Rigid and Flexible Frameworks
Author(s): B. Roth
Reviewed work(s):
Source: The American Mathematical Monthly, Vol. 88, No. 1 (Jan., 1981), pp. 6-21
Published by: Mathematical Association of America
Stable URL: http://www.jstor.org/stable/2320705 .
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RIGID AND FLEXIBLE FRAMEWORKS
B. ROTH
of Wyoming,
ofMathematics,University
Department
Laramie,WY 82071
Considera triangleor a squarein theplane R2whoseedgesare inextendible,
1. Introduction.
rodswhicharejoined but rotatefreelyat thevertices.The squareis said to be
incompressible
(or fallover)intoa familyof rhombi,as
flexiblein R2 sincethesquarecan movecontinuously
the
shownin Fig. 1. However,thetriangleis said to be rigidin R2sincethethreerodsdetermine
in R3 consistingof six rods,
a tetrahedron
relativepositionsof the threevertices.Similarly,
in the
is rigidin R3,whilea cube constructed
connectedbutfreelypivotingat thefourvertices,
shownin
same fashionis flexiblein R3. The collectionof rods and connectors(or framework)
witha commonedge,is rigidin R2butflexiblein R3 sinceone
of twotriangles
Fig. 2, consisting
will
trianglecan thenrotaterelativeto the otheralong thecommonedge. (Precisedefinitions
appearin Section3.)
Pi
P2
_0- -
P3
Pi
P2
P3
P4
_--0
P4
FIG. 1
FIG. 2
Is a givenframework
rigidorflexiblein a givenEuclideanspace RI? This,thebasic question
labeledas a problemin algebraor analysisor some
of thesubject,does notarriveconveniently
havebeenmadeto therigidity
problemfroma varietyofareas,
otherfield.In fact,contributions
linear
geometry,
differential
complexanalysis,projective
topology,
includingalgebraicgeometry,
Even the Inverseand ImplicitFunction
geometry.
algebra,graphtheory,and combinatorial
toolsforthestudyof rigidity
and flexibility,
Theoremsfromadvancedcalculusprovidepowerful
problemof someimportance
as we shallsoon see. The subjectsimplyarisesfroma real-world
mathematical
and leads to interesting
questionswhichone attacksby any availablemeans.
and others
mathematicians,
engineers,
The questionis hardlynew.For well overa century,
and mechanisms.Their
under such names as linkages,linkworks,
have studiedframeworks
labors producedgeneralresultsrangingfromthe theoremof Cauchy [9] on the rigidityof
surfacesof convexpolyhedrato theproofof Kempe [19] thatany algebraiccurvein theplane
in
in R2. But therewas widespreadinterest
framework
can be (locally)drawnby an appropriate
as well,rangingfromtheworkof Bricard[8], Bennett[5], and otherson
specificframeworks
[17,
flexibleoctahedrain R3,to thelinkageof Peaucellierdescribedby Hilbertand Cohn-Vossen
was also paid to thestaticsof
line segment.And attention
p. 273] whichtracesout a straight
of
throughtheformation
i.e., the resolutionof externalforceson theframework
frameworks,
A bibliography
on linkages
and tensionin the rods of the framework.
forcesof compression
compiledby Kanayama [18] in 1933, containingover threehundredentries,providesan
of ReeseProsser.Sincethenhe has beenat the
his Ph.D. in 1969underthedirection
The authorreceived
of
of California,
Davis, and at the University
of Wyoming,
exceptforleavesat the University
University
and
in ringsofcontinuous
idealsof differentiable
interested
functions,
He was originally
functions,
Washington.
is now
he becamefascinated
experience
distributions.
bygeodesicdomes,and aftersomepractical
Subsequently,
offrameworks.-Editors
a bookon therigidity
writing
6
7
RIGID AND FLEXIBLE FRAMEWORKS
appearanceof names
overtheyears.The frequent
pictureof thevariationin activity
interesting
indicatestheproblemswere
in theliterature
and Tchebychef
suchas Cayley,Maxwell,Sylvester,
and some
of interest
widelyknownand of broadappeal. Recentyearshave witnesseda rebirth
combinatorial
rangingfromresultsof Laman [21] concerning
excitingnew resultsin rigidity,
methodsin theplane to theworkof Bolkerand Crapo [6],[7] on bracinggridsof squaresand
is the flexiblepolyhedralsurfaceof
cubes. Perhapsthe most strikingmodem contribution
Connelly[10],[11],[12],an accountof whichappearsin Kuiper[20]. Yet anothersignificant
one ofwhoseprimary
Topology,
eventis therecentappearanceof theresearchbulletinStructural
themesis rigidity.
of theflavorof thesubjectby introducing
some
The presentpaperseeksto impartsomething
way and thenusingtheseto settleone natural
of itsconceptsand techniquesin an elementary
and basic question.We beginwitha brieflook at a fewsimpleexampleswhichmayexposethe
and describdefinitions
Afterformulating
uses of familiartheorems.
readerto someunfamiliar
ing a simplerigiditypredictorbased on the InverseFunctionTheorem,we focus on the
in a natural
followingproblem.A convex polyhedronC in R3 gives rise to a framework
of verticesand edges of C. Whichconvexpolyhedragiverigid
way-namely,the framework
Its
in R3?The answeris simplythosepolyhedraforwhicheveryfaceis a triangle.
frameworks
introduced
by Cauchyin 1813,remainamong
proofrelieson ideas which,althoughoriginally
in thesubject.
themostbeautifuland important
Finally,I wishto acknowledgemydebt to HermanGluck-much of the presentpaperis
simplyan expanded(and perhapsclarified)versionof his paper[15].
Examples.
2. Elementary
shownin Fig. 1, wherethe verticesinitially
2.1. Considerthe square framework
EXAMPLE
thesquarefrom
havecoordinates
pI = (0,1), P2 = (1, 1), p3 = (0,0), andp4= (1,0) in R2.To prevent
which
of thevertices
the
relative
and
rotations
translations
positions
-(in
the
movingin
planeby
whicharejoined by an edge,say thethird
do notchange),we fixtwoverticesof theframework
verticeswill be allowedto assumeany position
p4. The remaining
p3, and the fourth,
vertex,
withtheconstraints
imposedby theedges.
consistent
be xl and x2,
Let the coordinatesof the firstand second verticesof the framework
and let j j denotetheEuclideannormin R2.Thenthesetof solutionsof thesystem
respectively,
of edgeequations
.
Ix1-x212=
1
x-p312= IX12= 1
xI
Ix2-p412= =x2-
(2.1)
(1,0)12= 1
(In generalwe
is theset of possiblelocationsof thefirstand secondverticesof theframework.
anymechanicalproblemsthat
allowedgesto crosseach otherand verticesto coincide,ignoring
mightarisein thisway.)The familyof solutions
x(t) =(t,
t2),
x2(t) = (I + t, \i7'7_)
fort E[0,1]
givestheflexingof thesquareshownin Fig. 1 whichbeginsat (P1,P2)= (0, 1,1,1), i.e., satisfies
(Xl(0), X2(0)) = (P1,P2).
EXAMPLE2.2. We add an edge betweenthesecondvertex,
P2, and thethird,
p3, of Example
and
shownin Fig. 2. Againfixvertices
2.1,obtainingtheframework
p3 andP4 of theframework
let xl and x2 be thecoordinatesof thefirstand secondvertices.The systemof edge equations
withtheequation
nowconsistsof thethreeequations(2.1) together
= 2.
