afterward, we prove that these theorems account for all
achirally embeddable complete multipartite graphs.
Theorem 2.2 A complete multipartite graph γis achirally embeddable in S3if it has
the form 4G1+2G2+G3+Kq1,q2,q3where G1,G2,G3, and Kq1,q2,q3are (possibly
empty) complete multipartite graphs such that all of the partite sets in G2have even
size, all of the partite sets in G3have size divisible by 4, and one of the following
conditions holds:
(1) Two or three of the qi=0
(2) Two q iare odd and equal, and the third qi=0
(3) Two q iare odd and equal, the third qi1(mod 4), and G1=∅
(4) One qi2(mod 4), one qi1(mod 4), and the third qi=0
(5) Two q i2(mod 4), the third qi=0, and G1=∅
Proof Let Pbe a sphere in S3containing the origin Oand the point , and let be a
circle orthogonal to Pthat meets Pat Oand .Leth:S3S3be the composition
of a 90rotation around and reflection through P, and let Hdenote the group of
order 4 generated by h.
Let γdenote a multipartite graph of the form 4G1+2G2+G3+Kq1,q2,q3satisfying
one of the conditions (1)–(5). We will embed the vertices of γas a set Vin S3which
is setwise invariant under Hsuch that Hinduces a faithful action on γ. To embed the
edges of γwe will apply Lemma 2.1 after checking that hypotheses (a)—(d) of the
lemma are satisfied. Since S3is connected, hypothesis (d) is immediate. We will
never embed a pair of adjacent vertices at Oand , and hence hypothesis (a) will
always be satisfied. Thus in each case we only need to check hypotheses (b) and (c).
Furthermore, if no adjacent pair of vertices is contained in , then hypothesis (c) is
trivially satisfied.
Let Bdenote a ball which is disjoint from , and is small enough that
B,h(B), h2(B), and h3(B)are pairwise disjoint. We embed the vertices of one copy
of G1as a set V1of distinct points in B. We embed the vertices of the three other copies
of G1as h(V1), h2(V1), and h3(V1).Let4V1denote the set of embedded vertices of
4G1.
Next we embed the vertices of 2G2. Since each partite set in G2has an even number
of vertices, we embed half the vertices of each partite set of G2as a set of distinct
points W2in BV1. Then we embed the other half of the vertices of each of these
partite sets as h2(W2). Then V2=W2h2(W2)is the set of embedded vertices of
one copy of G2. We embed the vertices of the second copy of G2as h(V2).Let2V2
denote the set of embedded vertices of 2G2.
By hypothesis, each partite set in G3has size divisible by 4. We embed one fourth
of the vertices of each partite set of G3as a set W3of distinct points in B(V1W2).
Then embed the remaining vertices of each partite set in G3as the images of W3under
Hin such a way that each partite set is setwise invariant under H.LetV3denote the
set of embedded vertices of G3.
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Thus we have embedded the vertices of 4G1+2G2+G3as the set of points
4V12V2V3, which is setwise invariant under H. Furthermore, Hinduces a faithful
action on γ. For each qi,let Qidenote the set of qivertices in Kq1,q2,q3. In each of
the following cases, we embed Q1,Q2, and Q3and then use Lemma 2.1 to embed
the edges of γ. After embedding each set Qiwe will abuse notation and refer to the
embedded set of vertices as Qi.
Case 1a: All three qi=0
In this case, we have already embedded all of the vertices of γ. Observe that h2is
the only element of Hthat interchanges any pair of vertices. For v2V2V3,the
vertices h2(v) and vare in the same partite set, and hence are not adjacent. However,
for v4V1, the pair vand h2(v) are in different partite sets, and thus are adjacent.
In order to avoid violating hypothesis (b) of Lemma 2.1 we define an associated
graph γobtained from γby adding a vertex of valence 2 in the interior of every edge
vh2(v) with vV1h(V1). We will refer to these valence 2 vertices as auxiliary
vertices. For each vV1we embed a distinct auxiliary vertex vfor the edge vh2(v)
in −{O,∞}, and we embed an auxiliary vertex for the edge h(v)h3(v) as h(v).
Thus we have embedded the vertices of γsuch that Hinduces a faithful action on γ
and no pair of adjacent vertices is interchanged by an element of H. Thus hypothesis
(b) is satisfied.
