Proof of the Double Bubble Conjecture
Author(s): Frank Morgan
Source: The American Mathematical Monthly, Vol. 108, No. 3 (Mar., 2001), pp. 193-205
Published by: Mathematical Association of America
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Proof of the Double Bubble Conjecture
Frank Morgan
1. THE NEWS. In 2000, Hutchings,Morgan,Ritore,and Ros ([11]; see [4], [13])
announceda proof of the Double Bubble Conjecture,which says that the familiar
double soap bubbleof Figure 1 providesthe least-areaway to enclose and separate
two given volumes of air.The two sphericalcaps are separatedby a thirdspherical
cap, all meetingat 120 degreeangles;if the volumesareequal,the separatingsurface
is a flatdisc. Thisresultis the culminationof tenyearsof remarkableprogressby many
mathematicians,
includingundergraduates.
_~~~~~~~~~~~~~~~~~~~~~~~~~~~~-----
. l_l
- .W_-----__ ll
1.201
The standarddouble bubble provides the least-perimeterway to enclose and sepaFigure
MarchBJ
193
rate two prescribedvolumes. CopyrightJohn M. Sullivan, Universityof Illinois; color version at
www.math.uiuc . edu/_jms/
Images.
2. THEPLANARDOUBLEBUBBLE. It all startedwhen the 1990 WilliamsColresearchGeometryGroup[6] provedthe Planar
lege NSF "SMALL"undergraduate
DoubleBubbleTheorem:the standarddoublebubbleof Figure2ab providesthe leastperimeterway to enclose and separatetwo regionsof prescribedarea in the plane.
Perimetercountseverypiece of curveonce, whetherit is on the exterioror the interior.
The group leader, Joel Foisy, in his subsequent undergraduatethesis, apparently
gave the first statementof the Double Bubble Conjecturein R3 as a conjecture.
Plateau[14, pp. 300-301] had studiedthe doublebubbleovera hundredyearsearlier,
andBoys [1, p. 120] had spokenof the conjectureas a fact, as it was widely accepted
for many years.
b
a
B2
AlB
e
B
Bl.
~~~
.::
B2
B:~
fB42t~:l:K
~
24t$2
*....*....
4
Figure 2. The standardplanar double bubble (a and b), and not some exotic alternativewith disconnected
regions or empty chambers (c, d, or e), provides the least-perimeterway to enclose and separatetwo regions
of prescribedarea, as proved by the 1990 undergraduateresearchGeometry Group [6, Figs. 1.0.1, 1.0.2].
3. PROOF OF THE PLANARDOUBLEBUBBLE THEOREM. The majordifficultyin the proofis showingthateachregionandthe exterioris connected.Althoughit
may seem obviousthathavingseveralcomponentsscatteredthroughoutthe bubbleas
in Figure2cde couldnot minimizeperimeter,this turnsout to be not so easy to prove.
Of coursewe preferto prove, ratherthanto assume,thatthe regions are connected.
Besides, in higherdimensionsa minimizerwith disconnectedregionscould ariseas a
limitof regionsconnectedby thintubesas the tubesshrinkaway.
The originalingeniousproofhas undergoneseveralsimplifications,largelydue to
MichaelHutchings.The simplestargumentstartswith any competitoranddeformsit
to a standarddoublebubblewhile decreasingperimeter.Unlike the originalproof,it
does not needto assumethata nice minimizerexists;it does not dependon the general
existenceandregularitytheory.
Duringthe argumentit is convenientto workin the categoryof the "overlapping"
bubblesof the 1992 GeometryGroup[2, 3], smoothimmersionsof finite embedded
planargraphs,in whichthe faces may overlapas in Figure3.
Exterior
t.
Figure 3. In this "overlapping"bubble, components of the regions R1 and R2 overlap. Copyright2000 Frank
Morgan.
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In Section 4 we show thatfor prescribedareasthereis a uniquestandarddouble
bubble,consistingof threecirculararcsmeetingat 120 degrees;if the areasareequal,
the separatingcurveis a straightline segment.
GivenareasA1andA2, considerany,perhapsoverlapping,doublebubblewithareas
at least A1 andA2. Wemay assumethereis no suchbubbleof fewerfaces with smaller
perimeter.We may alwaysassumethatall verticeshavedegreeat least three.
Firstwe claim thatthereare no emptychambers(boundedcomponentsof the exterior).Otherwisewe could delete an edge, incorporatethe empty chamberinto the
neighboringregion,andreducethe numberof faces.