=IX212
Ix2-p3j12
8
[January
B. ROTH
The setof solutions(X1,X2) E R2x R2= R4of thissystemof equationsis thefiniteset
{(0, 1,1,1),(1,0,1,1),(0,-1, 1,- 1), (1,0,1,-1))
as can easilybe seen by drawingcirclesto representthe solutionsets of the variousedge
cannotcontinuously
equations.Sincethefirstand secondverticesof theframework
moveaway
in thesolutionsetof thesystem
fromtheirgivenpositionp = (p1,p2)= (0, 1,1,1) whileremaining
is rigidin R2.
of edgeequations,theframework
Thus it is the natureof the solutionset of the systemof edge equationsnear the given
whichdetermines
the rigidityor flexibility
locationof the verticesof the framework
of the
in theplane,one need only
or flexibility
of a framework
framework.
And to predicttherigidity
solve its systemof edge equations,at least near the initiallocationof its vertices.This very
elementary
approachto rigidity
generalizesto higherdimensionalspaces,althoughsome care
mustbe exercisedin choosingthe verticesto fix in orderto eliminatethe motionsof the
framework
as a rigidbody.For example,in R3one can fixthreenoncollinear
vertices,
all pairs
of whicharejoined by edges,i.e., fixa triangle.
The Inverseand ImplicitFunctionTheoremsprovidea somewhatmore sophisticated
It is convenient
at thisstageto adopta slightly
different
approachto rigidity.
pointof viewand
ratherthanthesystemof edge equations.The edge
of a framework
considertheedgefunction
function
of Example2.2 is thefunction
f: R4--R4 definedforx =(xI, x2) E R2x R2= R4 by
f(x) = (IxI - x212,
jx2 p312)
(2.2)
Ix -p312,jx2 P412,
wherep3 andp4 are thetwofixedvertices.The solutionset of thesystemof edge equationsis
precisely
f -'(f(p)) wherep = (P1,P2) = (0, 1,1, 1).
differentiable
function
The InverseFunctionTheoremsaysthata continuously
f fromRn to
Rn has a continuously
inversein a neighborhood
of anypointp ERRn forwhichthe
differentiable
derivativedf(p) is a nonsingularlinear transformation.
Therefore,among otherthings,the
of anyp E Rn for
InverseFunctionTheoremguaranteesthatf is one-to-onein a neighborhood
whichdf(p)has rank,n.
of theframework
in Example
EXAMPLE2.2 (continued).Let f: R4 -R4 be theedge function
2.2 givenin equation(2.2). The matrixwithrespectto thestandardbasis forR4 of thederivative
of thefour
df(p) off at P = (P1,P2) = (0, 1,-1,1) is obtainedby evaluatingthepartialderivatives
off atp. Ratherthanwriting
coordinatefunctions
df(p)as a matrixwithfourcolumns,it is very
convenient
forourpurposesto writethematrixdf(p) withjust twocolumnswheretheentriesin
each columnare actuallyvectorsin R2. Doing this,we find
[P1P2
df(p)= 2
PI
P1P3
0
0
P2-P1
P2-P4
P2-P3
withrespectto thecoordinatesof xl
wherethefirstcolumnis obtainedfrompartialderivatives
withrespectto thecoordinates
ofx2.One simple
and thesecondcolumnfrompartialderivatives
is to showthatits rowsare linearlyindependent.
geometrical
wayto see thatdf(p) is invertible
Considera linearcombinationof therowsof df(p) whichequals zero. Summingoverthefirst
of the firsttwo rowsof df(p) mustvanishsince
colunmof df(p),we see thatthe coefficients
vectorsin R2. Similarly,
summingoverthe
PI -P2 and PI -P3 are clearlylinearlyindependent
of thelasttworowsofdf(p)also vanishsince
secondcolunmofdf(p)showsthatthecoefficients
P2-p4 andP2 p3 are linearlyindependent.
U ofp =(p1,P2) in R4 such
By the InverseFunctionTheorem,thereexistsa neighborhood
thatf is one-to-oneon U. Therefore,
f-(f(p)) n U= (p}, i.e.,p is the only solutionof the
systemof edge equationsin U. The meaningof thisis picturedin Fig. 3 whereU1 and U2 are
RIGIDANDFLEXIBLE
FRAMEWORKS
1981]
9
ofp1 andp2,respectively,
suchthatU1x U2c U. The onlysolution(x1,x2)of the
neighborhoods
edgeequationswithxl E U1andx2E U2is givenby (XI,X2)= (P1,P), and thusitis notpossibleto
move the firstand second verticesof the framework
continuously
away fromtheirgiven
theedgelengthsof theframework.
positions
Therefore
theframework
pi andp2whilepreserving
is rigidin R2.
U2
U,
P3
P4
FIG. 3
of a tetrahedron
to verify
therigidity
in R3 in thesame
The readermightfindit instructive
thematrixdf(p) as havingjust one column
to fixa triangleand considering
way,remembering
withentriesthatare vectorsin R3.
The ImplicitFunctionTheorem,whichcan be viewedas providing
information
aboutsetsof
of
the formf '(f(p)), turnsout to be everybit as usefulfor establishingthe flexibility
frameworksas the Inverse Function Theorem forrigidity.Supposef: RI +
m-Rm
is continuously
and letp= (a, b) E Rn" I wherea ClRI and bE Rm.The ImplicitFunctionTheorem
differentiable,
thenthereexistsa neighborsays thatif thelast m c.olumnsof df(p) are linearlyindependent
differentiable
function
hood U of a in R" such thatthereis a uniquecontinuously
g: U-*Rm
satisfyingg(a) = b and
(x,g(x))E f-'(f(p))
forallxE U.
f:R4-11i3of the squareframework
2.1 (continued).The edge function
EXAMPLE
in Fig. 1 is
definedby
f(x)=(IxI
-x212, xP312
Ix2-p412)
for
X = (x1, x2) ER 2 X R2 = R4
thatthelastthreecolumnsofdf(p) are
wherep3andp4 are thefixedvertices.It is easyto verify
wherep = (p1,p2)= (0,1,1,1). Thus thereexistsa neighborhood
U of 0 in IR
linearlyindependent
and a unique functiong: U--R3 satisfyingg(O) = (1, 1,1) and (t,g(t)) ef -l (f(p)) forall t E U. Of
course,in thisparticular
example,we evenknowthat
g(t)=(
l-t2,1+t,
1-t2)
forallteUn[-l,1]
since g is unique and
(t,
l -t2,
+ t, V/1-t2 ) ef '(f(p))
for all t E[-l,
1].
However,theimportant
thingis that(t,g(t)) fortE U givesa flexingof the square satisfying
(0,g(O)) =P.
In general,theImplicitFunctionTheorem(ifapplicable)allowsone to varysomecoordinates
of some verticesin a neighborhoodU and the functiong then prescribesthe remaining
10
[January
B. ROTH
of theverticesin such a way as to remainin thesolutionset of thesystemof edge
coordinates
equations.Evenif U c RI wheren > 1, it is theneasyto producea continuouspathx beginning
at p and lyingin the solutionset of thesystemof edge equations,i.e., satisfying
x(0) =p and
x(t) ef -1(f(p)) forall tE [0,1]. This is essentially
thedefinition
of flexibility
we adopt in the
nextsection.
3. Definitions.
In thissection,definitions
of frameworks,
edge functions,
rigidity,
and flexiin Rn in orderto deal withthe mostinteresting
bilityare formulated
cases (n = 2 and n= 3)
simultaneously.
