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PROOF OF THE PLANAR TRIPLE BUBBLE CONJECTURE
WACHARIN WICHIRAMALA
Abstract. We prove the planar triple bubble conjecture that the standard triple bubble is the
unique least-perimeter way to enclose and separate three regions of given areas. We also prove
a bound on the number of convex components and a bound on number of all components in any
stable planar bubble.
Contents
1. Introduction 1
2. Preliminaries 2
3. Weak approach to the soap bubble problem 3
4. Bounds on the number of components 5
4.1. Variations of bubbles 5
4.2. Variations of convex components 7
4.3. Variations on long edges 8
4.4. Bounds on the number of convex components 9
4.5. Bounds for the total number of components 10
5. Geometry of planar bubbles 12
5.1. Elementary geometry 12
5.2. Reduction and decoration on bubbles 14
5.3. Geometry of triple bubbles 16
5.4. Symmetry of weakly minimizing bubbles 19
6. Proving the conjecture 28
6.1. Crucial variational argument 28
6.2. Eliminating nonstandard bubbles 30
6.3. The conjecture 41
References 41
1. Introduction
Humans have been fascinated by soap bubbles in many ways since ancient times. One aspect is
that they are believed to minimize surface area while enclosing fixed volumes of air. The planar
soap bubble problem is the mathematically analogous problem in two dimensions: the search for
the least-perimeter way to enclose and separate regions R1, . . . , Rmof given areas A1, . . . , Amon
the plane.
It has been known since the time of the ancient Greeks that a circle is the best way to enclose
a single given area, but this was not proved rigorously until much later in the late nineteenth
century. Intuitively, it seems clear that it should be best to keep each region in a single connected
1
2 WACHARIN WICHIRAMALA
Figure 1.1. A standard triple bubble.
component. But mathematically, this is the main difficulty: there is no a priori way to show that
it is not better to split one region into several smaller components.
The case of two areas was solved by a group of undergraduate students, Foisy, Alfaro, Brock,
Hodges and Zimba [FABHZ]. They found a new approach to eliminate the possibility of having
empty chambers. Another group of undergraduate students, Cox, Harrison, Hutchings, Kim, Light,
Mauer and Tilton [CHHKLMT] proved that the standard triple bubble in Figure 1.1 is shortest
among enclosures of connected regions (including the exterior). In his Ph.D. dissertation, Vaughn
showed the new approach is valid for the case of three areas and proved that any minimizing
triple bubble with equal pressures and without empty chambers is standard. The double bubble
conjecture was additionally proved in R3[HMRR] and in R4[RHLS]. Before the twentyfirst century
began, the triple bubble problem was open even for the case of three equal areas.
Our proof. We prove the bound on the number of convex components and then eliminate the
remaining triple bubbles using a variation argument developed from [HMRR]. The original proof
of the component bound uses explicit elementary calculation as we can see in [Wic2].
Content. Preliminary results are in section 2. In section 3, we discuss the new approach to the
bubble problem that automatically eliminates the possibility of having empty chambers. Section 4
contains the variation argument needed for proving the bound on the number of convex components
and, later, the conjecture. We use the bound for convex components to make the bound on the
total number of components. In section 5, geometric tools are developed and, at the end, we prove
a structure theorem for a certain class of bubbles. We finally prove the triple bubble conjecture in
section 6.
Acknowledgement. This work is based on the Ph.D. thesis [Wic1] of the author. Thanks
are due to his advisor John M. Sullivan. Helpful communication from Richard Vaughn, Michael
Hutchings, Angel Montesinos Amilibia, Manuel Ritor´e and Frank Morgan are also appreciated. In
addition, the author would like to thank Aimo Hinkkanen for his support.
2. Preliminaries
An m-bubble can be considered as an embedded graph on the plane where each face is labeled
by a number 1, . . . , m or 0.
Theorem 2.1. [Ble1, Mor, CHHKLMT] For A1, . . . , Am>0, there is a minimizing cluster of
areas A1, . . . , Am. Every minimizing cluster (1) is composed of finitely many circular/straight
edges separating different regions and meeting only in threes at 120◦angles. (2) All edges form a
connected graph (or every component is simply connected). (3) There are pressures p1, . . . , pm∈R
PROOF OF THE PLANAR TRIPLE BUBBLE CONJECTURE 3
such that every edge between Riand Rjhas curvature pi−pj(curves into the lower pressure region)
where p0is set to be zero.
A bubble is called stationary or equilibrium if it has no area-preserving variation that initially
decreases length. We found that [CHHKLMT] being stationary is equivalent to satisfying properties
(1) and (3).
Proposition 2.2. [CHHKLMT] For a bubble Bwith pressures p1, . . . , pm, and any variation {Bt}
of B, we have
dlength(Bt)
dt ¯
¯
¯
¯0
=
m
X
i=1
pi
dAi(t)
dt ¯
¯
¯
¯0
,
where Ai(t)denotes the area of the ith bounded region of Bt.
Proposition 2.3. [CHHKLMT] A stationary bubble of areas A1, . . . , Amand pressures p1, . . . , pm
has total length 2PpiAi.
We can easily see that the highest pressure must be positive and that every area-preserving
variation on a stationary bubble initially fixes length. A bubble is called stable if it is stationary
and has no area-preserving variation that decreases length in the second order.
Lemma 2.4. For an n-sided component of a bubble, the sum of all edges’ turning angles is 6−n
3π
if the component is bounded and is −6−n
3πif the component is unbounded.
Proof. This is clear using the Gauss-Bonnet equation. ¤
Proposition 2.5. [Ble3] For a minimizing bubble, any two components may meet at most once,
along a single edge.
Corollary 2.6. [FABHZ] A minimizing m-bubble has no two-sided component if m≥3.
3. Weak approach to the soap bubble problem
In the proof of the planar double bubble conjecture [FABHZ], a new approach was introduced
in order to rule out the existence of empty chambers. The authors proved that a minimizing
double bubble has no empty chambers and then proved that a minimizing double bubble without
empty chambers is standard. To see briefly how this approach works, consider a bubble with an
empty chamber. If we assign the chamber to be part of one of the neighboring components and
then remove the redundant edge, we would get a shorter cluster that encloses bigger areas. Later,
Vaughn showed [Vau] explicitly how this idea works for the problem of three areas.
For given v1, . . . , vm>0, let A(v1, . . . , vm) be the hyper-surface area Hn−1of the minimizing
cluster enclosing volumes v1, . . . , vmin Rn. Specifically for two dimensions, let L(A1, . . . , Am) be
the length of the minimizing cluster enclosing areas A1, . . . , Am.
Proposition 3.1. [Hut] The mappings Aand Lare continuous.
Lemma 3.2. For V1, . . . , Vm>0, the minimum minvi≥ViA(v1, . . . , vm)is attained.
Proof. As Ais eventually increasing to infinity in each vi, the statement is clear. ¤
The clusters that attain the minimum surface area as in the previous lemma are called weakly
minimizing bubbles or weak minimizers for the given volumes V1, . . . , Vm. By the first para-
graph of this section, a weak minimizer has no empty chambers. By definition, each weak minimizer
is a minimizing cluster for the volumes vithat it encloses.
4 WACHARIN WICHIRAMALA
Theorem 3.3. [Vau] The planar triple bubble conjecture holds if every weakly minimizing triple
bubble is standard.
This theorem suggests an approach to proving the planar triple bubble conjecture, which we
refer to as the weak approach.
Theorem 3.4. [Vau] A minimizing triple bubble with equal pressures and without empty chambers
is standard and has equal areas.
Although a standard triple bubble has equal areas if and only if it has equal pressures, the
previous theorem does not settle the planar triple bubble conjecture for equal areas. However, it
does rule out a certain case of bubbles. Hence, by the weak approach, we have to rule out only
triple bubbles without empty chambers and with unequal pressures.