Since no pair of adjacent vertices is embedded in , hypothesis (c) is trivially
satisfied. It follows that the hypotheses of Lemma 2.1 are satisfied by the embedded
vertices of γ. Hence we can apply Lemma 2.1 to obtain an embedding of the edges of
γsuch that the resulting embedding of γis setwise invariant under H. By omitting
the auxilliary vertices we obtain an achiral embedding of γ.
Case 1b: Precisely two qi=0
Without loss of generality we assume that q1= 0 and q2=q3=0. If q1=2m,
we embed mvertices of Q1in −{O,∞}, and embed the other mas the image of
the first munder h.Ifq1=2m+1, we embed 2mvertices as above and embed an
additional vertex at the point O. Now the embedded vertices of γare setwise invariant
under h, and Hinduces a faithful action on γ.
To satisfy hypothesis (b), we define the associated graph γas in Case 1a. For each
vV1we embed an auxiliary vertex vfor the edge vh2(v) in ({O,∞} ∪ Q1),
and we embed an auxiliary vertex for the edge h(v)h3(v) as h(v).Novertexin Q1is
interchanged with an adjacent vertex by an element of H, so no additional auxiliary
vertices are required. Since no pair of adjacent vertices is embedded in , hypothesis
(c) is again trivially satisfied. Now the hypotheses of Lemma 2.1 are satisfied, so again
we can embed the edges of γand omit the auxiliary vertices to obtain an achiral
embedding of γ.
Case 2: Two qiare odd and equal, and the third qi=0
Without loss of generality, we assume that q1=q2=2k+1 and q3=0. We
embed kvertices of Q1as distinct points in B(V1W2W3). Then embed the
other kvertices of Q1as the image of the first kvertices under h2. We embed the last
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vertex of Q1as a point v1on −{O,∞}. Now embed the vertices of Q2as h(Q1).
Observe that even though the sets Q1and Q2are interchanged by h, the only adjacent
vertices in Q1Q2that are interchanged by an element of Hare v1and h(v1).
Let Abe the arc in from v1to h(v1)which contains O. For each vV1we
embed an auxiliary vertex vfor the edge vh2(v) in (A∪{}), and we embed an
auxiliary vertex for the edge h(v)h3(v) as h(v). Finally, we embed the auxiliary vertex
corresponding to the edge v1h(v1)as the point O. Since v1and h(v1)are adjacent
to the auxiliary vertex at Oand are fixed by h2, we must check that hypothesis (c)
is satisfied. Observe that the two subarcs A−{O}satisfy this property. Now the
hypotheses of Lemma 2.1 are satisfied for γ, so again we can embed the edges of γ
and omit the auxiliary vertices to obtain an achiral embedding of γ.
Case 3: Two qiare odd and equal, the third qi1(mod 4), and G1=∅.
Without loss of generality, we assume q1=q2=2k+1 and q3=4j+1. We
embed Q1and Q2as in Case 2. We embed jvertices of Q3in B(W2W3Q1),
and embed the other 3 jvertices as their images under H.Weletv3denote the final
vertex of Q3, which we embed at .
Since G1is empty, v1and h(v1)are the only adjacent vertices interchanged by any
element of H. We embed a single auxiliary vertex for the edge v1h(v1)at O. Thus
hypothesis (b) is satisfied. Since each vertex on has been embedded between the
two vertices it is adjacent to, hypothesis (c) is also satisfied. Now the hypotheses of
Lemma 2.1 are satisfied, so again we can embed the edges of γand omit the auxiliary
vertex to obtain an achiral embedding of γ.
Case 4: One qisatisfies qi2(mod 4), one qisatisfies qi1(mod 4), and the
third qi=0
Without loss of generality, we assume that q1=4k+2,q2=4j+1, and q3=0.
First we embed kvertices of Q1and jvertices of Q2as disjoint sets of points in
B(V1W2W3). Then we embed the other three sets of jand kvertices of Q1
and Q2by taking the images under Hof the kand jvertices we have embedded in
B. We embed one more vertex of Q1as v1on −{O,∞}. Then we embed the last
vertex of Q1as h(v1). Finally, the last vertex of Q2we embed at the point O.