An (abstract)framework
G is a set V= {1,2,,...,, v) togetherwitha nonemptyset E of
two-element
subsetsof V. Each elementof V is referred
to as a vertexof G whileeach element
of E is called an edgeof G. For iE V, we let a(i)={ j V: {i,j)EE), theset of verticesof G
whichare adjacent(orjoinedby an edge) to thevertexi. Sincean (abstract)framework
is really
nothingotherthan an (abstract)graph,we occasionallyuse the languageof graphtheory.
However,our primaryinterestis not in abstractframeworks
but ratherin theirconcrete
realizationsin some Euclideanspace Rn. A framework
G(p) in Rn is an abstractframework
witha point
G=( V,E) together
P
(P1I ..* pv) E-RtnX ...X
RI'=Rnv.
We refertopAfori E V as a vertexofG(p) and theclosedlinesegment
[p,,pj]in RI for{ ij) E E
as an edgeofG(p). In otherwords,theframework
G(p) in RI is obtainedby locatingvertexi of
G at thepointPAE Rn.
For theremainder
of thepaper,we dispensewithfixingvertices;and thusour definition
of
now takesa slightly
different
form.Considera framework
the edge functionof a framework
Orderthee edges
G=(V,E) withv verticesand e edges,i.e., V= {1,... ,v) and E has e elements.
ifyouwish)and definef: Rnv-*R', theedgefunction
of G in someway(lexicographically,
ofG,
by
f(P) =ff(P i, ...,Pv)=
.'*.,PIA jy2,. ...)
where{i,j) E E, pkE Rnfor 1 ? k ? v, and I denotesthe Euclideannormin RI. If G(p) is a
in Rn,thenf(p) e RI consistsof thesquaresof thelengthsof thee edgesof G and
framework
such thatG(p) and G(q) have corresponding
thusf-(f(p)) is the set of q E RlRV
edge lengths
equal.
.
of Example2.2. shownin Fig. 2. Its abstractframeEXAMPLE
3.1. Considertheframework
workG is givenby V= { 1,2,3,4) and E ={{1, 2), { 1,3), {2,3}, {2,4), {3,4)) and we examined
theframework
G(p) in R2where
P = (P1,P2,P3,P4)= (O, 1,1,1, 0,,1, 0).
of G is themapf: R8'-R5 definedby
The edge function
f(q)=
(Iqi -
q212, ql - q312,Iq2- q312,1q2- q412,kq3i q412)
where
q = (ql, q2, q3,q4)
E R2X R2 X
R2x R2 =R8.
thesetf -1(f(p)) now includesall q E Rnvthatare obtainedby simplymoving
Unfortunately,
thevertices
p of the framework
G(p) aroundby translations,
rotations,
and, in general,rigid
motionsof R . Recall thata rigidmotionof Rn is a distancepreserving
map T: RI R i.e., a
map satisfying
ITx- TyI= Ix-yI forall x and y in R. Forp=(p1,I...,pV) and q=(q I ... qv) in
ifthereexistsa rigidmotionT of R' suchthatTpi= q, for
p and q are congruent
Rnv, we say that
thenthesetM= {q E R?v: q is
in Rn andf is itsedge function,
1 S i S v. If G(p) is a framework
are distancepreserving.
since
motions
of
a
subset
to
is
congruent p)
obviously
f `(f(p))
rigid
1981]
RIGID AND FLEXIBLE FRAMEWORKS
11
thepaper
top is a smoothmanifold(wherehereand throughout
The setM ofpointscongruent
Moreover,iftheaffinespan of thepointsp1,... ,pv,is
differentiable).
"smooth"meansinfinitely
Rn(whichmeans thatthe pointsPi, .., p,,do not lie on any hyperplanein Rn), thenM is
manifoldof orthogonal
sinceit arisesfromthen(n- 1)/2-dimensional
n(n+ 1)/2-dimensional
of
of Rin. Thus M is
manifold
translations
n-dimensional
the
of
Rn
and
transformations
forG(p) in R2. We are now in a positionto
forG(p) in R3 and 3-dimensional
6-dimensional
and flexibility.
definerigidity
andp E Rnv.
f is itsedgefunction,
withv vertices,
DEFIMNTION3.2. SupposeG is a framework
x: [0,I]->R V
G(p) in Rn is flexiblein Rn if thereexistsa continuousfunction
The framework
satisfying
(i) x(O)=p,
(ii) x(t) ef-(f(p)) forall te[O,1], and
top forall t E(0, 1].
(iii) x(t) is notcongruent
G(p) is rigidin Rn ifitis notflexiblein
Sucha pathx is calleda flexingof G(p). The framework
n
R
Condition(i) says thatthepathbeginsat p, (ii) thatedge lengthsremainconstant,and (iii)
thatforall te (0,1], G(x(t)) is notobtainedby simplymovingG(p) as a rigidbody.Thus G(p) is
moved fromp to
flexiblein Rn if and only if the verticesof G(p) can be continuously
Note that this
of
the
framework.
the
lengths
edge
while
preserving
positions
noncongruent
oneanotherduringa flexing.
allowsedgestopass through
definition
are invariantunder reasonablechanges in the
The conceptsof rigidityand flexibility
(or even real
differentiable
For example,requiringthatthe path x be infinitely
definitions.
top forjust somete(O, 1] leads to equivalentnotionsof
analytic)or thatx(t) be noncongruent
feature
(see Gluck[15] or Asimowand Roth[2]). However,thereis one disconcerting
flexibility
to meanthateveryq ef -1(f(p)) sufficiently
One wouldlikerigidity
of rigidity.
ofourdefinition
top (whichis theanalogof thebehaviorshownin Fig. 3 in the
close top is actuallycongruent
this.
guarantees
withno verticesfixed)and it is farfromclearthatourdefinition
presentsetting
of G(p)
close top butnotin M whileno flexing
Theremightexistpointsinf `(f(p)} arbitrarily
exists.However,generalresultsin algebraicgeometryregardingthe existenceof paths in
and itis indeedthecase that
algebraicsets(see Milnor[22,Lemma3.1])preventthisoccurrence,
G(p) is rigidin Rnif and onlyif M andf-(f(p)) coincidenearp.
flexibiltheconceptofinfinitesimal
and Statics.We nowintroduce
Flexibility
4. Infinitesimal
conditionsimposedby the
by focusingon thetangential
itywhicharisesfromthatof flexibility
withv vertices,
f is itsedge
SupposeG is a framework
of edgelengthrequirements.
preservation
function,and p e Rv. Let x = (xI-..., xe,)be a smooth functionon [0,1] satisfyingx(0) =p and
x(t)ef'-(f(p)) forall tE [0,1].Thus forall edges { ,j) of G, we have
Ix,(t)- x(t)12=IP,-pj12 forall t E [ 1].
and evaluatingat t= 0, one obtains
Differentiating
(xi(0) - xi(0))' (xi'(0)-
xy())
= (pi
p1).(x'(0) - x'(O)) = 0
x of G(p) assignsa velocityvectorp = xi'(O)e Rn
forall edges{ i,j) of G. Thus a smoothflexing
to each vertex
pi of G(p) in sucha way that
(Pi-Pi).(pA -Il)=0
forall edges{i,j} of G.
(4.1)
it is easy to see that =( lS, p,) E RnlVsatisfies(4.1) if and
In termsof theedge function,
df(p): RnvS8>Re.For by applying
onlyif,t is an elementofthekernelofthelineartransformation
the matrixof df(p) to It, one finds df(p)( It)= 0 if and only if
12
B. ROTH
[January
forall edges{i,j} of G.