Secondwe claim thatthereareno verticesof degreegreaterthanthree.Otherwise
two of the faces at the vertexboth belong to the first region, to the second region,
or to the exterior.You can alwayscombinethese faces while reducingperimeterand
maintainingareas,as suggestedin Figure4.
Ext.
Ext.
Figure 4. Vertices of degree four or more may be reduced. Copyright2000 FrankMorgan.
Considerthe dual graphformedby placing a vertexinside each face of the two
regions,with an edge betweenverticesof adjacentfaces. Becausethe exterioris connected,this dualgraphhas no cycles. Hence thereis some face F thatlies at an endpoint of the dual graph.This face F must have exactly two edges and exactly two
vertices,as in Figure5.
Unless the bubbleis alreadycombinatoriallythe standardbubble,take out F and
one neighboringedge e, flip it overend to end, andreinsertit as in Figure5. This operationproducesa possiblyoverlappingdoublebubblewith a vertexv of degreefour,
R
R2
(
R
1 )/~(
Figure 5. Flipping a face of two edges and an adjacentedge creates a vertex of degree four. Copyright2000
FrankMorgan.
March 2001]
195
a contradiction.We concludethatthe bubbleis combinatoriallythe standardbubble,
consistingof threeedges meetingat two points.
Replacethe threeedges by circulararcs,maintainingareas.By the isoperimetric
propertyof circlesandcirculararcs,anysuchreplacementreducesperimeter.Now we
may assumethatthe bubbleis embedded.
Next minimizeperimeterin the categoryof threecirculararcsenclosinggivenareas.
We claimthatthe edges mustmeet at 120-degreeangles.Otherwiseyou coulddeform
the bubblea bit so as to preserveareasanddecreaselength.Afterthis deformationthe
edges wouldnot be arcsof circles,butyou couldreplacethemwithcirculararcs,so as
to decreaselengtheven more.
Now we havea standardbubbleof areasatleast A1 andA2. Sinceby reducingeither
areayou can reducethe perimeter,the standardbubbleof areasexactlyA1 andA2 has
even smallerperimeter.
Thus startingwith any othercompetitor,we havereducedperimeterand arrivedat
the standarddoublebubble,whichmustthereforebe perimeterminimizing.
Beforegoing further,we verifythatthereis a uniquestandarddoublebubblein any
Rn(n > 2).
4. THE STANDARD DOUBLE BUBBLE. For prescribed volumes v, w, there is a
unique standard double bubble in Rn consisting of three spherical caps meeting at 120
degrees as in Figure 1.
Proof. Considera unit spherethroughthe origin and a congruentor smallersphere
intersectingit at the origin (andelsewhere)at 120 degreesas in Figure6. Thereis a
uniquecompletionto a standarddoublebubble.Varyingthe size of the smallersphere
yields all volumeratiospreciselyonce. Scalingyields all pairsof volumesprecisely
once.
Figure 6. Varyingthe size of the smaller component yields all volume ratios precisely once. Copyright 2000
FrankMorgan.
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5. EXISTENCE AND REGULARITY.Proving the existence of a perimeterminimizingdoublebubbleof prescribedvolumesin R' is no easy matter.Classical
spacesof surfacesare not compact,especiallyif thereis no a priori boundon topoone cannotconsider
logical complexityor how the pieces fit together.Furthermore,
only bubbleswith connectedregions,becausethey might in principledisconnectin
the minimizinglimit as thinconnectingtubesshrinkaway.
Geometricmeasuretheory (see [13]), as developedin the 1950s and 1960s by
L. C. Young,E. De Giorgi,H. Federer,andW. Fleming,providesvery generalcompact spaces of surfacesof boundeddiameterin RW.Soap bubblesprovideadditional
difficulties,becausethereis no easy a priori boundon the diameter.In 1976 F. Almgren provideda generalproof using geometricmeasuretheory of the existence of
soapbubbleclusters,andwith J. Taylorprovedthatin R3 they
perimeter-minimizing
surfacesmeetingin threesat 120 degrees
consist of smoothconstant-mean-curvature
along curves,which in turnmay meet in fours at equal angles of about 109 degrees
(cos-l (-1/3)).
The followingkey symmetrytheoremis basedon an idea of BrianWhite,written
up by Foisy [5, Thm. 3.4] andHutchings[12, Thm.2.6]. Since it reducesthe Double
BubbleConjectureto questionsaboutcurvesin the plane,it was the mainreasonfor
mistakenlyconsideringthe mattersettled.