Let Abe the arc in from v1to h(v1)which contains O. For each vV1we
embed an auxiliary vertex vfor the edge vh2(v) in (A∪{}), and we embed an
auxiliary vertex for the edge h(v)h3(v) as h(v).NovertexinQ1Q2is interchanged
with an adjacent vertex by an element of H, so hypothesis (b) is satisfied without any
additional auxiliary vertices. Since vertices v1and h(v1)are adjacent to the vertex at
Oand are fixed by h2, we have to check hypothesis (c) of Lemma 2.1. However, the
subarcs A−{O}satisfy the required property. Now the hypotheses of Lemma 2.1 are
satisfied, so again we can embed the edges of γand omit the auxiliary vertices to
obtain an achiral embedding of γ.
Case 5: Two qisatisfy qi2(mod 4), the third qi=0, and G1=∅
Without loss of generality, we assume q1=4k+2,q2=4j+2, and q3=0. We
embed the vertices of Q1and 4 jvertices of Q2as in Case 4. The last two vertices of
Q2are embedded at Oand .
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Since Q1and Q2are setwise invariant under h, none of their vertices are inter-
changed with adjacent vertices by any element of H. Now because G1is empty,
hypothesis (b) is satisfied without any auxiliary vertices. Since each vertex on is
embedded between the two others it is adjacent to, hypothesis (c) is also satisfied. It
then follows from Lemma 2.1 that we can embed the edges of γto obtain an achiral
embedding of γ.
Theorem 2.3 A complete multipartite graph γis achirally embeddable in S3if it has
the form G +Kq1,q2,q3,q4where G and Kq1,q2,q3,q4are (possibly empty) complete
multipartite graphs, all of the partite sets in G have even size, and one of the following
conditions holds:
(1) All qi=0
(2) One qiis odd, and all other qi=0
(3) Two q i=1, one qi=0, and the last qiis either odd or 0
(4) All qi=1
Proof Let h:S3S3be an inversion through the origin O, which fixes Oand .
Then his an orientation reversing homeomorphism whose fixed point set is {O,∞}.
Let Hdenote the group of order 2 generated by h.LetPdenote a sphere that contains
{O,∞}.Letγdenote a complete multipartite graph of the form G+Kq1,q2,q3,q4
satisfying one of the conditions (1)–(4).
Since S3−{0,∞} is connected, hypothesis (d) of Lemma 2.1 holds for any embed-
ding of the vertices of γ. Hypothesis (a) must also hold because Hhas only one
nontrivial element. Additionally, since fix(h)={O,∞}, hypothesis (c) will hold as
long as we do not embed a pair of adjacent vertices at Oand . Thus we only need
to check hypothesis (b) for each embedding of the vertices.
Each partite set in Ghas an even number of vertices, so we embed half the vertices
of each partite set in one component of S3Pas W. Next we embed the other half
of the vertices of Gas h(W). Then V1=Wh(W)is the set of embedded vertices
of G. Note that each partite set in V1is setwise invariant under h,sonovertexinV1is
interchanged with an adjacent vertex by hand thus the vertices in V1satisfy hypothesis
(b).
For each i,letQidenote the partite set with qivertices. After embedding each Qi
we will abuse notation and refer to the embedded set of vertices also as Qi.
Case 1: All qi=0
In this case, we have already embedded all of the vertices of γ. Since all the
hypotheses of Lemma 2.1 hold, it follows that there is an embedding of the edges
of γsuch that the resulting embedding of γis setwise invariant under H. Thus γis
achirally embeddable.
Case 2: One qiis odd and all the other qi=0
Without loss of generality, we assume q1=2k+1 and q2=q3=q4=0. We
embed kof these vertices in one component of S3(PV1)and another kvertices
as the image of the first kvertices under h. The final vertex is embedded at O. Thus
Q1is setwise invariant under h. Furthermore, Hinduces a faithful action on γ, and
hypothesis (b) holds. Thus Lemma 2.1 provides an achiral embedding of γ.
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Case 3: Two qi=1, one qi=0, and the last qiis either odd or 0
If precisely one qi=0, we assume q1=2k+1,q2=q3=1, and q4=0. We embed
Q1as in case 2. Embed the single vertex of Q2as a point v2on P−{O,∞} and
embed the single vertex of Q3as h(v2). Consider the associated graph γobtained
from the abstract graph γby adding an auxiliary vertex on the edge v2h(v2), which we
embed at . Then Hinduces a faithful action on γ, and γsatisfies the hypotheses
of Lemma 2.1. Thus we can embed the edges of γand then omit the auxiliary vertex
to obtain an embedding of γthat is invariant under Hand is therefore achiral.
If two qi=0, we assume q1=q4=0, and we embed Q2and Q3as above.