Recall nowthata flexing
x(t) of G(p) beginsat p, preserves
edgelengths,
and also does not
belongto themanifoldM of pointscongruent
top forall t> 0. In thesame spirit,we require
thatan infinitesimal
flexingu of G(p) instantaneously
preserveedge lengths,
i.e., satisfy(4.1),
and also notbelongto thetangentspace T. to themanifoldM at thepointp. Note thatTpc
kerneldf(p),sinceifx = (xi,. . . ,x))R--M is a smoothpathwithx(O)=p thenforall i andj we
have
jx,(t) - xj(t)12= IPi-p12 forall tE R
lead to thefollowing
and hence,A= x'(0) satisfies(4.1). These observations
definition.
DEFINITION
withv vertices,
4.1. SupposeG is a framework
f is itsedgefunction,
andp e"lo.
The framework
G(p) is infinitesimally
rigidin R' if Tp= kerneldf(p)and infinitesimally
flexiblein
RI otherwise.
Elementsof kerneldf(p)- Tp are called infinitesimal
flexingsof G(p).
How are infinitesimal
flexibility
and flexibility
related?In Section5, we showthatflexibility
impliesinfinitesimal
flexibility,
and thefollowing
exampleestablishesthatinfinitesimal
flexibilityis a strictly
weakernotion.
EXAMPLE
4.2. Considerthe degeneratetriangleG(p) in R2 shownin Fig. 4 withcoUinear
vertices,sayPI = (0,0), P2= (1,0), and P3= (2,0). Then G(p) is infinitesimally
flexiblebut not
flexiblein R2. For if we let = 3= 0 and ft2= (0, )where c 0, then(4.1) is satisfiedbut
is clearlynotthederivative
at t = 0 of a smoothmotionof G(p) as a rigidbodyin
,
IL = ( 1 t,ju3)
R2.However,itis easyto see thatno flexing
of G(p) exists.
P2
P3
FIG. 4
notionsof rigidity
and the
Historically,
theexistenceof thesetwocloselyrelatedbut distinct
of bothhave combinedto createwidespreadconfusion.Even
absenceof adequatedefinitions
textbookscontainmaterialthatis, at best,misleadingand, at worst,
today,manyengineering
just plain false in connectionwith rigidity.Griinbaumand Shephard[16] providea vivid
thatengineers
The evidencestrongly
areprimarily
situation.
accountof thisunfortunate
suggests
This preference
is explainedby theclose
in infinitesimal
ratherthanrigidity.
interested
rigidity
and thestudyof staticsforframeworks
betweeninfinitesimal
connections
rigidity
(whichhave
We now embarkon a very
as staticrigidity).
led someauthorsto referto infinitesimal
rigidity
in R3 as actual
shortcourseon staticsin R3 wherethe readershould thinkof frameworks
stiffrods thatare connectedby articulated
physicalobjectswhoseedgesare straight,
jointsat
thevertices.
Certainsystemsof internalforcesin a framework
providea convenientbeginningforthis
discussion.Considera framework
G(p) in R3 wherep E R3v and supposethevariousrodsof the
are subjectto forcesof compressionor tensiondirectedalong the rods. More
framework
suchthatwy,11(Pi-p1)is
precisely,
supposethereis associatedwitheach rod [pI,pj]a scalarwC{11
p and wy,j1(pj
theforceexertedby therod on thevertex
-p,) is theforceexertedby therodon
of theforceper unitlength.If wy,j1
thevertex
<0,
pj. The scalarwy,j)thusgivesthemagnitude
to as a compression.
in therod; whileifw{gjj> 0, theforceis referred
theforceis calleda tension
Note that the rod exertsforceson the verticesPAand pj whichare equal in magnitudebut
oppositein direction.In general,a vertexof G(p) will be incidentwithseveralrods and our
interest
focuseson thesituationin whichthesum of theforcesexertedon each vertexequals
zero.A stressof a framework
G(p) in R3 is a collectionofscalarsw{,11,one foreach edge[p,p1]
of G(p), suchthat
Ea(iJ}(Pi)=O
j E a(i)
forl<i6v.
1981]
13
RIGID AND FLEXIBLE FRAMEWORKS
Lettingw{ =0 forall edgesgivesthetrivialstress,and we say a framework
is stressfreeif it
admitsonlythetrivialstress.The degenerate
triangleG(p) ofExample4.2 shownin Fig. 4 has a
= -2 and
= 1. One can imaginethisstressarising
nontrivial
stressgivenby w(1,21
=(2,3
(I,3)
froman attempt
to construct
thedegenerate
triangle
withrod[P1,P3]slightly
too long,creating
a
in thisrod and tensionsin theothertworods.
compression
in termsof theedge function
Stresseshave a convenient
description
of a framework.
Since
theedgefunction
G withv verticesand e edgesis definedforq E R3V
f:R3v * R' of a framework
by
f(q) = (. . . ,qi -
qYl
where{i,j} is an edge of G, we see thatthe matrixof df(p) has 3v columns,each tripleof
ofa vertexp,of
withrespectto thethreecoordinates
columnsarisingfromthepartialderivatives
G(p). And thematrixofdf(p)has e rows,each arisingfroman edge {i,j} of G. The {i,j} rowof
of Iqi- qj12and evaluatingatp, has onlytwotriples
df(p),obtainedby takingpartialderivatives
of nonzeroentries,
namely,2(p,-pj) in thecolumntriplecorresponding
top, and 2(pj -p,) in
topj. Thus theex 3v matrixof df(p) has theform
thecolumntriplecorresponding
vertex ...
1
. . .
O
edge {i,j}
vertex
i
2(pi-pj)
vertex
...
*. .
jI
2(pj-pi)
...
. . .
vertex
0
In lightof this,it is clear thata stressof a framework
G(p) in R3 is nothingotherthanthe
an
of a lineardependenceamongtherowsof df(p) (or, equivalently,
collectionof coefficients
elementofthekernelofdf(p)': Re->R3v, thetransposeofdf(p)).Thusa framework
G(p) in R3 iS
stressfreeifand onlyifrankdf(p)= e, thenumberof edgesof theframework.
If a set
The nextstagein our studyof staticsallowsexternalforcesto act on theframework.
one forceforeach vertex,
thenforces
ofexternalforcesis appliedto theverticesofa framework,
arisein therodsof theframework.
Can theframework
of tensionand compression
presumably
resolvethe set of externalforcesin the sense thatat each vertexthe sum of all the forces,
is zero?If so, is theresolution
internaland external,
unique?
on external
is in statics,we concentrate
Considera framework
G(p) in R3. Sinceourinterest
forcesof the followingtype. A vector F=(F1,...,Fv)ReR3V is an -equilibrium
forcefor p
=(PI,..,pV) if
v
2 F1=0 and
i-1
v
Ep,xF,=0
i-1
wherex denotesthe crossproductin R3. The firstconditionmerelysays thatthesum of the
forcesFi is zerowhilethesecondconditionsaysthatthesumof themoments(or torques)piX Fi
theoriginof theforcesF, appliedat thepointspi is zero.Together,
about anyaxis through
the
conditions
implythatthesumof themomentsaboutanyaxisis zero.On theotherhand,we say
thatF= (F1,..., Fv)E=R3V is a resolvable
forceforG(p) if thereexistscalarswy.), one foreach
edge[p,,P1]of G(p), suchthat
Fj+
E (0(iA,-}Pi-)=OforI <i
jea(i)
<v.
This conditionmeansthatthe sum of all the forcesat everyvertexis zero. In theseterms,a
stressof G(p) is a resolution
of thetrivialforceF=(0,...,0) forp.