Next we will show that the weak approach is also valid for the planar m-bubble problem with
m≤6 and for the m-bubble problem in Rnwith m≤n+ 1. In fact, the approach also works for
seven and eight areas.
Proposition 3.5. A weak minimizer for areas A1, . . . , Amhas no empty chamber. Its pressures
p1, . . . , pmare nonnegative. Each region Rihas area exactly Aiunless pi= 0.
Proof. Let Wbe a weak minimizer. If Whas an empty chamber, we can reassign it to be part of
one of the neighboring components and then remove the redundant edge. Then we get a shorter
cluster, contradicting the minimality of W. If pi<0, as we increase the area of Ri, the length
decreases by Proposition 2.2, again giving a contradiction. If the area of Riis greater than Aiand
pi>0, as we decrease the area of Ri, and length again decreases. Hence we get a shorter cluster
that still contains areas bigger than A1, . . . , Am, a contradiction. ¤
The previous proposition is also valid in higher dimensions.
Theorem 3.6. For m≤6, the planar m-bubble conjecture holds if every weak minimizer is stan-
dard.
Proof. Let Wbe a weak minimizer for areas A1, . . . , Am; it is standard by the assumption, meaning
that the regions are connected. First we show that Whas areas A1, . . . , Am. Suppose region Ri
has area greater than Ai. Hence pi= 0 which is the lowest pressure, so the turning angle for each
edge of Riis nonpositive. Since m≤6 and the regions are connected, Rihas at most six sides, and
then by Lemma 2.4, Riis hexagonal. Hence m= 6 and Ritouches all other regions including R0.
Thus all neighboring regions have pressure zero, contradicting the fact that the highest pressure
must be positive. So Whas areas A1, . . . , Am, and is therefore a minimizer. Now let Mbe any
minimizer. Since l(M) = l(W), Mis also a weak minimizer. By the assumption, Mis standard.
¤
This theorem shows that, to prove the planar soap bubble conjecture for m≤6, it suffices to
consider nonstandard clusters without empty chambers and with nonnegative pressures, and to
show they are not weakly minimizing.
Theorem 3.7. The m-bubble conjecture in Rn, where m≤n+ 1, holds if every weakly minimizing
m-bubble is standard.
Proof. Let V1, . . . , Vmbe prescribed volumes. Assume that every weakly minimizing m-bubble is
standard. Let Wbe a weak minimizer. By the assumption, Wis standard. By [Mon], Whas
positive pressures since every bounded component meets the exterior region through a strictly
convex, spherical surface. By Proposition 3.5 and the remark following it, then Whas volumes
PROOF OF THE PLANAR TRIPLE BUBBLE CONJECTURE 5
V1, . . . , Vm, so Wis a minimizer. Let Mbe a minimizer. Since Mhas the same surface area as W,
the cluster Mis also a weak minimizer. By assumption, Mis standard. Therefore, the m-bubble
conjecture holds. ¤
Local renovation can show a bubble Bis not weakly minimizing as in the following example.
Suppose for a 3-dimensional bubble that regions R1and R2have interface Iof surface area Aand
the components Ciof R1meeting Ihave volume V. If Ais greater than the area of the sphere Sof
volume V, then we can construct a cluster of greater volumes but less surface area by reassigning
Cito be part of R2, removing I, and then adding Sto be part of R1floating away from the other
part.
4. Bounds on the number of components
One of the few most important things on the way to prove the soap bubble problem is knowing the
bound of the number of components. Bounds for the number of components for minimizing double
bubbles in Rnhave been proved for only small n[HLRS]. Unfortunately there is no previously
known bound for a minimizing planar triple bubble even for the case of equal areas. In this section,
we will find a bound for the number of nonhexagonal convex components for planar bubbles. From
this bound, we will establish the first bound for the total number of components for planar bubbles.
4.1. Variations of bubbles. In this subsection, we will discuss variations of planar bubbles fol-
lowing [HMRR]. We will find a formula for the second derivative of the perimeter of a bubble.
Consider a planar cluster Bwith smooth interfaces Eij between Riand Rj. Let Nij be the unit
normal vector on Eij from Rjinto Ri. Consider a continuous variation V={Bt:B→R2}|t|<ε of
Bthat is smooth on each Eij up to the boundary. The associated vector field is X:= dBt/dt|0.
The scalar normal component of Xfrom Rjinto Riis uij := X·Nij . Let kij be the curvature
of Eij ; this is nonnegative if Rihas higher pressure. Note that Nij , uij , kij are skew-symmetric in
their indices. Let N , u, k be the disjoint union functions `i<j Nij ,`i<j uij ,`i<j kij . Note that at
a vertex, the three values of N, u, k are not equal. The normal component of Xis uN, where
uN denotes the pointwise product of the functions uand N. Given a scalar or vector valued
function f=`fij defined on the interfaces of B, we define a function Y(f) on the vertices of B
by Y(f)(p) = fij(p) + fj h(p) + fhi(p) if Ri, Rj, Rhmeet at p(in that order counterclockwise). If
Y(f)(p) = 0, we say for fij agree at p. For a bubble, since Nij agree at pfor any Xand the
associated normal component u, we have Y(u)(p) = X·Y(N)(p) = 0. Hence uij agree at p. Initially
the area of Richanges at the rate PjREji uji . We say Vhas character (x0, . . . , xm)∈Rm+1 if
the areas the of Riinitially change with rates xi. We also say skew-symmetric real-valued functions
˜uij on Eij have character (x0, . . . , xm) if PjREji ˜uji =xi. If uij is the scalar normal component
of V, then the character of uij is equal to the character of V. We say Vis steady if each area of
Bchanges at a constant rate.
We will calculate the second variation formula after the first variation formula. We let T(p) be
the sum of the unit tangent vectors to the edges meeting at p. Note that T(p) = ∗Y(N)(p).
Lemma 4.1. [HMRR] (First variation of length for a planar cluster) For a cluster Bwith smooth
interfaces Eij and a variation with associated vector field Xand scalar normal component uij , the
initial first derivative of length is
−X
i<j ZEij
kij uij −X
vertex p
X(p)·T(p) = −ZB
ku −X
vertex p
X(p)·T(p).
6 WACHARIN WICHIRAMALA
For smooth scalar normal components uand v, we define a symmetric bilinear form
Q(u, v) = ZB
(u0v0−k2uv)−X
vertex p
Y(quv)
where u0is the derivative of uwith respect to arc length along edges and qij =kih +kjh
√3at a vertex
pwhere Ri, Rj, Rhmeet. We write Q(u) for Q(u, u).
Lemma 4.2. Given a C2variation on a bubble with initial velocity Xand normal component u,
the first variation of curvature is
dk
dt ¯
¯
¯
¯0
=u00 +k2u.
Proof. This is clear using elementary calculation. ¤
Lemma 4.3. [HMRR] For a variation on a bubble, at a vertex p, we have
dT (p)
dt ¯
¯
¯
¯0
=Y((qu +u0)N).
Proposition 4.4. [HMRR] (Second variation of length for planar bubble) For a bubble with inter-
faces Eij and a steady variation, the initial second derivative of length is Q(u).
Remark 4.5. Since each of Vand Xuniquely determine u, we will also write Q(V) or Q(X) to
refer to Q(u).
Let u={uij :Eij →R}be such that uij =−uji (that is, the uij are skew-symmetric in their
indices). We say uis admissible if uij is in the Sobolev space W1,2(the set of all functions in L2
with derivative also in L2) on Eij and uij agree at every vertex (uij +ujh +uhi = 0 where Ri, Rj, Rh
meet). Hence every scalar normal component is an admissible function. Recall that the character
of uis (x0, . . . , xn) where xi=PjREji uji. Furthermore, if xiis zero, positive, or negative, we
say upreserves, increases, or decreases the area of Rirespectively. Indeed, we can see that the
character is a linear map on admissible functions. Let Fbe the set of admissible functions with
zero characters. Hence Fis clearly a vector space over R. We extend the definition of Qfor all
admissible functions. For smooth uand v, by integration by parts and the fundamental theorem of
calculus, we have Q(u, v) = −R(u00 +k2u)v−PY((qu +u0)v). For piecewise C2functions uand
v, we then have Q(u, v) = −R(u00 +k2u)v−PY((qu +u0)v)−P[u0v] where [u0v] is the jump of
u0vand indeed we sum over discontinuities of u0vother than vertices.
