Barrio “San Francisquito”
5. DISCUSIÓN Y ASPECTOS A CONSIDERAR
Although Lov´asz’ conjecture was specific to simple graphs, the definition of a cir- cle representation extends without change to multigraphs. Working with multi- graphs is actually somewhat easier because there are a very limited number of ways in which a loop or digon can be represented. In this section, we give a 12-vertex multigraph that does not have a circle representation from which we construct simple quartic planar graphs with the same property.
84 CHAPTER 5. CIRCLE REPRESENTATIONS
Theorem 5.8. The multigraph H shown in Figure 5.6(a) does not have a circle representation.
(a) (b)
Figure 5.6: A non-simple counterexample
The idea of the proof is to first show that any circle representation of this graph must have a particular structure forced by the neighbouring digons, indicated in Figure 5.6(b). This corresponds to a particular configuration of circles in the plane with restrictions imposed by certain pairs that must be kissing circles. The problem is then reduced to showing geometrically that this configuration is not realisable. We will handle this geometric aspect first.
Lemma 5.9. Suppose we have four circles {Ci(ri, ti)}i=1,2,3,4 in the plane with
radii ri >0 and which touch the x-axis at points (ti,0) respectively, and assume
the circles are numbered so that t1 < t2 < t3 < t4. In addition, suppose that C1
is tangent to C2 and C4, and C3 is also tangent to C2 and C4. Let N =t4−t1,
M =t3−t2, L=t2−t1 and R=t4 −t3. Then
(i) M N =LR, and
(ii) the lengthsLand Rare in fN+(M), fN−(M) (not necessarily in that order) where fN+(M) = N−M+ √ (M−N)2−4M N 2 and f − N(M) = N−M−√(M−N)2−4M N 2 for
some fixed N and 0< M ≤(3−2√2)N. In addition, (iii) fN−(M) is increasing, and
(iv) fN−1(M)> fN−2(M) whenever 0< N1 < N2 and 0< M ≤(3−2
√ 2)N1.
Proof. From Figure 5.7, we observe that
(t4−t1)2(t3−t2)2 = (4r1r4)(4r2r3) = (4r1r2)(4r3r4) = (t2−t1)2(t4−t3)2
5.2. COUNTEREXAMPLES TO LOV ´ASZ’ CONJECTURE 85
t1 t2 t3 t4
L M R
N
Figure 5.7: Configuration of circles in Lemma 5.9.
For (ii) we can write L= 1−M −R, and so from the first part we have (N −M −R)R =M N R2+ (M −N)R+M N = 0 R = N −M ± p (M−N)2−4M N 2 .
One of the solutions isR whilst the other is L, noting that we could equivalently have used the substitution R= 1−L−M.
If we let f(N, M) := fN−(M), then ∂ ∂Mf =− 1 2 2(M −N)−4N 2p(M −N)2 −4M N + 1 !
and the inequality ∂M∂ f >0 has as a solutionM < (3−2√2)N and N >0. This implies (iii). Similarly, one can verify that ∂N∂ f < 0 on the specified interval to establish (iv).
Note that the circles described in the previous lemma are not necessarily disjoint. In particular, we do not exclude the possibility thatC1 andC3 orC2and
C4 cross. These basic results allow us to show the impossibility of the following
circle configuration. Take two sets of circles,{C1, C4, C5, C8}and{C2, C3, C6, C7},
satisfying the conditions of the previous lemma so that we have eight circles in total, all tangent to the x-axis and within the sets some circles are tangent to each other as described above. Assume additionally that there are no tangencies
86 CHAPTER 5. CIRCLE REPRESENTATIONS 1 2 3 4 5 6 7 8 (a) 1 2 3 4 5 6 7 8 (b)
Figure 5.8: Impossible circle configurations.
between circles of different sets. It is convenient to view one set of four circles, say {C1, C4, C5, C8} as being above the x-axis, and the other below. Moreover,
adopting the notation for radii and tangent points with the axis defined before, we stipulate that t1 < t2 < . . . < t8. The tangencies are illustrated in Figure 5.8a in
which all numbered simple closed curves should be interpreted as circles, tangent points between circles are marked in red, and the order of the tangent points with the axis may be inferred from the numbering.
Lemma 5.10. The configuration described is impossible.