14
B. ROTH
[January
Let & and 6Rbe thecollectionsofall equilibrium
and resolvableforcesforG(p), respectively.
Both& and 6A are subspacesof R3Vsince& is thekernelof thelinearmap L: R3v-+R6defined
by
L(F1. . . Fv)= (2 FF, piX F,)
and 6Ais theimageof thetransposedf(p)': Re-+R 3v ofdf(p). It is easyto showthat6Rc& since
F= (F1,..., Fv) EiR. meansthatF is a linearcombination
of therowsof df(p) and each rowof
theconditionsdefining
df(p) satisfies
E. That is,p, -pj andpj -p, are theonlynonzeroentries
in the {i, j} rowof df(p); so we have E:Fi=0. And :p, XFi =0 sincep, x(pi -pj) +pj X(pj pi)=O.
The dimensionof the vectorspace 6A of resolvableforcesfor G(p) is rank df(p) since
A= imagedf(p)t.IfPi,...,pv are notcoplanarin R3,then& has dimension
3v-6 sinceL maps
R3V onto R6 in thiscase. To see this,supposeP2 -P P3-p, and P4-PI span R3 and consider
vectorsx andy in R3.ThereexistscalarsAl,A2,and X3 suchthat
arbitrary
3
x=
Let F=(F1,...,Fv)=(-S,y,A1
I
i=l
/\-(A+li-Pi).
that
y,A2y,A3y,0,..
.,0). Thenit is easyto verify
L(F) = (IFi,Y pi X Fi) = (0,x Xy).
Therefore
imageL containsall vectorsof theform(0,z) forz E R3. SinceimageL also contains
all vectorsof theform(z;p1x z) forz E-R3, L is clearlyonto.
We are now in a positionto relateinfinitesimal
to theresolvability
of equilibrium
rigidity
forces.
PROPOSITION
4.3. Suppose G(p), p = (pl, ... ,pv)E-R3V, is a frameworkin R3 whereP . .. ,p,vare
notcoplanar.ThenG(p) is infinitesimally
rigidin R3 ifandonlyifeveryequilibrium
forceforp is a
resolvable
forcefor G(p). Moreover,each equilibrium
forceis uniquelyresolvableif and onlyif
G(p) is infinitesimally
rigidand stressfree.
Proof.Sincedimension63 = rankdf(p),dimensionE = 3v-6, and 6Rc &, we have E c 6Rif
and onlyif rankdf(p)= 3v-6. But the tangentspace Tpto themanifoldM is 6-dimensional,
sincethepointspl,... ,pv are notcoplanar,and thusG(p) is infinitesimally
rigidin R3 ifand only
if dimensionkerneldf(p)= 6, whichis equivalentto rankdf(p)= 3v-6. The uniquenessresult
fromthe factthatG(p) is stressfreeif and onlyif kerneldf(p)tis
followswithoutdifficulty
whichis equivalentto theuniqueresolvability
of thetrivialequilibrium
force.
trivial,
5. A RigidityPredictor.Definition3.2 and the subsequentdiscussionmake clear thatthe
or flexibility
of a framework
rigidity
G(p) in RI is determined
by the natureof theinclusion
nearp of themanifoldM of pointscongruent
top in thealgebraicsetf `(f(p)). Iff `(f(p))
happensto be a manifoldof knowndimensionnearp, thentherigidity
or flexibility
of G(p) is
This is preciselythesituationwe directour
governedby thedimensionsof thetwomanifolds.
attention
to here.
For a smoothmapf: Rna+Rm, let k = max{rankdf(x):x e RI}, themaximumof therankof
the derivativeof f. We say thatp E R' is a regularpointoff if rankdf(p)= k. The Implicit
FunctionTheoremimpliesthatf `(f(p)) is an (n - k)-dimensional
smoothmanifoldnearp
providedp is a regularpointoff (see Auslanderand MacKenzie [4, Implicit-Parametrization
Theorem,p. 32]).
Considera framework
G(p), p=(pI,... pv)e RnV,in Rn with edge functionf: Rnv-+Re.
in Rn and let M be the n(n+ 1)/2-dimensional
SupposepI, ... ,pv do not lie on a hyperplane
to p. Since thetangentspace Tpto M at p is a subsetof kernel
manifoldof pointscongruent
df(p) by thecommentprecedingDefinition
4.1,we have
rank df(p)= nv-dimension kerneldf(p) < nv- n(n+ 1)/2.
(5.1)
1981]
15
RIGID AND FLEXIBLE FRAMEWORKS
Ifp is a regularpointoff whererankdf(p) = k, thenmuchmorecan be said. For in thiscase,
f - l(f(p)) is a (nv-k)-dimensionalmanifoldnearp, and thusM andf -'(f(p)) agreenearp if
are equal. SinceG(p) is rigidin RI ifand onlyifM andf - I(f(p))
and onlyiftheirdimensions
coincidenearp,we concludethatG(p) is rigidin Rn ifand onlyifrankdf(p)=nv-n(n+ 1)/2.
whichwas introduced
by Gluck [15] and extensively
This givesthefollowing
rigidity
predictor
used in Asimowand Roth[2].
point
in R n where
p = (p ...pv) e Rv is a regular
PROPOSITION 5. 1. Let G(p) be a framework
in Rn. ThenG(p) is rigidin Rn
of theedgefunction
f andPI..-., p do notlie on a hyperplane
ifand onlyifrankdf(p)=nv-n(n+ 1)/2 and G(p) is flexiblein Rn if and onlyif rankdf(p)
<nv-n(n+ 1)/2.
One applicationof therigidity
leads to thenotionof the"generic"behaviorof an
predictor
in R . Can rigidity
in Rn be considereda property
of an abstractframework
abstractframework
in Rn? The two
G ratherthanjust a propertyof particularrealizationsof the framework
but the
in the plane shownin Fig. 5 are givenby the same abstractframework,
frameworks
(b)
(a)
FIG. 5
in (a) is flxible in R2 (itsbottomedgesfloparound)whiletheframework
in (b) is
framework
is not determined
of its bottomedges).Therefore,
rigidity
rigidin R2 (due to the collinearity
thelocationof theverticesmustalso be taken
of theframework;
solelyby theabstractstructure
in (a) is rigidin R2 while
intoaccount.Anotherexampleappearsin Fig. 6 wheretheframework
(b)
(a)
FIG. 6
are the
abstractframeworks
theframework
in (b) is flexiblein R2, eventhoughtheirunderlying
same,However,thereadermayhavenoticedthatthevertexlocationin (b) in each case is rather
thetypicalbehaviorof the
carefully
contrived
while,in some sense,(a) in each case represents
frameworkin R2.
it is usefulto observethatthesetof regularpointsoff is a
To understand
thisphenomenon,
denseopen subsetof Rn".For x is a regularpointoff ifand onlyifP(x)#O0,whereP(x) is the
ofall k x k submatrices
ofdf(x).
polynomial
givenby thesumof thesquaresofthedeterminants
Sincetherigidity
p (withaffinespan Rn) of a given
predictorsays thattheregularrealizations
16
[January
B. ROTH
framework
are eitherall rigidin Rinor all flexiblein Rip,we see thateveryabstractframework
has a typicalor genericbehaviorin Rip.In otherwords,givenan abstractframework
G, we have
eitherG(p) rigidin Rinfora denseopen setofp E RwlVor G(p) flexiblein Rinfora denseopenset
of p EHRnv.
The rigidity
predictorand thenotionof a regularpointalso serveto clarifytherelationship
betweenrigidity
and infinitesimal
rigidity.