Next we will show that Q(u+v) = Q(u)+ Q(v) when the supports of uand vare almost disjoint.
Lemma 4.6. Let uand vbe admissible functions. If at every point u= 0 or v= 0, then Q(u, v) = 0
and Q(u+v) = Q(u) + Q(v).
Proof. By definition Q(u+v) = Q(u) + Q(v) + 2Q(u, v). We need to show only that Q(u, v) = 0.
By definition and the assumption, Q(u, v) = R(u0v0−k2uv)−PY(quv) = Ru0v0. Since u0v0has
support in the intersection of the boundaries of the supports of uand vwhich has zero length, we
have Q(u, v) = 0. ¤
We can easily show that a smooth admissible function is the initial velocity of some steady
smooth variation, and that if there is an admissible function u∈ F with Q(u)<0, then the bubble
is unstable.
We will prove the main lemma of this subsection in Lemma 4.11. First we need some lemmas as
follows.
PROOF OF THE PLANAR TRIPLE BUBBLE CONJECTURE 7
Lemma 4.7. Let Kbe a Killing vector field on a bubble and set u=K·N. Then u00 +k2u= 0
and at any vertex Y((uq +u0)v) = 0 for any admissible function v.
Proof. This follows from Lemmas 4.2 and 4.3. ¤
We denote the interior of a set xby x◦.
Lemma 4.8. [HMRR] In a bubble B, let S⊂Bbe closed with ∂S disjoint from the vertices, and
let K6= 0 be a Killing vector field such that K·Nvanishes on ∂S . If u=K·Non Sand u= 0
elsewhere, then uis admissible, u00 +k2u= 0 on B\∂S,Y((qu +u0)v) = 0 for any vertex and any
admissible function v, and Q(u) = 0.
From the previous lemma, if K6= 0 is a rotation vector field around p, we say urotates S
around p(see examples in Figure 6.1). If pis at infinity, then Kis constant and we additionally
say utranslates S. We call Sthe support of u. Note that Sis closed and uvanishes on ∂S
which is disjoint from the vertices. An edge of Sthat meets ∂S is called a tentacle of S.
Lemma 4.9. If a function u: (0,1) →Ris zero on a nondegenerate interval and u00 +k2u= 0 for
a real number k, then u= 0.
Proof. This is clear using the elementary theory of differential equations. ¤
Lemma 4.10. [HMRR] In a stable bubble, let u∈ F be such that Q(u) = 0. Then uis smooth and
u00 +k2uis constant on each edge.
The next lemma is based on the key proposition of [HMRR]. We say a point pis the center
of a circular arc eif eis on a circle centered at p. In case eis straight, we say ehas its center at
infinity.
Lemma 4.11. Let Bbe a stable bubble and uibe rotating functions of Siaround pi. Suppose that
the Siare disjoint and u=Pαiui∈ F for some αi6= 0. Then each piis the center of every
tentacle of Si.
Proof. First we will prove that Q(u) = 0. Let vi=αiui. By Lemma 4.8, Q(vi) = 0 and v00
i+k2vi= 0
on B\∂Si. By Lemma 4.6, Q(u) = Q(Pivi) = PiQ(vi) = 0. We have u00 +k2u= (Pivi)00 +
k2Pivi=Pi(v00
i+k2vi) = 0 on B\ ∪∂Si. By Lemma 4.10, u00 +k2u= 0 on B.
Fix iand let tbe a tentacle of Sicontained in an edge e. Hence e\ ∪jSj6=∅and e∩Si6=∅.
Since u= 0 on e◦\ ∪jSj6=∅, by Lemma 4.9, u= 0 on e. Since vi=uon Si, we have vi= 0 on
e◦∩Si6=∅. Let Ki6= 0 be the rotational vector field such that Ki·N=uion Si. By Lemma 4.7,
(Ki·N)00 +k2(Ki·N) = 0 on B. Since Ki·N=ui= 0 on e◦∩Si6=∅, by Lemma 4.9, Ki·N= 0 on
e. Hence Kiis tangent to eand thus the center of eis pi. In case piis infinity, eis a line segment
in the direction of constant Ki. Therefore piis the center of t.¤
Remark 4.12. The previous proposition is also true under weaker conditions. To have Q(u) = 0,
we have to show that Q(vi, vj) = 0. Hence we just want ∂Sjto be a subset of Sior to be disjoint
from S◦
i, so u0
iuj= 0 on ∂Sjwhich is the set of discontinuities of u0
iuj. Now, instead of having
e∩Si6=∅, we need e∩Si\ ∪j6=iSj6=∅to make sure that vi= 0 on some nondegenerate interval.
4.2. Variations of convex components. Since a convex six-sided component must have straight
sides, by Lemma 2.4, a nonhexagonal convex component has at most five sides.
For a bubble Bwith a component C, let Vbe a steady variation on Bthat has inward normal
component identically 1 on the edges of Cand vanishes elsewhere. We say this variation uniformly
8 WACHARIN WICHIRAMALA
Figure 4.1. Curvatures of edges around the vertex between edges Eiand Ei+1.
shrinks C. Hence the second variation of length of Bis
Q(V) = −ZC0s edges
k2−X
C0s vertices
Y(q).
Let Eibe the edges of Cand ki, libe their curvatures and lengths. Since, at the vertex between Ei
and Ei+1,ki+1 −kiis the curvature of the incident edge, Y(q) is ki+1 +(ki+1 −ki)
√3+ki+(ki−ki+1)
√3=ki+1+ki
√3
and thus
Q(V) = −Xk2
ili−2
√3Xki.
Hence Q(V) is nonpositive if Cis convex and, moreover, is negative if Cis not six-sided. We say
this variation uniformly shrinks Cat speed 1. The character of this variation is (x0, . . . , xm) where
xi=−lwhere lis (the length of) the perimeter of Cand, for j6=i,xjis the total length of the
edges where Cmeets Rj. We will call (x0, . . . , xm) the character of C. We call the scalar normal
component of a uniform variation a uniform function.
4.3. Variations on long edges. An edge is long if it has absolute turning angle greater than π.
The long-edge variation on a long edge Eis the variation Vthat moves only Eusing circular
edges and fixing both end points. In addition, if ais the area of the domain bounded by Eand the
line segment joining the end points of E, then Vmust have da/dt = 1. If land kare the length
and curvature of the varying edge, we have derivatives at t= 0,
l00 =l00a0−l0a00
a03=(l0
a0)0
a0=(dl
da )0
a0=d2l
da2=dk
da
since dl
da =k. This formula shows l00 <0 since kis strictly decreasing in a. This is because Eis
long. Hence a steady variation Vhas second variation Q(V)<0. If Ehas absolute turning angle
exactly π, the same variation has zero second variation since ksmoothly decreases as achanges.
We say this variation changes the areas around Eat speed 1. If Eis between Riand Rj
with pi> pj, the character of this variation is (x0, . . . , xm) where xi= 1, xj=−1, and xk= 0 for
k6=i, j. We will call this character the character of E. We call the scalar normal component of
a long-edge variation a long-edge function.
PROOF OF THE PLANAR TRIPLE BUBBLE CONJECTURE 9
4.4. Bounds on the number of convex components. In this subsection, we discuss bounds on
the number of nonhexagonal convex components. This subsection ends with a bound for minimizing
triple bubbles.
We will prove a bound on the number of convex components. First we need the following lemma,
proposition and corollary.