6. SYMMETRY THEOREM. A perimeter-minimizing double bubble B in RWis a
surface of revolution about a line.
Proofsketch,case n = 3. Firstwe claimthattherearetwo orthogonalplanesthatsplit
bothvolumesin half. Certainly,for every0 < 0 < 7r, thereis a verticalplaneat angle
0 to the xz-plane that splits the first region in half. These planes can be chosen to
varycontinuouslybackto the originalposition,now with the largerpartof the second
regionon the otherside. Hencefor some intermediate0, the planesplitsbothvolumes
in half. Turningeverythingto makethis planehorizontalandrepeatingthe argument
yield a second plane, as desired.Hence we may assumethat the xz- and yz-planes
bisectbothvolumes.
Actually,we need to modify the process a bit. After obtainingthe firstplane, reflectingthe half of B of least areayields a bubbleof no more area.It must therefore
have the same area,and eitherhalf would work.If the originalB were not a surface
of revolution,neitheris some suchhalf plus reflection.Hence we may assumethat B
is symmetricunderreflectionacrosseach plane, and hence undertheircomposition,
rotationby 180 degreesaboutthe z-axis.Henceeveryplanecontainingthe z-axis splits
bothvolumesin half.
We claim that at every regularpoint, the bubble B is orthogonalto the vertical
plane.Otherwisethe smalleror equalhalf of B, togetherwith its reflection,wouldbe
a minimizerwith an illegal singularity,whichcouldbe smoothedto reduceareawhile
maintainingvolume.Now it follows thatB is a surfaceof revolution.
Hutchingsrealizedthatthe symmetryproof could be generalizedto provethe following monotonicityresult,whichis not as obviousas it sounds.
7. MONOTONICITY [12, THM. 3.2]. The least area A(v, w) of a double bubble
of volumes v, w in RWis a nondecreasing function in each variable.
Proof sketch, case n = 3. We prove that for fixed w0, A(v, wo) is nondecreasing in v.
If not, thenthereis a local minimumat some vo.For simplicity,we treatjust the case
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of a strictlocal minimum.Considera minimizingdoublebubbleB of volumesv0, w0.
By the SymmetryTheorem6 and its proof, B is a surfaceof revolutionabouta line
L = Pfl nP2, where P1 and P2 are planes thatdivideboth regionsin half. Choose a
planeP3 nearP2 thatdividesthe secondregionin halfbutdoes notcontainL. Weclaim
thatit divides the firstregionin half. Otherwise,the half with smalleror equal area,
reflectedacrossthe plane,wouldyield a bubbleof no moreareaandslightlydifferent
volume,contradictingthe assumptionthat vo is a strictlocal minimum.ThereforeP3
splitsbothvolumesin half. Now as in the proofof Symmetry6, B is symmetricabout
the line L' = P1 n P3 as well as about L. It follows that B consists of concentric
spheres,whichis impossible.ThereforeA(v, w0) mustbe nondecreasingas desired.
8. COROLLARY(CONNECTEDEXTERIOR) [12, THM. 3.4]. An area-minimizing double bubble in Rnhas connected exterior ("no empty chambers").
Proof. If theexteriorhas a second,bounded,component,removinga surfaceto makeit
partof one of the two regionswouldreduceareaandincreasevolume,in contradiction
to Monotonicity7.
SymmetryandConnectedExteriorarethe primarylemmasfor the followingstructuretheorem.
9. HUTCHINGSSTRUCTURETHEOREM[12, THM. 5.1]. An area-minimizing
double bubble in Rn is either the standard double bubble or another surface of revolution about some line, consisting of two round spherical caps with a toroidal innertube,
successively layered with more toroidal innertubes, as in Figure 7. The surfaces are
all constant-mean-curvaturesurfaces of revolution, "Delaunay surfaces", meeting in
threes at 120 degrees.
Commentson theproof. Once the exterioris known to be connected,there are not
manypossibilitiesfor a doublebubbleof revolution.A contractionargumentgivenby
Foisy [5, Thm.3.6] showsthatthe bubblemustintersectthe axis.
.06.
doublebubblein Rn consistsof a centralbubblewith layersof
area-minimizing
Figure 7. A nonstandard
toroidalbands.Copyright2000 YuanY.Lai.
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Therecan be at most two sphericalcaps andat most one toroidalinnertubedirectly
on the sphericalcaps. Indeed,if therewere a bandof a thirdsphereS betweentwo
toroidalinnertubesin the firstlayer,the rest of the bubblewould consist of two big
pieces, whichcouldbe rolledaroundS to toucheachotherandcreatean illegal singularity.
finitenessfollows fromthe followingbound.