Considera framework
G(p) in RI wherep E
is a
regularpointof theedge function
f.In thiscase,f-'(f(p)) is a manifoldnearp whosetangent
space atp is kerneldf(p).SinceM andf-'(f(p)) agreenearp ifand onlyiftheirtangentspaces
Tpand kerneldf(p) at p are equal and the latteris preciselythe definitionof infinitesimal
we concludethatG(p) is rigidin Rinif and onlyif G(p) is infinitesimally
rigidity,
-rigidin Rin.
Thus at regularpoints,rigidity
(flexibility)
and infinitesimal
rigidity
(infinitesimal
flexibility)
are
equivalent.Moreover,thefollowing
proposition
showsthatinfinitesimal
rigidity
occursonlyat
regularpoints.
HfV
PROPOSITION 5.2. SupposeG(p), p = (p1,...
in Rin and theaffine
span
.,pj) E H"R,is a framework
ofpj,. . . ,p,vis Rn. ThenG(p) is infinitesimally
rigidin Rinifand onlyifp is a regular
pointoffand
G(p) is rigidin Rip.
Proof.In lightof-theabove observations,
it sufficesto show thatif G(p) is infinitesimally
rigidin Rin,thenp is a regularpoint.The intersection
oftwodenseopensetsgivesa regularpoint
(fl@
that
such
lie
do
not
on
a
in Rin.If G(p) is infinitesimally
qv)e
lV
hyperplane
q=(q
qj,..., qv
rigidin Rin,thenTp= kerneldf(p) whichgivesn(n+ 1)/2= nv- rankdf(p).By inequality(5.1)
and thefactthatq is a pointof maximumrank,one obtains
rankdf(q) <nv- n(n+ 1)/2= rankdf(p) < rankdf(q)
whichsaysthatp is a regularpointoff.
6. Frameworks
Givenby ConvexPolyhedrain R3. Let C be a convexpolyhedron
in R3,i.e.,
theconvexhullof a finitesetof noncoplanarpointsin Hi3.A vertexof C is a pointwhichis the
intersection
of C witha supportplane of C, whilean edgeof C is a closedlinesegment
whichis
the intersection
of C witha supportplane of C. Suppose C has v verticeswithcoordinates
P
,pvEHi3.Let V { ,...,v) and
E={{i,j}:[AiPj1]isanedgeof
C).
in Hi3givenby C.
Thenwe referto G(p) whereG = (V,E) andp=(p ,... ,pv)as theframework
in R3and whichgiveflexibleframeworks
Whichconvexpolyhedrain R3giverigidframeworks
in H3?
C in Hi3.We now showthat
SupposeG(p) is theframework
givenby a convexpolyhedron
G(p) is stressfree,i.e., rankdf(p)= e wheref: R3tv--Reis theedge functionof G and e is the
numberof edgesof G. This impliesthatmax{rankdf(x):xE R3) =e and thusp is a regular
point of f. Consequently,the rigiditypredictor(Proposition5.1) with n= 3 allows one to
or flexibility
of G(p) by the simplestimaginableprocedure-justcount
therigidity
determine
theresultscan evenbe
thenumberofedgesof C and comparetheresultto 3v -6. Furthermore,
if desired,since rigidity(flexibility)
and infinitesimal
interpreted
"infinitesimally"
rigidity
amountto thesamethingat regularpoints.
(infinitesimal
flexibility)
in Cauchy'sproof[91of
The proofthatrankdf(p)= e has twoparts,bothof whichoriginate
facescongruent
and "arrangedin
thefactthattwoconvexpolyhedrain Hi3withcorresponding
thehypothesis
of Cauchy'sTheorem
thesameway" are themselves
congruent.
(More formally,
4i betweenthe sets of verticesof the two
says thatthereexistsa one-to-onecorrespondence
ifand onlyif+(S) is the
suchthatS is thesetofverticesofa faceofone polyhedron
polyhedras
themap 4,preservesdistancesbetween
set of verticesof a face of theotherand, furthermore,
verticeson corresponding
faces.)One partis of a topologicalnatureand deals withgraphson a
1981]
RIGID AND FLEXIBLE FRAMEWORKS
17
while the otheris of a geometricalnatureand relieson the convexityof the
polyhedron,
of ideas used hereis due to Alexandrov[1] and Gluck
arrangement
The particular
polyhedron.
[15],and relatedresultsappearin Dehn [14] and Weyl[23].
C and supposethereexistsa
G(p) in R3 givenby a convexpolyhedron
Considera framework
nontriviallinearcombinationof the rows of df(p) whichvanishes;say fiJJ denotes the
Summingeach columntripleofdf(p),we
of the{i,j} rowin thislinearcombination.
coefficient
findthat
jea(i)
wpj)iv(pi-pj)=O forl< i <v
(6.1)
wherea(i)= {j: [pi,,p]is an edge of G(p)}, theset of verticesadjacentto vertexi.
The signsof thecoefficients
W(i,)are now used to attachthesymbols+ and - to someof
theedgesof C. If w{ij) >0, thenthe{ij} edgeof C is marked+ whileifwfi} <0, thenthe{i,j)
edgeof C is marked-. The edge {i,j} is leftunmarkedif w(ij= 0. ConsiderthegraphG' on
thesurfaceaC of C inducedby themarkededgesof C, whichmeansthattheedgesof G' are the
edgesof C marked+ or - and theverticesof G' are theverticesof C incidentwithat leastone
edgemarked+ or -. For each vertex
pi of G', theedgesof G' incidentwithp, can be cyclically
on thesurfaceof C as thevertexPi is circledonce (say,in
orderedaccordingto theiroccurrence
directionwithrespectto an outwardnormalof aC). The indexofpi is the
a counterclockwise
in thiscycleof edgesaroundthevertexp,and theindex
of
of
number changes signencountered
I is the sumof theindicesof theverticesof G'. The topologicalpartof theproofdeals with
whereeach edge in the
graphsinducedby some subsetof the edges of a convexpolyhedron
subsetis marked+ or LEmmA6.1. The indexsatisfies
I S 4v' -8
of G'.
ofvertices
wherev' is thenumber
Proof.Let e' be the numberof edges of G' and f' the numberof regions(or topological
of edgesof G'. Letf"'be
of aC - at. Each suchregionhas a boundaryconsisting
components)
the numberof regionswithexactlyn boundaryedges wherean edge is countedtwicefora
regioniftheregionlies on bothsidesof theedge.Clearlyf = 0 andf2 is nonzeroonlywhenG'
hasjust one edge.SinceI=0 = 4v'-8 in thiscase, we assumef2= 0. Then
2e'= I
n>3
nfn and f'= 2 fn
n>3
We now computethe indexI by circlingregionsratherthan vertices.Since the numberof
theboundaryof a regionwithn edgesis an evennumberlessthan
as one traverses
sign-changes
or equal to n,we have
1 < 2f3+ 4f4+ 4f5+ 6f6+ 6f4+*<
=2 2 nfn-4
n>3
i
n>3
?
n>3
(2n- 4)fn
fn=4e'-4f'.
of thegraph
v'- e' +f' = 1 + N > 2 whereN is thenumberof components
By Euler'sformula,
G'. Therefore
I < 4e'-4f' < 4v'- 8.
of C
whichrelieson the convexity
partof the argument
Next,we presentthe geometrical
withequation(6.1).
together
LEMmA6.2. Theindexofeveryvertexof G' is greaterthanor equal tofour.