Lemma 4.13. For a bubble B, let χ1, . . . , χn+kbe the characters of nonhexagonal convex com-
ponents C1, . . . , Cn. Suppose the components are disjoint and there are nonzero αisuch that
Pαiχi= 0. Then Bis unstable and thus not minimizing.
Proof. From such αi, we will make a variation that uniformly shrinks each convex component Ci
at speed αi. For a component Ci, let Vibe the steady variation that uniformly shrinks Ciat speed
1. Hence Vihas character χi. Let V=PαiViwhich is steady. Since Pαiχi= 0, the character of
Vis zero. Since Q(Vi)<0, we have Q(V) = Pα2
iQ(Vi)<0. Therefore Bis unstable and thus not
minimizing. ¤
We say two components of a region are identical if there is an isometry from one to the other,
preserving labels of surrounding components.
Proposition 4.14. A stable bubble may not have two identical nonhexagonal convex components.
Proof. Suppose to get a contradiction that a minimizing bubble Bhas two identical components.
Let χ1and χ2be characters of the two components. Since χ1−χ2=χ1−χ1= 0, Bis not
minimizing by Lemma 4.13, a contradiction. ¤
Corollary 4.15. A region of a stable bubble may not have two convex three-sided components with
the same surrounding regions.
Proof. Since the shape of a three-sided component is uniquely determined by its three curvatures
by Lemma 5.8, any two three-sided components with same surrounders must be isometric and then
can be ruled out by the previous proposition. ¤
The regions involved in a component Care the regions surrounding Cand the region that C
belongs to.
Theorem 4.16. Let Bbe a stable bubble. Suppose some nnonhexagonal convex components are
disjoint. Then there are more than nregions involved in those components.
Proof. Let C1, . . . , Cnbe the components and V1, . . . , Vnthe variations uniformly shrinking the
Ciof speeds 1. Let each Viinitially change areas with rate Xi. Let Nbe number of nonzero
coordinates of all Xi. Suppose to get a contradiction that there are at most nregions involve in Ci.
Hence N≤n. Thus there are real numbers α1, . . . , αnsuch that PαiXihas at most one nonzero
coordinate. Hence PαiViis a variation that preserves all Nbut one area. Since Vchanges only
areas of the nregions while initially fixing n−1 of them, it follows that Vpreserves all areas.
Hence PαiXi= 0 and thus Bis not stable by Lemma 4.13. ¤
Corollary 4.17. A stable m-bubble has at most mdisjoint nonhexagonal convex components.
Proof. This follows from Theorem 4.16. ¤
Remark 4.18. When using the weak approach, a variation is allowed to initially increase areas of
bounded regions with zero pressures since length does not change in first order.
Remark 4.19. If a region has a hexagonal component, this implies that it has the same pressure
as at least two surrounding regions. Hence, conversely, if a convex component of a region meets at
most one region of equal pressure, then it is not hexagonal.
10 WACHARIN WICHIRAMALA
Figure 4.2. We have that Rimeets U=∪j<iRjalong Diedges of Ri. As Fi, Ei
and Ei−1are the numbers of the outside edges of Ri, Ri∪Uand U, we can see that
Ei=Ei−1+Fi−2Di.
Corollary 4.20. For a stable triple bubble with no empty chamber, the highest pressure region has
at most three components. Of these, at most two are internal and at most one is three-sided.
Proof. Note that the highest pressure is positive. If three bounded regions have the same pressure,
then the cluster is standard [Vau] and hence the highest pressure region, say R, has only one com-
ponent. If not, Rmeets at most one region of equal pressure. By Remark 4.19, Rhas no hexagonal
convex component. Therefore Theorem 4.16 implies that Rhas at most three components. Of
these, at most two are internal. By Corollary 4.15, there is at most one three-sided bubble among
the components of R.¤
4.5. Bounds for the total number of components. In this section, we establish bounds on
the total number of components of a minimizing planar m-bubble.
Theorem 4.21. An m-bubble with exactly Cdisjoint convex components may have at most
C
2(3m+1 −2m−1) components.
Proof. Let p1≥p2≥. . . ≥pm+1 be the pressures of the m+ 1 regions including the exterior
region. Let A1be the number of components (of any region) that do not meet any component of
a lower-numbered region. Hence these A1components are convex and include all components of
R1. Hence A1≤C. Next we reassign all these components to R1. Note that the new cluster is
no longer a bubble. For i= 2, . . . , m + 1, let Aibe the number of (remaining) components of Ri.
Hence the number of all components, including those of the exterior region, is Pm+1
1Ai. For each i,
let Ri1, . . . , RiAibe the components of Ri. These components are not adjacent to each other since
the original bubble has no redundant edges. Let Nij be the number of the edges between Riand
Rj. Note that Nii = 0. Let Fibe the number of edges of Ri, and Eibe the number of outside edges
of ∪j≤iRj. Hence E1=F1,Fi=PjNij and Ei=Pj≤iPk>i Nj k . Let Dibe the number of those
edges of Rithat meet ∪j<i Rj. Hence Di=Pj <i Nji. Each component of Ri,i > 1, meets ∪j<iRj
because otherwise it would have been included in R1. Hence Di≥Ai. Since R1jare convex, we
have E1≤6A1. Because these E1edges form loops and each loop may meet R2along at most half
of its edges, we have A2≤E1
2. Similarly, for i > 1, we have Ai+1 ≤Ei
2.
We will prove in general that Ei=Ei−1+Fi−2Diand Pj<i Fj=Ei−1+2 Pj<i Dj. For the first
equation (see Figure 4.2), we have Ei=Pk>i Nik +Pj<i Pk>i Njk and Ei−1=Pj <i(Pk >i Njk +
Nji ) = Pj<i Pk>i Njk +Pk<i Nik . We also have Fi=Pk>i Nik +Pk<i Nik and Di=Pk<i Nik .
PROOF OF THE PLANAR TRIPLE BUBBLE CONJECTURE 11
Hence Ei−1−Fi−2Di=Pj<i Pk>i Nik +Pk>i Nik =Ei. For the second equation, we have
Pj<i Fj=Pj<i(Pk>i−1Nkj +Pk≤i−1Nkj ) = Ei−1+Pj<i Pk<i Nkj =Ei−1+ 2 Pj<i Pk<i Nkj =
Ei−1+ 2Di.
To find a bound for E2, we need the following estimate. For each j, let t2jbe the sum of
the absolute turning angles of the edges between R2jand R1. By Lemma 2.4, the sum of the
turning angles of the edges of R1equal to 6A1−F1
3π. Hence Pt2j≤6A1−E1
3π. For a component
R2j, let n−be the number of its edges facing R1and n+be the number of its other edges. Also
let t−and t+be the sums of the absolute turning angles of the n−and n+edges respectively.
By Lemma 2.4, t+−t−=6−n+−n−
3πif R2jis bounded, or −6g−n+−n−
3πif R2jis the unbounded
component and gis the number of connected subclusters. Thus, the number of the edges of R2j
is n++n−≤6+3t−−t+
π≤6 + 3
πt2j. Hence F2≤6A2+3
πPt2j≤6A2+ 6A1−E1. Therefore
E2=E1+F2−2D2≤6A2+ 6A1−2D2≤4A2+ 6A1.
In the general case for i > 1, we can bound Eias follows. First we have to bound Fi. For each j,
let tij be the sum of the absolute turning angles of the edges between Rij and U=R1∪. . . ∪Ri−1.
Note that the outside edges of Uare convex. The sum sof the turning angles of the outside edges of
Uis equal to the sum of the signed turning angles of the edges of R1, . . . , Rn−1. By Lemma 2.4, we
have s=Pi−16Aj−Fj
3π. As Ptij is a partial sum of s, we have Ptij ≤6
P
i−1Aj−
P
i−1Fj
3π. Hence
Fi≤6Ai+ 6 Pi−1Aj−Pi−1Fj. As D1= 0, we then have Ei=Ei−1+Fi−2Di≤Ei−1+6 PiAj−
Pi−1Fj−2Di=Ei−1+ 6 PiAj−Ei−1−2Pi−1Dj−2Di= 6 PiAj−2PiDj≤4Pi
2Aj+ 6A1.