Case 4: All qi=1
Embed the single vertices of Q1and Q3as distinct points v1and v3on P−{O,∞}
such that h(v1)= v3. Then embed the single vertices of Q2and Q4as the points
h(v1)and h(v3). Then hinterchanges Q1with Q2, and Q3with Q4,so Hinduces a
faithful action on γ.
Consider the associated graph γobtained from the abstract graph γby adding
auxiliary vertices on the edges v1h(v1)and v3h(v3). We embed the auxillary vertices
at Oand . Then Hinduces a faithful action on γ, and γsatisfies the hypotheses of
Lemma 2.1. Thus we can embed the edges of γand then omit the auxiliary vertices
to obtain an embedding of γthat is invariant under Hand is therefore achiral.
For the next result we need the following definition. A planar graph is called out-
erplanar if its join with a single vertex is still a planar graph.
Theorem 2.4 A graph γis achirally embeddable in S3if it has the form G P+Q
where one of the following conditions holds:
(1) GPis a planar graph and Q is K1,1,K2,2, or a set containing an even number of
vertices.
(2) GPis an outerplanar graph and Q is a set containing an odd number of vertices.
Proof Let Pdenote a sphere passing through the origin Oand , and let denote a
circle orthogonal to Ppassing through Oand . Define has a reflection across the
sphere P, so the fixed point set of his P.LetHdenote the group of order 2 generated
by h.
Case 1: GPis a planar graph and Qis K1,1,K2,2, or a set containing an even
number of vertices.
Embed GPin P.IfQis a set of 2kvertices, embed kvertices in one component
of −{O,∞} and the other kas their image under h.Letdenote the vertices of γ
together with the edges joining Qand GP. Observe that Hinduces a faithful action
on , and satisfies all four hypotheses of Lemma 2.1. This yields an embedding
of the edges between Qand GPthat is setwise invariant under h. The union of this
embedding of and our original embedding of GPin Pis an achiral embedding of
γ.
If Qis K1,1or K2,2, we embed the vertices of Qand the edges joining Qand GP
as we did above when Qwas even, making sure in the K2,2case that the vertices of
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the two partite sets alternate around . The remaining edges of Qare contained in .
Thus again we have an achiral embedding of γ.
Case 2: GPis an outerplanar graph and Qis a set containing an odd number of
vertices.
If Qhas 2k+1 vertices, we embed the join of GPand one vertex of Qin the plane
P. Then embed the remaining 2kvertices of Qand the edges as we did when Qwas
even. This gives us an achiral embedding.
An immediate consequence of Theorem 2.4 is that every complete bipartite graph
is achirally embeddable, since any set of vertices is an outerplanar graph.
We now prove the converse of Theorems 2.2,2.3, and 2.4.
Theorem 2.5 Let γbe a complete multipartite graph that has an achiral embedding
in S3. Then γcan be expressed in one of the forms given in Theorem 2.2, Theorem 2.3,
or Theorem 2.4.
Proof If γis not 3-connected, then either γ=Kn,1,1for some n1, or γhas fewer
than three partite sets. In either case, γcan be expressed as GP+Qwhere GPis an
outerplanar graph and Qis a (possibly empty) set of vertices. Thus γhas the form
given in Theorem 2.4.
Thus we assume that γis a 3-connected graph which has an achiral embedding 1
in S3. Then there is an orientation reversing homeomorphism h1of the pair (S3,
1).
Since γis 3-connected, it follows from Flapan [4] that there is a possibly different
embedding 2of γin S3such that (S3,
2)has a finite order, orientation reversing
homeomorphism h2. We may express order(h2)=2ab, where a,bZand bis
odd. Note that a1 because h2is orientation reversing. Let h=(h2)b. Then
order(h)=2a, and his orientation reversing because bis odd. By Smith Theory [11],
fix(h) is either two points or a sphere.
Suppose that hpointwise fixes a sphere P. We aim to prove that γhas the form
given in Theorem 2.4.Let Aand Bdenote the two components of S3P. Since his
orientation reversing, hmust interchange Aand B. Observe that hsetwise fixes any
edge that passes through P, so if two adjacent vertices are in separate components,
they are interchanged by h(and so their corresponding partite sets are interchanged as
well). It follows that a vertex in one component can be adjacent to at most one vertex
in the other component.