18
[January
B. ROTH
p of G' and let a'(i) = {j: [p,,pj1is an edge of G'}. By equation
Proof.Considerany vertex
(6.1),we have
j E at(i)
W
w{,2}(p,-Pj)=O
(6.2)
sincethecoefficients
wij) ofedgesof G butnotG' arezero.First,theindexofp, cannotbe zero
since the scalarswtijl forjEa'(i) are eitherall positiveor all negativein thiscase. By the
C onlyatp; say an equationof the
convexity
of C, thereexistsa plane in R3 whichintersects
planeis n *(P - x) = 0, wheren E R3 is a normalto theplane.Sinceall theverticesof G' except
PAlie on one side of theplane,n * (P -p1) is eitherpositiveforallje a'(i) or negativeforall
j E a'(i). Therefore,
I
jEa'(i)
(A{jj.s[n-(APi-j)]:#-0
whichis impossiblesince(6.2) gives
jea'(i)
wg(ij}
[ n-(pi -pj)] = n
-
(i)
W{ij (pi -pj)] = O.
showsthattheindexofPAcannotbe two,sincein thiscase thereis
a similarargument
Moreover,
a setofedgesof G' marked+ followedbya setofedgesmarked- in thecycleofedgesaround
of C, thereexistsa plane through
p, withtheedgesof G' incidentwithpi
PA.By theconvexity
marked+ on one side of theplane and thosemarked- on theotherside of theplane. If an
equationofthisplaneis n (pi - x) =0, we haven'(pi -pj) ofone signforall theedgesmarked+
and of theoppositesignforthosemarked-. Thus by (6.2)
O= n-[
j E a'(i)
i(Pi-pp)
1
=
O(i*)i
j E a'(i)
[ n-(pi--pj)] #0.
completestheproofof Lemma6.2.
This contradiction
C and
in R3 givenbya convexpolyhedron
THEOREM 6.3. Let G(p), p E R3v, be theframework
of G. Then
supposef is theedgefunction
rankdf(p)= e,
thenumber
ofedgesof C.
We
Proof.We supposethatG(p) admitsa nontrivialstressand arriveat a contradiction.
attachthesymbols+ and - to some of theedgesof C accordingto thesignsin a nontrivial
stressand let G' be thegraphinducedby themarkededgesof C. By Lemmas6.1 and 6.2 we
have
I < 4v'-8 <4v' <I
showsthatG(p) is
whereI is theindexand v' thenumberofverticesof G'. This contradiction
stressfreeand thusrankdf(p)= e.
of a framework
therigidity
arising
or flexibility
in lightof therigidity
predictor,
Therefore,
is determined
bya simplecomparisonofe and 3v-6. However,this
froma convexpolyhedron
C in R3
way. Considera convexpolyhedron
same comparisonarisesin anotherquitedifferent
e edges,and f facesof whichf, have exactlyn edges.By Euler'sformula,
withv vertices,
3v- 6 = 3(v -2) = 3(e -f) = e + (2e - 3f).
But
3f=3 1 fn< I
n>3
n>3
nfn=2e
withequalityif and onlyiff=f3, i.e., everyface of C is a triangle.Thereforee < 3v -6 with
whichleads to thefollowing
corollary.
equalityifand onlyifeveryface of C is a triangle,
1981]
19
RIGID AND FLEXIBLE FRAMEWORKS
C is rigidin R3 ifand only
COROLLARY 6.4. Theframework
G(p) givenbya convexpolyhedron
ifevery
face ofC is a triangle.
Proof.by Theorem6.3,rankdf(p)= e wheree is thenumberof edgesof C andf is theedge
functionof G. Therefore,
p = (P ,p-,) is a regularpointof f and clearlyP ,...pv are not
coplanar.By therigidity
predictor,
G(p) is rigidin R3 ifand onlyife = rankdf(p)= 3v -6. But,
as wejust observed,'
e = 3v -6 if and onlyifeveryfaceof C is a triangle.
7. Concluding
arisingfromTheorem6.3 and the
Remarks.The e = 3v -6 testforrigidity
In fact,theinaccuracies
rigidity
predictor
has certainly
not escapedtheattentionof engineers.
thatmar manyaccountsof rigiditystemfromattemptsto apply thissimpleformulato all
frameworks
in R3.It is not difficult
to findexamplesshowingthisis inappropriate.
with
For instance,
considertheframework
G(p) in R3shownin Fig. 7, whichis a tetrahedron
a triangle
in theplaneofitsbase. A simplegeometrical
argument
usingtheveryspeciallocation
of thethreeverticesin theinterior
of thebase of thetetrahedron
showsthatG(p) is rigidin R3
even thoughe < 3v -6. (Note Theorem6.3 is not applicablesince G(p) is not the framework
givenby a convexpolyhedron
in R3in thesensedefinedin Section6.) However,it is quiteeasy
to showthattheframework
G (in fact,anyframework
withe < 3v -6) is generically
flexibleor,
equivalently,
alwaysinfinitesimally
flexiblein R3.
FIG. 7
FIG. 8
withe = 3v -6 whichare notonlyflexiblebuteven
On theotherhand,thereexistframeworks
flexiblein R3.For example,theframework
G(p) shownin Fig. 8 is flexiblein R3
generically
rotatesrelativeto theotheraroundthedottedline shown,and
sinceone halfof theframework
thisis clearlythe typicalbehaviorof G in R3. However,forplanargraphswithat least three
it can be shownusingSteinitz'sTheoremthate=3v-6 is a necessaryand sufficient
vertices,
in R3. Of course,thisdoes notprecludetheexistenceof (a smallset
conditionforgenericrigidity
e = 3v-6. For instance,Bricard[8] describes
of a planargraphsatisfying
of) flexiblerealizations
all of the flexiblerealizationsof the framework
arisingfroman octahedronin R3. Since the
one can concludethattheframework
surfacesofall theseflexibleoctahedraare self-intersecting,
R3.
The
octahedron
is
in
embedded
conjecture,a versionof whichEuler
givenby any
rigid
surfacesare rigid,has recently
been settledby
proposedin 1766,thatall embeddedpolyhedral
20
B. ROTH
[January
of Connelly[10],[11],[12]. This flexiblepolyhedralsurfaceis a
an ingeniouscounterexample
collectionof triangles
joined together
along commonedgesin sucha way as to forma closed
in R3. Its remarkable
(butnotconvex)polyhedron
withoutself-intersections
property
is thatthe
and edgesof thetriangles
formsa flexibleframework
collectionofvertices
in R3.
WhileCorollary6.4 tellsus thattheframework
givenby a cube is flexiblein R3, it givesno
information
about the "braced" cube obtainedby addingnew diagonaledges acrosssome(or
are surelyof interest,
perhapsall) of the six faces of the cube. Since such frameworks
we
whicharisein variousmoregeneralwaysfrom
concludewitha shortdiscussionof frameworks
thisis notmeantto implythatframeworks
convexpolyhedrain R3. Incidentally,
arisingin one
or important
ones fromeithera
wayor anotherfromconvexpolyhedraare theonlyinteresting
mathematical
or a structural
pointof view-these are simplytheonlyframeworks
aboutwhich
muchis known.
For a framework
G(p) obtainedfroma convexpolyhedron
C by firstaddingnewverticesin
the interiorof edges of C (so each edge of C is now subdividedby edges of G(p)) and then
addingnewnoncrossing
diagonaledgesacrosseach faceof C, Alexandrov[1, Chapter10]shows
forsuch
thatG(p) is stressfree,i.e., rankdf(p) equals the numberof edges of G. Therefore,
frameworks,
G(p) is rigid(or,equivalently,
infinitesimally
rigid)in R3 ifand onlyif G(p) forms
of the surfaceof C. A versionof Alexandrov'sproofappearsin Asimowand
a triangulation
Roth[3].