Thus PnEi≤4PnPi
2Aj+ 6nA1≤2PnPi−1Ej+ 6nA1= 2 Pn−1PiEj+ 6nA1. Let
xn=PnEi. Therefore xn≤2Pn−1xi+ 6nA1. Next recursively define Xn= 2 Pn−1
i=1 Xi+ 6nA1
where X1= 6nA1. We can see that xi≤Xiby induction. We have Pm+1 Ai≤1
2Pm+1
2Ei−1+A1=
1
2PmEi+A1≤Xm
2+A1. Now we solve the recurrence relation for Xn. Let Yn=Pn−1Xi. Hence
Yn=Xn−1+Yn−1and Xn= 2Yn+ 6nA1= 2Xn−1+ 2Yn−1+ 6nA1. Thus
µXn
Yn¶=µ2 2
1 1¶µXn−1
Yn−1¶+µ6nA1
0¶.
Let Abe the matrix µ2 2
1 1¶,Wn=µXn
Yn¶, Un=µ6nA1
0¶. Hence Wn=AWn−1+Un=
A2Wn−2+AUn−1+Un=An−1W1+An−2U2+. . . +Un. Since W1=µX1
Y1¶=µ6A1
0¶=U1, we
have Wn=Pn
1An−iUi. Since the characteristic polynomial of Ais λ2−3λ, the eigenvalues of Aare
λ1= 3 and λ2= 0 with eigenvectors V1=µλ1−1
1¶=µ2
1¶and V2=µλ2−1
1¶=µ−1
1¶. Since
µ1
0¶=1
3(V1−V2), we have Akµ1
0¶=1
3(λk
1V1−λk
2V2) = 3k−1V1. Hence Wn=Pn
16iA13n−i−1V1
and thus
Xn= 4A1
n
X
1
i3n−i= 4A1(3n−1+ 2 ·3n−2+. . . +n) = 4A1
n
X
1
(3i−1+ 3i−2+. . . + 1)
= 4A1
n
X
1
3i−1
2= 2A1(3n+1 −1
2−1−n) = A1(3n+1 −2n−3).
Hence the number of all components is Xm
2+A1≤A1
2(3m+1 −2m−3)+ A1≤A1
2(3m+1 −2m−1) ≤
C
2(3m+1 −2m−1). ¤
12 WACHARIN WICHIRAMALA
Figure 5.1. The line lis in the middle of eand gand perpendicularly bisects f.
Remark 4.22. If all pressures are nonnegative, by setting the exterior region to play the role of
Rm+1, we find that the number of bounded convex components excluding empty chambers is at
most Xm−1
2+A1.
Theorem 4.23. A stable m-bubble without hexagonal convex components has less than m
23m+1
components.
Proof. By Corollary 4.17, without hexagonal components, there are at most mdisjoint convex
components. From the previous theorem, we have A1≤mand the number of components is at
most m
2(3m+1 −2m−1) <m
23m+1.¤
Remark 4.24. We found that the variational argument fails to bound the number of hexagonal
convex components. The author expects that local renovation will give us such a bound.
5. Geometry of planar bubbles
In this chapter, we discuss geometric properties of planar soap bubbles. In the last section, we
focus on triple bubbles.
The main goal is to understand the geometry of weakly minimizing triple bubbles with unequal
pressures. Hence we will focus on a minimizing bubble without empty chambers and with unequal
nonnegative pressures.
5.1. Elementary geometry. The next lemma will discuss the geometry of consecutive edges of
a component where it has internal angles 120◦.
Lemma 5.1. Let e, f, g be consecutive edges of a component. Suppose eand ghave the same
(signed) curvature and they have different centers pand q. Let lbe the line that perpendicularly
bisects the segment pq. Then lperpendicularly bisects f(see Figure 5.1).
Proof. Using Figure 5.1, this is obvious. ¤
Remark 5.2. For the previous lemma, if eand gare cocircular, then there are infinitely many
choices of l. Hence lis not necessarily perpendicular to f.
We say two circular arcs are parallel if they have the same center. Hence two cocircular arcs
are parallel.
PROOF OF THE PLANAR TRIPLE BUBBLE CONJECTURE 13
Figure 5.2. An arc meets eand fat angles 120◦.
Figure 5.3. We reflect hacross land get h0. If gand h0are not cocircular, then l06=l.
Lemma 5.3. For a given pair of nonparallel arcs eand f, and a given curvature k, there are at
most two possible arcs with curvature kthat cut eand fwith angles 120◦(see Figure 5.2). In
particular, if there are two possibilities, one can be reflected through the line lperpendicular to e
and fto be cocircular with the other.
Proof. Note that eand fare not cocircular. Hence there is a unique line lperpendicular to eand
f. Indeed lis the line through the centers of eand f. Suppose there are two possible arcs gand
hwith curvature kand meeting angles 120◦, Let h0be the mirror image of hacross l(see Figure
5.3).
Suppose to get a contradiction that gand h0are not cocircular. Let l0be the line that perpendic-
ularly bisects the segment between the centers of gand h0. Since g6=h, we have l06=l. Thus l0is
not perpendicular to either eor f. By the previous lemma, gand h0are cocircular, a contradiction.
Hence any two possible arcs are cocircular. Therefore, there are at most two possibilities. ¤
Remark 5.4. For the previous lemma, if the pair of arcs eand fare parallel, any solution can be
rotated around their common center to produce uncountably many solutions.
Remark 5.5. It is known that two circles (constant curvature lines) cut at most twice in the
extended plane. Note that three points uniquely determine a circle.
Lemma 5.6. (Trichotomy law of cutting) Given points pand qand two arcs from pto qmeeting
at angle θ < 180◦, suppose a circle Ccuts both arcs at angles 120◦. Then Cpasses through por q
if θ= 60◦,Cseparates pfrom qif θ > 60◦, and Cleaves both pand qoutside or inside if θ < 60◦
(see Figure 5.4).
Proof. We first map qto infinity by a M¨obius transformation and thus the two arcs become straight
(see Figure 5.5). If θ= 60◦, then Cbecomes straight and thus Cmeets q. If θ < 60◦, then Ccuts
14 WACHARIN WICHIRAMALA
Figure 5.4. Three ways for Cto meet eand f.
Figure 5.5. M¨obius images of two arcs are straight. The image of Chas three
possibilities depending on θ.
each arcs twice and thus Cleaves both pand qoutside or inside its interior. If θ > 60◦, then C
cuts each arc once and thus Cseparates pfrom q.¤
Lemma 5.7. Given two arcs eand fmeeting at internal angles 120◦, there is at most one arc of
prescribed curvature kmeeting eand fat 120◦(see Figure 5.6).
Proof. Suppose there are two such arcs. By Lemma 5.3, one of them can be reflected to be cocircular
with the other (see Figure 5.6). This contradicts Lemma 5.6. ¤
Lemma 5.8. [Vau] A three-sided component is uniquely determined by its three vertex positions
and orientation. It is also uniquely determined, up to rigid motion, by its three curvatures.
5.2. Reduction and decoration on bubbles. Here we show how to simplify bubbles by reducing
one component at a time. This method helps us to understand the hidden geometry of bubbles.
[CHHKLMT], Lemma 8.1, states that in any two-sided component, its two incident edges are
cocircular. Thus we can remove any two-sided component of a bubble and then connect its two
incident edges to get another bubble. We call this method reduction of a two-sided component.
Similarly, the next lemma will tell us that we can reduce a three-sided component and get another
bubble.
Lemma 5.9. For a three-sided component, if its three incident edges are prolonged into the com-
ponent, they will meet at a point satisfying the cocycle condition.