Suppose that there is a vertex vin Athat is adjacent to a vertex win B.Letvbe
in the partite set Vand let wbe in the partite set W. Then hinterchanges vand was
well as Vand W, so neither set can have any vertices in P. Furthermore, since vand
wcan be adjacent to at most one vertex in the complementary component of S3P,
the partite set Wcan have no additional vertices in B, the partite set Vcan have no
additional vertices in A, and no partite set other than Vor Wcan have vertices in
AB. Thus Vcan have at most one additional vertex, and it must be in B, in which
case Whas another vertex embedded in A. Thus Vand Wboth have one vertex or both
have two vertices, and the vertices from the remaining partite sets must be embedded
in P. It follows that γcan be expressed in the form GP+K1,1or GP+K2,2, where
GPis a planar graph.
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If instead ABdoes not contain a pair of adjacent vertices, then ABcan contain
only vertices from a single partite set that is setwise invariant under h,soγ=GP+Q
where GPis a planar graph and Qis a set of vertices. If Qhas an odd number of
vertices, then one must be embedded in P, in which case GPmust be an outerplanar
graph. In either case, γhas the form in Theorem 2.4.
For the remainder of the proof we assume that fix(h) is two points. As a consequence,
hcannot fix two adjacent vertices, since if it did, it would pointwise fix the edge
between them.
Suppose that order(h)=2. We aim to prove that γhas the form in Theorem 2.3,so
toward a contradiction suppose that it does not. Then one of the following apply: (i)
γhas at least two odd partite sets each with at least three vertices, (ii) γhas precisely
one odd partite set with at least three vertices, and either exactly one or at least three
partite sets with one vertex, or (iii) γhas at least five partite sets with only one vertex.
Suppose that (i) holds, then let V1and V2be odd partite sets with at least three
vertices. If each Viwas setwise invariant under h, then one vertex in each set would
be pointwise fixed by h. As this would violate our assumption that hdoes not fix two
adjacent vertices, hmust instead interchange V1and V2. However, for each of the
vertices vV1,itnowfollowsthathfixes the midpoint of the edge vh(v), which
contradicts the assumption that |fix(h)|=2.
Suppose that (ii) holds, then let V1be a partite set whose size is odd and contains
at least three vertices. No other set has the same number of vertices, so V1must be
setwise invariant under h, and we let v1V1be a vertex which is fixed by h. Then
no vertex in any other partite set can be fixed, since hcannot fix a pair of adjacent
vertices. The partite sets with one vertex must be interchanged in pairs by h, so there
must be an even number of them (specifically, there cannot be just one). Then if there
are at least three partite sets that have one vertex, there are at least two such pairs of
interchanged vertices. In this case, hfixes the midpoints of these two edges as well as
v1, which again contradicts |fix(h)|=2.
Finally, suppose that (iii) holds, then either all of the single vertex sets are inter-
changed in pairs, or one is pointwise fixed while the rest are interchanged. If all are
interchanged in pairs, then hfixes the midpoint of the edge between each pair, of
which there are at least three. If instead one vertex is fixed, then halso still fixes the
midpoints of at least two pairs, so again hfixes three points. Thus, if order(h)=2,
then γhas the form in Theorem 2.3.
From now on, we assume that the order of his a multiple of 4. We aim to prove that
γhas the form given in Theorem 2.2.Let=fix(h2). Then is a circle since h2is
orientation preserving and not the identity. Furthermore, for every i1,fix(h2i)=
.
Every partite set in γis either setwise invariant under h, is interchanged with
another partite set by h, or is cycled by hwith order a multiple of 4. We consider
these types of partite sets one at a time beginning with partite sets which are cycled
by hwith order a multiple of 4. Let U1,U2,...,Urbe representatives of each orbit of
these partite sets under h. Then each Uiis cycled by hwith order 4ki. For each i,let
Ui=Uih(Ui)∪···hki1(Ui)and let g=h2a2. Then ghas order 4 and cycles
the vertices in each
Uiwith order 4.
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Now define V1=
U1···∪
Ur. It follows that the partite sets which are cycled
with order a multiple of 4 are contained in V1g(V1)g2(V1)g3(V1). We define
G1γas the complete multipartite graph with vertices in V1. Then the subgraph of
γspanned by vertices in V1g(V1)g2(V1)g3(V1)can be expressed as 4G1.Note
that h2a1=g2has order 2 and for each vV1g(V1), the vertices vand g2(v) are
adjacent and interchanged by g2. Hence g2fixes the midpoint of each edge vg2(v).