If new verticesare also allowedin theinterior
of facesof C beforetheadditionof thenew
noncrossing
edgesin thefacesof C, thentheresulting
framework
mayadmitnontrivial
stresses.
A stress{wfi,j:{i,j} an edgeof G) of a framework
G(p) is called a facialstressifthereexistsa
whichdo notlie in theface.Whiteley
faceof C suchthatw1,,,= 0 forall edges[pI,p1]
[24]proves
of Alexandrov'sresultwhichstatesthat,forsuchframeworks,
a significant
generalization
every
stressis a sum of facial stresses.Relatedresultsbased on the studyof infinitesimal
flexings
ratherthanstressesappearin Connelly[13].
offacesof C beforethefaces
Finally,whathappensifnewverticesare presentin theinterior
of C are triangulated
by noncrossing
edges?In thiscase, theframework
G(p) is not infinitesiin theinterior
of a face,assignvectorswhichare perpendicular
mallyrigidin R3 (to the&vertices
verticesin orderto obtainan infinitesimal
to theface and assignzero vectorsto theremaining
stresses.
flexing),
p is not a regularpointof the edge function
f, and G(p) admitsnontrivial
thesecondderivative
of
Nevertheless,
Connelly[13]showsthatG(p) is rigidin 3 by examining
In otherwords,everytriangulation
ofedgelengthconditions.
ofthesurfaceofa
thepreservation
in R3 givesa rigidframework
in R3.
convexpolyhedron
References
Berlin,1958.
Akademie-Verlag,
KonvexePolyeder,
1. A. D. Alexandrov,
ofgraphs,
Trans.Amer.Math.Soc.,245(1978)279-289.
2. L. Asimowand B. Roth,Therigidity
The rigidityof graphsII, J. Math. Anal. Appl., 68 (1979) 171-190.
3. __
to Differentiable
Manifolds,
Dover,NewYork,1977.
and R. E. MacKenzie,Introduction
4. L. Auslander
Proc.London.Math.Soc. (2), 10(1912)309-343.
Deformable
octahedra,
5. G. T. Bennett,
frameworks
II, SIAM J.Appl.Math.,36 (1979)491-508.
Bracingrectangular
6. E. D. Bolker,
I, SIAM J.Appl.Math.,36 (1979)473-490.
frameworks
7. E. D. Bolkerand H. Crapo,Bracingrectangular
J.Math.PuresAppI.(5), 3 (1897)113-148.
surla theorie
de l'octaedre
articule,
Memoire
8. R. Bricard,
19 (1813)87-98.
J.EcolePolytechnique,
SecondMemoire,
etpolyedres,
9. A. L. Cauchy,Surles polygones
forpolyhedra,
Inst.HautesEtudesSci. Publ.
conjecture
to therigidity
A counterexample
10. R. Connelly,
Math.,47 (1978)333-338.
11.
12.
_,
,
1 (1978) 130-131.
A flexiblesphere,Math. Intelligencer,
The rigidityof polyhedralsurfaces,Math. Mag., 52 (1979) 275-283.
ofarbitrarily
and thesecondorderrigidity
triangulated
ofcertain
cabledframeworks
, Therigidity
13.
Adv.in Math.(to appear).
convexsurfaces,
Math.Ann.,77 (1916)466-473.
konvexer
Polyeder,
14. M. Dehn,UberderStarrheit
1981]
PLUCKER EQUATIONS FOR CURVES
21
Topology,
LectureNotesin
arerigid,
in Geometric
connected
closedsurfaces
15. H. Gluck,Almostall simply
Berlin,1975,pp. 225-239.
Math.,no.438,Springer-Verlag,
mimeographed
notes,Univ.ofWashingLectures
in lostmathematics,
andG. C. Shephard,
16. B. Grunbaum
ton,Seattle.
Chelsea,NewYork,1952.
Geometry
and theImagination,
17. D. Hilbert
and S. Cohn-Vossen,
oflinkages,
TohokuMath.J.,37 (1933)294-319.
Bibliography
on thetheory
18. R. Kanayama,
Proc.London
method
planecurvesofthenthdegreebylinkwork,
ofdescribing
19. A. B. Kempe,On a general
Math.Soc.,7 (1876)213-216.
dansE3, d'apresRobertConnelly,
Bourbaki,
514
Semminaire
flexibles
20. N. H. Kuiper,Spherespolyedriques
(Feb. 1978)1-22.
J.Engrg.Math.,4 (1970)331-340.
ofplaneskeletal
structures,
21. G. Laman,On graphsandrigidity
no.61,Princeton
Ann.ofMath.Studies,
University
PointsofComplexHypersurfaces,
22. J.Milnor,
Singular
Press,Princeton,
N. J.,1968.
S.-B. Preuss.Akad.Wiss.(1917)
der Eiflachenund konvexen
Polyeder,
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250-266.
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Infinitesimally
rigidpolyhedra,
24. W. Whiteley,
PLUCKER EQUATIONS FOR CURVES
J.R. QUINE
FL 32306
Tallahassee,
TheFloridaStateUniversity,
ofMathematics,
Department
For a smoothclosed curvein the plane or the projectiveplane,the most
1. Introduction.
points.There are several
are cusps and self-intersection
evidenttopologicalcharacteristics
thenumberand natureof these.We willcall thesePluckerequations,after
equationsdescribing
theones givenin thelastcenturyby Pluckerforalgebraiccurvesin complexprojective2-space.
we believethat,by looking
originalequationsarepartofalgebraicgeometry,
AlthoughPliicker's
space,one can understand
at theanalogoussituationforsmoothclosedcurvesin realprojective
in real projective
The
situation
is
thatthe contentof Pliicker'sequations mainlytopological.
algebraiccurves.We
pointforunderstanding
space is easyto pictureand makesa good starting
will begin thispaper withthe simplestPluckertheoremfor curvesin the plane, called the
and end withtheclassicalPluckerequations.
Umlaufsatz,
Plucker
One featurethatwe would like to emphasizeis use of the dual correspondence.
thenumberof inflection
equationscan be appliedto thedual curvegivingequationsinvolving
pointsand double tangentsof the originalcurve.Our firstgoal will be to writea Pliicker
equationforreal curveswhichcan be appliedto the dual curveand whichwill be a natural
between
thedual relationship
of theUmlaufsatz.Thiswillinvolveunderstanding
generalization
is proved,we showthat
number."Once sucha theorem
"windingnumber"and "tangentturning
theclassicalPluckerequationscan be provedin exactlythesame way,withthehelp of some
withtechniquesof global analysiscan
global analysis.We hope that the readerunfamiliar
howtheyare a naturalextensionof theconceptof thedegreeof a map.
understand
of as theset of linesthrough
The realprojectiveplane RP2 can be thought
2. Preliminaries.
ofF. W. Gehring.
He
in 1971underthedirection
ofMichigan
hisPh.D.at theUniversity
J.R. Quinereceived
Institute.
Since1971,he hasbeenon themathematics
atTuskegee
academicyearteaching
hadspentthepreceding
is basedon a series
thisarticle
an associateprofessor;
wherehe is currently
oftheFloridaStateUniversity,
faculty
between
complex
in exploring
connections
He is interested
seminar.
oftalksgiventherein theadvancedtopology
and differential
topology.-Editors
geometry,
algebraic
analysis,