Proof. By scaling and translating horizontally, we may assume that t2 = 0 and
t7 = 1. Let M =t6−t3, M0 = t5−t4 and N0 = t8−t1. Since t3 < t4 < t5 < t6
we know that M0 < M. Also, as t1 < 0 and t8 = t1 +N0 > 1 we must have
N0 > 1. Lemma 5.9(ii) tells us that the lengths t4−t1 and t8−t5 are given by 1
2(N
0−M0±p
(M0−N0)2−4M0N0), so using our earlier notation the length of
the shorter interval is fN−0(M0). Applying Lemma 5.9(iv) then (iii), we find that fN−0(M
0
)< f1−(M0)< f1−(M).
On the other hand, we know thatf1−(M) = min(t3−t2, t7−t6), sofN−0(M0)< t3−t2
and fN−0(M0)< t7−t6. Now if fN−0(M0) =t4−t1, then we would have
t4 =t1+ (t4−t1)<0 +fN−0(M0)< t3−t2 =t3
which is a contradiction. Similarly, if fN−0(M0) = t8−t5 then t5 =t8−(t8−t5)>1−fN−0(M0)>1−(t7 −t6) =t6 so we reach a contradiction in both cases.
5.2. COUNTEREXAMPLES TO LOV ´ASZ’ CONJECTURE 87
Corollary 5.11. The configuration shown in Figure 5.8b cannot be realised by circles.
Proof. It is enough to show that it is possible to adjust the circles to create new touching points while preserving their order along the axis, so that we obtain the configuration in the preceding lemma. Below the axis, fix C2 and C6 and replace
C3 and C7 with two new circles C30 and C70 that are both tangent to C2, C6 and
the line, as shown in Figure 5.9. Then r03 > r3, meaning t6 −t3 = 2
√
r3r6 <
2pr03r6 < t6 −t03, so t03 < t3 < t4. Also, we know that t2 < t03 by choice of C30
being tangent toC2 and C6, so this replacement preserves the order of the circles.
Similarly, we haver07 < r7 from which we deduce thatt6 < t07 < t7 < t8. Since the
inequalities are strict at each step, no circle above the axis is tangent to any circle below the axis. Doing the same thing above the axis by fixing C4 and C8 and
replacing C1 and C5 produces the desired configuration, and then we are done by
Lemma 5.10.
2
6
Figure 5.9: Replacing C3 and C7 by the circles C30 and C70 with dashed edges.
Proof of Theorem 5.8. We begin by making two observations. Firstly, any pair of digons sharing exactly one vertex must be realised by two circles that are tangent at that shared vertex. This is because if one of the digons is produced by crossing circles, then the edges of the neighbouring digon are realised by arcs of the same two circles which is only possible if the two digons share both of their vertices. We also observe that for a quartic planar graph, if a vertexv of a digon is not a cut-vertex, then in any embedding of the graph the parallel edges must be consecutive in the cyclic ordering atv.
The first observation implies that any circle representation of H must have one circle representing each digon. One of the circles on which v1 lies therefore
corresponds to a digon, so the remaining edges e1 and e8 must lie on the same
88 CHAPTER 5. CIRCLE REPRESENTATIONS
point. The same argument applies at each vertex vi for i= 1,2, . . . ,8 so we find
that e1, e2, . . . , e8 all lie on the same circle and hence the cyclee1, e2. . . , e8 must
be represented by a single circle. In addition, all vertices have now been shown to be tangent points, so no circles cross. Now if a circle represetnation of H exists, it must have one circle C0 corresponding to the 8-cycle, and then 8 more circles
C1, . . . , C8 labelled so that vi is the tangent point ofCi withC0, and these points
occur in the cyclic order around C0. In addition, (C1, C4), (C2, C3), (C5, C8) and
(C6, C7) are pairs of kissing circles, and no other circles kiss.
These tangency relationships are the same as those in the configuration in Figure 5.8b if we takeC0to be the line. Indeed, if we view the circle representation
on the sphere, rotate so that a point on C0 strictly betweenv1 andv8 is the North
Pole and then apply a stereographic projection, the resulting planar drawing satisfies the conditions of Corollary 5.11. As that configuration was not realisable by circles, no circle representation of H can exist.