Thus the midpoints of 2|V1|edges must be contained in .
Next, we consider all those partite sets whose size is even which are interchanged
with another partite set by h.LetT1,T2,...,Tsbe representatives of each orbit of these
partite sets under h.LetV2be the union of these sets of vertices. Now all partite sets
whose size is even which are interchanged with another partite set by hare contained
in V2h(V2). Thus the subgraph of γspanned by vertices in V2h(V2)can be
expressed in the form 2G2.
Now let G3denote the subgraph spanned by all vertices in partite sets that are
setwise invariant under hwhose size is a multiple of 4. This leaves odd partite sets
that hinterchanges with another partite set and partite sets which are invariant under
hwhose size is not a multiple of 4. As we will show, both of these types of partite sets
must have some of their vertices embedded on .
Let W1and W2denote odd partite sets such that h(W1)=W2and |W1|=|W2|=
2k+1. Then |W1W2|=4k+2, so W1W2contains a cycle of length 2. Thus h
must interchange some vertex w1W1with some vertex w2W2. These vertices
are fixed by h2and hence are embedded on . Thus the edge w1w2is also contained
in . If there were an additional pair of odd partite sets interchanged by h, there would
be another two adjacent vertices w3and w4on . However, K1,1,1,1does not embed
in a circle, so four mutually adjacent points cannot be embedded on . Thus there is
at most one pair of odd partite sets interchanged by h. If there is such a pair of partite
sets, let V4denote the union of the vertices in this pair, otherwise let V4=∅.
Next, let Qdenote a partite set which is invariant under hand whose size is not a
multiple of 4. Then Qhas 4k+qvertices, where 1 q3. Since the order of the
cycles in Qmust divide order(h), every cycle in Qhas order 1, 2, or a multiple of 4. If
q=1, then hfixes one vertex of Q.Ifq=2, then heither fixes or interchanges two
vertices of Q.Ifq=3, then hfixes one vertex in Qand interchanges the two others.
In any of these cases, at least qvertices of Qare embedded on .
Since at most three mutually adjacent points can be embedded in a circle, there are
no more than three such partite sets Qiwhich are invariant under hand whose size is
not a multiple of 4. If there were exactly three Qi, then each qi=1, since K1,1,1is
the only tripartite graph that embeds in a circle. However, each Qiis setwise invariant
and hcannot fix a pair of adjacent vertices, so at most one qi=1. Therefore there are
at most two Qi.
If there are two Qi, then either both qi=2, or one qi=1 and the other qi=2.
This is because at most one qi=1, and K1,1,K1,2,and K2,2are the only bipartite
graphs that embed in a circle. Thus, one of the following holds:
There are two Qi, and both qi=2, or one qi=2 and one qi=1.
There is only one Qi, and qi=1,2,or 3.
There are no Qi.
123
J Math Chem (2013) 51:1853–1863 1863
To prove that γhas one of the forms given in Theorem 2.2, we will identify which
of the above configurations of the Qiare possible, according to whether V1and V4
are empty. In the case that both V1and V4are empty, all of the above configurations
of the Qiare possible, since no edges or vertices of γbesides those in the Qimust be
embedded on . In this case, γsatisfies condition (1), (4), or (5) in Theorem 2.2.
Suppose that V1is empty and V4is nonempty. Then there are two adjacent vertices
w1,w
2V4which are embedded on . Since can contain at most three mutually
adjacent vertices and K1,1,1is the only tripartite graph that embeds in a circle, there
can be at most one Qiand it can have only one vertex on (i.e. qi=1). In this case,
γsatisfies condition (2) or (3) in Theorem 2.2.
Suppose that V1is nonempty and V4is empty. Then there cannot be two Qiwith
both qi=2 because K2,2is homeomorphic to a circle, leaving no space on for the
midpoints of the 2|V1|edges. Thus γsatisfies condition (1) or (4) in Theorem 2.2.
Finally, suppose that both V1and V4are nonempty. Again since V4=∅there are
two adjacent vertices w1,w
2V4on . Hence there can be at most one Qiand
it would need to have qi=1. If there were such a Qi, then would contain three
mutually adjacent vertices, and the edges between them would leave no space on for
the midpoints of the 2|V1|edges. Thus there can be no Qi, and so γsatisfies condition
(2) in Theorem 2.2.
It follows that γcan be expressed in one of the forms given in Theorems 2.2,2.3,
or Theorem 2.4.
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123
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