Proof. Given a three-sided component C, we can prolong the incident edges at its three vertices as
in the left of Figure 5.7. Next we map all arcs with the M¨obius map that sends the three vertices
of Cto be the three vertices of a equilateral triangle with the same orientation (see Figure 5.7).
Hence the interior of the image of Cis the image of the interior of C. By Lemma 5.8, the image
of Cis a three-sided component with equal edges. Hence the images of all arcs are composed of
three overlapping symmetric two-sided components and three lines (see the right of Figure 5.7).
PROOF OF THE PLANAR TRIPLE BUBBLE CONJECTURE 15
Figure 5.6. The arcs eand fare cut by an arc at 120◦. If there are two such
cutting arcs, one can be reflected to be cocircular with the other, contradicting
Lemma 5.6.
Figure 5.7. A three-sided component with prolonged edges is mapped by a M¨obius
map to an equal-sided three-sided component.
As M¨obius map exchanges circular arcs, the three incident edges of Cmeet inside Csatisfying the
cocycle condition. ¤
In the opposite way, given any bubble, we can add a three-sided component at any vertex or add
a two-sided component on any edge to get another bubble. We call this method decoration.
In any bubble except a single round circle, each edge is a circular arc with two distinct end points.
Decorating such an edge by a two-sided component would create two components with more than
one meeting, hence produce a nonminimizing bubble by Proposition 2.5. Recall that a bubble B
may be considered as a connected, trivalent embedded graph with numbered faces. We will define
the dual graph Gof Bto be the graph with vertices corresponding to faces of B, and if components
Cand Dmeet in B, the corresponding vertices of Cand Dhave an edge between them. Note
that the corresponding vertex of R0has a link to the corresponding vertex of any component that
meets R0. Hence decorating a two-sided component or a three-sided component corresponds to
adding a vertex and edges to the dual graph (see Figure 5.8). As we saw, a decorating process that
introduces some two-sided component would give a nonminimizing bubble. Since we are studying
minimizing bubbles, we may assume that every step of decoration or reduction uses a three-sided
component. We can also assume that we are dealing with a bubble without two components that
meet twice, and with a bubble other than two trivial bubbles: a circle and a standard double bubble.
Hence a standard triple bubble is the smallest bubble. It is clear that a bubble with a three-sided
component can be reduced. Hence we can say a bubble is irreducible if it has no three-sided
16 WACHARIN WICHIRAMALA
Figure 5.8. Decorating a two-sided component on an edge and decoration a three-
sided component at a vertex are equivalent to adding new vertices and edges in the
dual graphs.
Figure 5.9. (left) Cis barbequed between Dand E. (right) Cis also fat.
component. Since decorating a bubble Bdoes not remove any component, every component in a
bubble corresponds to a well-defined component in any decoration.
A component is external if it meets R0and is internal if not. An edge is external if it is an
edge of R0and is internal if not. A vertex is external if it is a vertex of R0and is internal if
not.
5.3. Geometry of triple bubbles. In this section, we make necessary geometric conditions on
weakly minimizing triple bubbles with unequal nonnegative pressures. Hence we assume each
bubble Bhas no empty chamber, i.e., R0is connected. For pressures p1, p2, p3of B, we also assume
p1≥p2≥p3≥0 and p1> p3. By Proposition 2.5, we may also assume that any two components
meet at most once, meaning the edge between any two components is unique.
Definition 5.10. A component Cin a bubble Bis barbequed between two components Dand
Eif Bis a decoration of some bubble B0where components C0, D0, E0(corresponding to C, D, E)
have a common vertex (see the left of Figure 5.9). If Dand Ebelong to regions Rand S, we also
say Cis barbequed between Rand S. Furthermore, Cis fat if there is a line perpendicular to
the two edges of Cthat meet Dand E(see the right of Figure 5.9). Of course, Cis barbequed if
C, D, E meet at a vertex. In this case, we say Cis trivial. Any two barbequed components of a
region are compatible if they are between the same pair of regions.
PROOF OF THE PLANAR TRIPLE BUBBLE CONJECTURE 17
Figure 5.10. The line lis perpendicular to both edges of Rand passes through
the center cof the external edge of S.
Figure 5.11. The center cof the external edge of Sis not in R. (Left) When Sis
as big as R. (Middle) When Sis smaller than R. (Right) When Ris much bigger,
its external edge approaches to a straight line and its internal edge approaches c.
Note that the lines through the center of a circle care exactly the lines perpendicular to c. We
adopt the convention that the center of a straight line lis at infinity and lines through its center
mean those perpendicular to l.
Lemma 5.11. Given a standard double bubble of regions Rand S, let cbe the center of the external
edge of S. Then cis not in Rand there is a unique line lperpendicular to both edges of R(see
Figure 5.10). Moreover, lpasses through c.
Proof. Suppose Ris not bigger that S. Then the internal edge eof Ris convex and cuts out less
of the circle of Sthan in the equal-pressure case, thus eis on the right of c. Hence cis not in R.
Suppose Ris bigger than S. Note that we can scale the bubble so that the curvature of the external
edge of Sis 1. As Rgets bigger and bigger, the internal edge is moving to the right approaching
cand the external edge is approaching a vertical line. Hence Rnever contains c(see the right of
Figure 5.11). Note that a line perpendicular to an edge must pass through the center of the edge.
A standard double bubble has a line lof symmetry perpendicular to all three of its edges, and thus
passing through their colinear centers. Hence a line perpendicular to both edges of Rmust be l.
¤
By a cut and paste technique, we can show that a minimizing bubble may not have two compat-
ible, fat barbequed components.
Lemma 5.12. Let Cbe a three-sided component of a bubble, with vertices v, s, q. Let ebe the
incident edge to Cat vas in Figure 5.12. The center oof elies outside of Cand is colinear with
sand q.
Proof. First prolong the three edges incident to vto form a standard double bubble (see the left
of Figure 5.13 where Cis a subset of the left region). By Lemma 5.11, ois not in C. Let lbe
18 WACHARIN WICHIRAMALA
Figure 5.12. A three-sided component has o,s,qcolinear.
the line through vertex qand perpendicular to e. Hence lcontains o. The M¨obius map fixing p, q
and sending rto infinity maps arcs as in Figure 5.13. Thus the image of line lcontains vertex s.
Therefore lpasses through qand sas desired. ¤
Lemma 5.13. For a triple bubble, any three-sided component of R3is fat barbequed between R0
and R1and between R0and R2. Also any three-sided component of R2is fat barbequed between R0
and R3.
Proof. Note that any three-sided component is trivially barbequed (in three ways). Let Cbe a
three-sided component of R3. Let vbe an external vertex of Cand ebe the incident edge on Cat v
as in Figure 5.12. Hence eis an external edge of R1or R2. Suppose ebelongs to R1. From Figure
5.12, the edge qs of Cis nonconvex since p3≤p2. Hence the angle between the edge vq and the
line segment qs, called θ, is at least 120◦. Since θ >90◦, there is a line lthrough operpendicular
to the edge sv. Since the centers of these three edges at vare colinear (see Lemma 5.11), then lis
also perpendicular to the edge qv. Therefore Cis fat between R0and R1. Similarly, a three-sided
component of R3is fat between R0and R2, and a three-sided component of R2is fat between R0
and R3.¤
Lemma 5.14. Any external barbequed four-sided component of a triple bubble is fat.
Proof. Let Cbe a barbequed four-sided component of a triple bubble. If we rotate the bubble so
that Chas its external edge on the top, the left and right edges of Cmeet the same region. We
will show that Cis symmetric vertically and hence is fat. First prolong the top and bottom edges
to form a two-sided component (see Figure 5.14). The left and right edges are not cocircular by
Lemma 5.6. Hence, by Lemma 5.1, Cis symmetric as wanted. ¤
Lemma 5.15. All vertices of a four-sided component of a bubble are colinear or cocircular with
the same order they appear on the boundary of the component.