Combinatorial
10. HUTCHINGS COMPONENT BOUND. Consider a minimizing double bubble
of volumes v, 1 - v in Rn.Then the number k of components of thefirst region satisfies
A(v)k1ln < 2A(v,
1 - v)
-
A(1)
-
A(1
-
(1)
v).
Here A(v) is the surface area of single bubble (roundsphere) of volume v, and
A(v, 1 - v) is the (unknown)areaof a minimizingdoublebubble,which fortunately
is boundedaboveby the areaof the standarddoublebubble.
Proofsketch. The decompositionsuggested by Figure 8 gives a lower bound on
A(v, 1 - v) when the firstregionhas a small componentof volumex. Furthermore,
HutchingsgeneralizesMonotonicity7 to showthatA(v, w) is strictlyconcavein both
variables.Now the desiredcomponentboundfollows by just a little algebra.
2s(t)=
x+
+9
Figure 8. This decomposition gives a lower bound on A(v, 1 - v) when the firstregion has a small component
of volume x. Each piece of surface on the left occurs exactly twice on the right. Drawing by James F. Bredt,
copyright 2000 FrankMorgan.
Remark.Mathematicagraphsof the bound(1) on the numberof componentsk for R22,
R3, R4, andR', with A(v, 1 - v) replacedby the largeror equalareaof the standard
doublebubble,appearin Figure9. Some resultsare summarizedin Table10. In particular,in R2 both regionsare connected,from which the Double BubbleConjecture
6.625
6
5
4.06
n=5
3
n=4
2.5
2
n=
1.57
1
0
0.5
1
Figure 9. A bound on the numberof components of the first region in a minimizing double bubble of volumes
v, 1 - v in R2 throughR5. Figure by Ben W. Reichardt[10, Fig. 2].
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TABLE 10. Bounds on the number of components in a perimeter-minimizing
double
bubble.
Bounds on number of components in
larger or equal region
Bounds on number of components in
smaller region
R2
R3
R4
Rs
R
1
1
1
2
3
1
2
4
6
2n
in 12 follows easily. The Rn bounds are elegantly deduced from the Hutchings Bound
10 by the 1999 Geometry Group [10, Prop. 5.3].
The following conjecture [10, Conj. 4.10] would provide an elegant way to prove
the numerical bounds rigorously, a task otherwise quite awkward.
11. CONJECTURE.
In Rn, let Ho, H1, H2, respectively, denote the mean curvature
of a sphere of volume w, a sphere of volume w + 1, and the exterior of the second
region of the standard double bubble of volumes 1, w, as suggested by Figure 11. Then
2H2 > Ho + H1.
120
(
w
Ho
)
t
w+1
H1
w
H2
Figure 11. Conjecture:the pictured curvaturessatisfy H2 > (Ho + HD)/2. If true, this conjecture would provide an elegant way to prove the bounds of Table 10. Copyright2000 FrankMorgan.
Note that the Hutchings Bound implies that for equal volumes, each region of a
minimizing double bubble in R3 is connected. In 1995 Hass and Schlafly exploited
this information to prove this case of the Double Bubble Conjecture by computer.
12. THEOREM ([71, [9]). For equal volumes in R3, the standard double bubble
uniquely minimizes perimeter
Proof sketch ([9] and [8]). For equal volumes in R3, the Hutchings theory says that
both regions are connected (Table 10) and that a minimizer consists of a central bubble with a toroidal innertube around it as in Figure 12. The innertube need not be
centered around the waist, but there is at most a 2-parameter family of possibilities.
Hass and Schlafly used a computer to recursively divide the parameter domain into
cases and subcases, always seeking some contradiction. Often the volumes fail to be
equal. Sometimes there are unstable pieces of surface, as in Figure 13. Sometimes the
pieces just do not fit together. In all computations, the values are bounded above and
below by exact computations in integer arithmetic. If no contradiction appears with the
required accuracy, the case is subdivided into finer subcases. If eventually every case
terminates in a contradiction and the computer program halts, the theorem is proved.
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Figure 13. Candidateswith large oscillations may be eliminated as unstable. CopyrightJoel Hass, Jim Hoff. html.
man, and Roger Schlafly. Color version available at www. math. ucdavis . edu/-hass/bubbles
March2001]
201
In fact,in 1995it finishedon a PC in abouttwentyminutes,havingcomputed200,260
numericalintegrals.