Proof. Let Fbe a four-sided component with vertices v1, v2, v3, v4in counterclockwise order. We
may assume that the viare not colinear. Let abe half of the signed turning angle of the edge of F
from v4to v1, let bbe half of the angle of the edge v1v2, let cbe half of the angle of the edge v2v3, let d
be half of the angle of the edge v3v4. Let Gbe the 4-gon with directed edges connecting v1, v2, v3, v4
in that order. Let A, B, C, D be the internal angles of Gat vertices v1, v2, v3, v4respectively. Hence
a, b, c, d are the angles between Fand G.
We will divide into cases according to the way the viform G. Note that some three vertices may
be colinear.
CASE Ghas counterclockwise edges.
Hence Fand Gare illustrated by Figure 5.15 where a, b, c, d > 0. We have 120◦=a+A+b=
PROOF OF THE PLANAR TRIPLE BUBBLE CONJECTURE 19
Figure 5.13. A three-sided component is mapped by a M¨obius transformation to
be vertically symmetric.
b+B+c=c+C+d=d+D+a. Hence a+A+b+c+C+d= 240◦=b+B+c+d+D+a.
Thus A+C=B+Dand therefore the four vertices are on a circle with counterclockwise order.
CASE Ghas clockwise edges.
Without loss of generality, we may assume Fand Gare illustrated by Figure 5.16 where a, b, c < 0
and d > 0. We have 120◦= 360◦−A+a+b= 360◦−B+b+c=d−C+c=d−D+a. Hence
−A−B=−C−Dand therefore the four vertices are on a circle with clockwise order.
CASE Gis twisted (or is bowtie-shaped).
Without loss of generality, we may assume A < 180◦and C > 0◦as Fand Gare illustrated
by Figure 5.17 where a, b, d < 0 and c > 0. We have 120◦= 360◦−A+a+b=c−B+b=
c+C+d=d+D+aand A+B=C+D. Hence 360◦−A+C=−B+D=A−Cand thus
A= 180◦+C > 180◦, a contradiction. Therefore this case is impossible. ¤
Remark 5.16. A four-sided component is called circular if two of its edges are cocircular. By
Lemma 5.15, only one pair of opposite edges can be cocircular. Figure 5.18 shows all possible types
of circular four-sided components. We may have the component outside of the circle. We may have
two edges colinear. Or we may have the component inside the circle.
Remark 5.17. We will now discuss the case of a four-sided component Cwith two parallel (but
not cocircular) edges. Since the two parallel edge share the same center, they do not meet at
120◦. Hence the two edges are opposite. Suppose Clooks like the left one in Figure 5.19 where
the two edges are on different circles. If we draw two lines land l0from the common center to
both ends of the right edge e, we would get that land l0meet eat angles 30◦. This contradicts
that the two radii are different. So Cmust look like the component in Figure 5.20 where it is
composed of two opposite, parallel edges with opposite signs of curvatures (or both straight), and
two opposite, equal curvature, convex edges. We say this kind of four-sided component, with two
parallel, noncocircular edges, is annular , instead of parallel.
5.4. Symmetry of weakly minimizing bubbles. A four-sided component is bisymmetric if it
has two line symmetries. Since it has four vertices, the two symmetry lines must be perpendicular.
20 WACHARIN WICHIRAMALA
Figure 5.14. An external barbequed four-sided component is fat since it is symmetric.
Figure 5.15. A four-sided component with counterclockwise directed 4-gon and
signed angles a, b, c, d > 0.
As we study weakly minimizing triple bubbles, we are dealing with only triple bubbles without
empty chambers. The next lemma will discuss the symmetry and uniqueness of some types of
four-sided components.
Lemma 5.18. Consider a weakly minimizing triple bubble. Let Cbe a four-sided component of
a type in Figure 5.21. Then Cis symmetric as shown and the side edges of Care not cocircu-
lar. Moreover, if Cis external (has one of the first three types), then the shape of Cis uniquely
determined by its curvatures and the length of either side edge.
Proof. Suppose Cis of the first type. Suppose the side edges of Care cocircular, say on a circle O.
Consider the construction of Cby clipping O. Since p1≥p2, to make the bottom edge, we have to
clip out the perimeter of Ofor angle at least 2
3π(see Figure 5.22). Since p3≥0, to make the top
edge, we have to clip out the perimeter of Ofor angle at least 4
3π(see Figure 5.22). Hence we have
no perimeter of Oleft to make C. So the side edges of Care not cocircular. Therefore, by Lemma
5.1, Cis symmetric vertically.
Because of the curvatures of their edges, a circular four-sided component of the second and third
types may be cocircular but lie outside the circle as in Remark 5.16.
Suppose Cis of the second type. Suppose the side edges of Care on a circle O(which is not
straight since p1> p3). Consider its construction by clipping O. Since p2≤p1, to make the bottom
edge, we have to clip out the perimeter of Ofor angle at least 2
3π(see Figure 5.23). To make the
top edge (over O), we have to take the perimeter of Ofor angle less than 2
3π(see Figure 5.23).
Hence it is impossible to have Ccocircular (see Figure 5.23). Therefore Cis symmetric vertically.
Suppose Cis of the third type. Suppose the side edges of Care cocircular. Since p2≥p3, the
bottom edge would have turning angle at least 2
3π(see Figure 5.24). Hence the convex component
of R1under Cis a three-sided component by Lemma 2.4. So the three-sided component has a
PROOF OF THE PLANAR TRIPLE BUBBLE CONJECTURE 21
Figure 5.16. A four-sided component with clockwise directed 4-gon and signed
angles a, b, c > 0 and d > 0.
Figure 5.17. A four-sided component with twisted directed 4-gon and signed angles
a, b, d < 0 and c > 0.
redundant incident edge, a contradiction. Hence Cis not cocircular. Therefore Cis also symmetric
vertically.
Suppose Cis of the fourth type. By similar argument of the first type using the fact that p2≥p3,
we also have that Cis symmetric vertically.
Suppose Cis of the fifth (last) type. Since the side edges are concave and the top and bottom
are convex, the side edges are not cocircular, thus it is also symmetric vertically.
Next we will show that the shape of each of the first three types is uniquely determined by
curvatures and a side length. Note that the two side edges have the same length by symmetry.
If Cis of the first type, its top and bottom edges are not cocircular by the same reason why its
left and right edges are not cocircular, and they are not parallel by Remark 5.17. If Cis of the
second or third type, by Remark 5.17 and the signs of the curvatures of its edges, the top and the
bottom edges can not be parallel (otherwise the side edges must be strictly convex). Hence, given
four curvatures of edges and a side length, Lemma 5.3 forces the shape of Cto be unique. ¤
Proposition 5.24 will describe the structure of bubbles that include a chain of four-sided compo-
nents as in Figure 5.25. To prove it, we will need the following lemmas.
Lemma 5.19. In a sequence of four-sided components (see Figure 5.25), if uiand liare cocircular
for some i, then uiand liare cocircular for every i.
Proof. Suppose, according to Figure 5.25, that u1and l1are cocircular. Hence we also have u2and
l2cocircular. By induction, the lemma statement is clear. ¤
22 WACHARIN WICHIRAMALA
Figure 5.18. Three types of circular four-sided components; when the component
is outside the circle, when the two edges are colinear, and when the component is
inside the circle.
Figure 5.19. (left) A four-sided component with parallel edges hanging between
two circles. (right) This four-sided component is not possible since the radii are
different.
Figure 5.20. An annular four-sided component with two parallel edges (having the
same center) and two edges as caps for both ends.
Figure 5.21. Five types of four-sided components that are symmetric.
PROOF OF THE PLANAR TRIPLE BUBBLE CONJECTURE 23
Figure 5.22. In order to make a circular four-sided component of the first type,
we have to clip out the circle to make the edge meeting R2and the edge meeting
R0.