.i~~~~~~~~~~~~~~~U4
__
'1_4
1~~~~~~~~~~~~~~~~~~~~~?
Figure 14. A minimizing double bubble in R3 has at most three components. Copyright John M. Sullivan,
University of Illinois (color version at www. math. uiuc . edu/'-j ms/Image s); second image from [11].
For arbitraryvolumesin R , the HutchingsBound(Table10) says only thata minimizing doublebubblehas at most threecomponents,as in Figure14, muchtoo complicateda family of possibilitiesto be handledby currentcomputertechnology.The
202
2000
?_HproofAHMTCLASCAIO
uses a new instabilityargument.
FAEIA[otl0
computer-free
13. THEOREM [11]. For given volumes in R3, the standard double bubble of Figure 1 uniquely minimizes perimeter.
candidateis unstable;thatis,
Proofsketch. The idea is to show thatany nonstandard
it can be deformedso as to reduceareawhile preservingboth volumes.The desired
is obtainedby rotatingdifferentpieces of the doublebubble,suchas the left
instabillity
and righthalves of Figure 15, in differentdirectionsor at differentrates wi arounda
carefullychosenaxis A.
In general,maintainingthe volumeconstraintsrequirestwo extradegreesof freedom,requiringfourratherthantwo pieces. To avoidrippingthe bubbleapart,the bubble must be tangentialto the rotationalvectorfieldwherethe pieces meet. Figure 15
couldbe dividedinto the fourquadrants.
Of course,if every eo, = 1, the whole bubblejust rotates,andthe secondvariation
of perimetervanishes.If, as we assumeto obtaina contradiction,the bubbleis stable,
so thatno choice of coi'syields negativesecondvariation,it follows thateverychoice
of o1i'syields vanishingsecondvariation,correspondingto the solutionto some partial differentialequation.If thereare (at least) four pieces, then the ratesof rotation
(0, 3, (04 can be chosento maintainthe two volumes,with at least one butnot all
WI ,
of the o,i's equalto 0. Since one wi vanishes,by uniquecontinuationfor solutionsto
Figure 15. Rotating the two halves of this bubble in opposite directions about a new axis A stretches the top,
shrinks the bottom, and reduces area. Copyright2000 FrankMorgan.
nice partialdifferentialequations,all thevariationsmustvanishandtherelevantpieces
of bubblemustbe spherical,whichleadsto a contradiction.
Can you always find an axis A of rotationso thatthe curveswhere the bubbleis
tangentialdividethe bubbleinto fourpieces?
to the axis of rotationalsymmetry,the
Wecarefullychoosethe axis A perpendicular
x-axis. The surfaceis automaticallytangentialin the planenormalto A, so we always
haveat least two pieces, top andbottom.
If as in the case of equal volumesthe bubblealwayshadjust two componentsas
in Figure 16, a suitableaxis A is providedby the perpendicularbisectorof the two
verticesV1,V2:at the closest andthe most distantintermediatepoints,Pi and P2, the
radiusvectorfrom A meets the bubbleorthogonally,so thatthe bubbleis tangential
at each pi andat the whole circularorbitof pi aroundthe x-axis. The bubbledivides
into the fourrequisitepieces. This provesthe DoubleBubbleConjecturefor the case
of equalvolumes.
P2
A
1
Figure 16. If the bubble has just two components, a central bubble and toroidal innertube,a suitable axis of
" where
instability A is provided by the perpendicularbisector of the two vertices VI, V2. The places "
the rotational vectorfield is tangent to the surface divide the bubble into four pieces. Copyright 2000 Frank
Morgan.
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203
A-----A---
--
Figure 17. If the bubble has three components, one has to find the axis A of instability case by case, according
to the relative position of parts of the bubble. Drawing by James F. Bredt, copyright 2000 FrankMorgan.
For the general case of three components, the Euclidean geometry is much more
complicated but manageable. One considers cases according to the relative position of
parts of the bubble, as in Figure 17.
14. HIGHER DIMENSIONS. In an amazing postscript, the 1999 Williams College
"SMALL"Geometry Group, consisting of Ben Reichardt, Cory Heilmann, Yuan Lai,
and Anita Spielman, has extended the proof of the Double Bubble Conjecture to R4
and certain higher dimensional cases.
15. THEOREM [15, THMS. 9.1, 9.21. In R4, the standard double bubble is the
unique minimizerIn Rn, for prescribed volumes v > 2w, the standard double
bubbleis the uniqueminimizer
Proof sketch. In these cases, the Hutchings theory implies that the larger region is connected. Since there is no bound on the number of components of the second region, the
case-by-case analysis of Figure 16 must be broken down into more general arguments
about constitutive parts.
16. OPEN QUESTIONS. HIutchingset al. [11, Intro.] conjecture that the standard
double bubble in Rn is the unique stable double bubble. Although the conclusion of
their proof shows the final competitor unstable, earlier portions such as Symmetry 6
assume area minimization, so that there could be unsymmetric stable bubbles, for all
we know.
Sullivan [16, Prob. 2] has conjectured that the standardk-bubble in Rn (k ' n + 1)
is the unique minimizer enclosing k regions of prescribed volume. There is also the
very physical question in R3 of whether the standarddouble bubble is the unique stable
double bubble with connected regions; by [11, Cor. 5.3], it would suffice to prove
rotational symmetry.
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ACKNOWLEDGMENTS. This article is largely based on the new Chapter 14 of the 2000 edition of [13],
where furtherdetails and references may be found. This work was partially supportedby a National Science
Foundationgrant.
REFERENCES
1. C. V. Boys, Soap-Bubbles, Dover, New York, 1959.
2. ChristopherCox, Lisa Harrison,Michael Hutchings, Susan Kim, JanetteLight, Andrew Mauer,and Meg
Tilton, The shortest enclosure of three connected areas in R2, Real Anal. Exchange 20 (1994/95) 313335.
3. ChristopherCox, Lisa Harrison,Michael Hutchings, Susan Kim, JanetteLight, Andrew Mauer,and Meg
Tilton, The standardtriple bubble type is the least-perimeterway to enclose three connected areas, NSF
"SMALL"undergraduateresearchGeometry Groupreport,Williams College, 1992.
4. Barry Cipra,Why double bubbles form the way they do, Science 287 (17 March2000) 1910-1911.
5. Joel Foisy, Soap bubble clusters in R2 and R3, undergraduatethesis, Williams College, 1991.
6. Joel Foisy, Manuel Alfaro, JeffreyBrock, Nickelous Hodges, and Jason Zimba, The standarddouble soap
bubble in R2 uniquely minimizes perimeter,Pacific J. Math. 159 (1993) 47-59.
7. Joel Hass, Michael Hutchings, and Roger Schlafly, The double bubble conjecture, Electron. Res. Announc. Amer.Math. Soc. 1 (1995) 98-102.
8. Joel Hass and Roger Schlafly, Bubbles and double bubbles,Amer.Scientist, Sept-Oct, 1996, 462-467.
9. Joel Hass and Roger Schlafly, Double bubbles minimize, Ann. Math. 151 (2000) 459-515.
10. Cory Heilmann, Yuan Y. Lai, Ben W. Reichardt, and Anita Spielman, Component bounds for areaminimizing double bubbles, NSF "SMALL"undergraduateresearch Geometry Group report,Williams
College, 1999.
11. Michael Hutchings, Frank Morgan, Manuel Ritore, and Antonio Ros, Proof of the double bubble
conjecture, Electron. Res. Announc. Amer. Math. Soc. 6 (2000) 45-49. Preprint of full paper at
www. ugr . es/-ritore/bubble/bubble
. htm.
12. Michael Hutchings, The structureof area-minimizingdouble bubbles, J. Geom. Anal. 7 (1997) 285-304.
13. FrankMorgan,GeometricMeasureTheory:a Beginner's Guide, thirded., Academic Press, Boston, 2000.
14. J. Plateau, Statique Experimentale et Th6orique des Liquides Soumis aux Seules Forces Moleculaires,
Paris, Gauthier-Villars,1873.
15. Ben W. Reichardt,Cory Heilmann, YuanY. Lai, and Anita Spielman, Proof of the double bubble conjecture in R4 and certainhigher dimensions, Pacific J. Math., to appear.
16. John M. Sullivan and Frank Morgan, ed., Open problems in soap bubble geometry, Internat.J. Math. 7
(1996) 833-842.
FRANK MORGAN works in minimal surfaces and holds one of the first MAA national Haimo awards for
distinguished teaching. He has some recent books: Geometric Measure Theory:a Beginner's Guide; Riemannian Geometry: a Beginner's Guide; Calculus Lite; and The Math Chat Book, based on his column and TV
show, available at www.maa. org. He is second Vice-Presidentof the MAA.
Departmentof Mathematicsand Statistics, WilliamsCollege, Williamstown,MA 01267
[email protected]
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