Apolysegment is a piecewise-straight path with two ends.
Lemma 5.20. Given a polysegment Pof at least two segments, let d≥0be less than the distance
between the ends of P. Then there is a continuous movement that maintains the length of each
segment and decreases the distance between the ends such that finally the distance is d, or the angle
between some consecutive segments is zero.
Proof. Suppose the vertices of Pare p0, . . . , pn. Let sibe the segment pi−1piof length li. Let dij be
the distance between piand pj. We will show that there is a continuous movement that maintains
li, decreases d0n, and finally makes zero angle at some pi. If, during this move, d0nreaches dor
some angle reaches zero, we are done. Hence we may assume that d0nnever reaches d.
We will prove this by induction on n. For n= 2, we can rotate the shorter segment to lie on the
longer segment (see Figure 5.26). This movement decreases d02 and finally makes zero angle at p1.
For n > 2, we may assume that the piare distinct. Otherwise we can treat Plike it has less
than nsegments as we can maintain the overlapping of pialong any movement. We will make a
movement that fixes the relative position of p2, p3, . . . , pn. In particular, we fix d2n. We will divide
into cases depending on the comparison of l1, l2, d2n.
CASE l1+l2< d2n.
We can rotate s1and s2together around p2until p0is colinear with p2and pn(see Figure 5.27).
Then we can move p0in the direction to pnuntil p1is colinear with p2and pn. Hence the angle at
p1is 180◦and then we can treat Pas it has n−1 segments.
CASE l2+d2n< l1.
We can use a similar movement as in the previous case to put first pnand then also p2on s1
while fixing d2n(see Figure 5.28). Hence the angle at p1is zero.
CASE l1+d2n< l2.
Let Cbe the circle of radius l1centered at p1and Dbe the circle of radius d2ncentered at p2.
Hence we can move p0and pntowards each other along Cand Duntil they are on s2(see Figure
5.29). For this, we move p0along Cto be closest to pnand then move pnalong Dto be closest to
p0. As we repeat this process, we have p0and pnconverge to s2. Hence the angle at p1is zero.
CASE l1+l2≥d2nand l2+d2n≥l1and l1+d2n≥l2.
Let Cand Dbe defined as in the previous case. Hence Cmeets Das shown in Figure 5.30. Thus
we can move p0and pntowards each other along Cand Duntil they meet as follows. Since the pi
are distinct, then p0is not the center of D. Hence we can alternately move p0and pntowards each
other as in the previous case. Therefore at some time during the moving, the distance d0nreaches
d.
This finishes the induction and, hence, the proof. ¤
24 WACHARIN WICHIRAMALA
Figure 5.23. To make a circular four-sided component of the second type, we have
to clip out the circle for angle at least 2
3πto make the edge meeting R2. The external
edge (top) of the component covers the circle for angle less than 2
3π. Hence those
edges (top and bottom) overlap.
Lemma 5.21. Let two concentric circles and a polysegment of at least two segments with ends on
the two circles be given. Then there is a continuous movement that maintains the length of each
segment and keeps each end on its circle such that finally either the two ends are on the same ray
from the center or the angle between some consecutive segments is zero.
Proof. Let dbe the difference of the two radii. If the distance between the two ends is d, then the
two ends are on a ray from the center. Hence we may assume that the distance between the two
ends is greater that d. By the previous lemma, there is a continuous movement Mthat maintains
the length of each segment and decreases the distance between the ends such that finally either two
ends are on the same ray from the center or the angle between some consecutive segments is zero.
It is clear that Mcan keep each end on its circle since the distance between the ends is decreasing.
Therefore Mis the desired movement. ¤
Lemma 5.22. Let Bbe a bubble consisting of a sequence of at least three circular four-sided com-
ponents in between two components of some region R(see Figure 5.31). Then Bis not minimizing.
In particular there is a continuous variation fixing length and areas that creates a vertex of degree
4.
Proof. Let C1, . . . , Cnbe the circular four-sided components. For each i, let Dibe the circle
containing the two cocircular edges of Ci, and piand ribe the center and the radius of Di, and si
be the segments pipi+1 with length li. Hence siand piform a polysegment Pof n−1 segments.
By Lemma 5.19, the two external edges of Rare also cocircular. Let Dbe the circle that contains
these two edges with center p. Here we may think that the external component is circular and,
together with Ci, makes the sequence of four-sided components into a closed loop. Let Eand Fbe
the circles centered at pthrough p1and pn. We have that D1and Dnmeet Dwith external angles
60◦, and Diand Di+1 meet with external angles 120◦. Let e0be the edge between Rand C1and
enbe the edge between Rand Cn. For each i, let eibe the edge between Ciand Ci+1. We will
show that a movement of Pfixing liand keeping p1and pnon Eand Fis equivalent to a variation
of Bthat fixes the total length and the areas of B.
Suppose the pihave an infinitesimal change in their position on the plane, but still have the
original lengths liand p1and pnare on Eand F. It suffices to show that this change is equivalent
to a slight change of Ciand Rthat fixes the total length and the areas of Ciand R. We will make
PROOF OF THE PLANAR TRIPLE BUBBLE CONJECTURE 25
Figure 5.24. To make a circular four-sided component of the third type, the (bot-
tom) edge meeting R1has angle at least 2
3π. Hence the lower neighbor is a three-
sided component of R1.
Figure 5.25. A sequence of three four-sided components.
a new sequence of circular four-sided components from these new p0
i. First, for each p0
i, draw a
circle D0
iaround p0
iof radius ri. Hence D0
1and D0
nmeet Dwith external angles 60◦, and D0
iand
D0
i+1 meet with external angels 120◦. We then remove parts of D, D1, . . . , Dnas follows (see Figure
5.32). We remove the parts of D1and Dnoutside Dand the parts of Dinside D1and Dn. For
each i, we remove the overlapping parts of Diand Di+1 (the part of Diinside Di+1 and the part of
Di+1 inside Di). Next we will make new edges e0
0, . . . , e0
n. It is clear that there is a unique way to
put an edge e0
0between the two vertices where Dand D1meet. Similarly we can make e0
1, . . . , e0
n.
This completes a new bubble B0with a sequence of ncircular four-sided components C0
iin between
other two components R0
1and R0
2that have their external edge cocircular. It is obvious that Ci
and C0
ihave the same area. Hence R0
1and R0
2together have the same areas as R. The total length
of the external edge of Ris equal to the total length of the external edges of R0
1and R0
2. The total
length of the two edges of Cimeeting Ris equal to the total length of the two edges of C0
imeeting
R0
1and R0
2. The edge eiis isometric to e0
i. Hence Band B0have the same length.
By the previous lemma, there is a continuous movement Mof Pthat fixes liand keeps p1and
pnon Eand Fsuch that finally p1and pnare on the same ray from p, or the angle at some piis
zero. Hence we have an equivalent variation on Bthat fixes the total length and the areas of Ci
and R, and makes a vertex of degree 4. Therefore Bis not minimizing. ¤
Remark 5.23. By an argument similar to that in the previous lemma, we can make a standard
bubble that has a multi-parameter variation that fixes length and areas. This bubble is composed
of a closed sequence of circular four-sided components surrounding a central region. When the
sequence does not maximize the central area, there is a variation that fixes length and areas and
has many degrees of freedom. This bubble is an example to show that a given combinatorial
26 WACHARIN WICHIRAMALA
Figure 5.26. We rotate the segment s1around p1until p0is on s2.
Figure 5.27. We can rotate and move so that finally p0and p1are colinear with
p2and pn.
Figure 5.28. We can rotate and move so that finally p2and pnare on s1.
Figure 5.29. We can rotate so that finally p0and pnare on s2.
Figure 5.30. Two circles meet in three ways. (left) They intersect at two points.
(middle) They touch outside to outside. (right) They touch inside to outside.
PROOF OF THE PLANAR TRIPLE BUBBLE CONJECTURE 27
Fig