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From Foucault’s Pendulum to the Gauss–Bonnet Theorem Orlin Stoytchev∗ Abstract Wepresentaself-containedproofoftheGauss-Bonnettheoremfortwo-dimensionalsurfaces embedded in R3 using just classical vector calculus. The exposition should be accessible to ad- 7 vanced undergraduate and non-expert graduate students. It may be viewed as an illustration and 1 exercise in multivariate calculus and a motivation to go deeper into the fields of geometry and 0 2 topology. n a J 1 Introduction 6 ] The Gauss–Bonnet theorem states that the total curvature of a closed two-dimensional oriented O surface (i.e., the integral of the Gaussian curvature over that surface) is equal to the Euler char- H acteristic of the surface multiplied by 2π. This is a beautiful result relating a geometric quantity . h – the curvature – to a purely topological one – the Euler characteristic. Given the right intuition t a about geodesics and parallel transport, one can prove the Gauss-Bonnet theorem for embedded m surfaceswithlittlemorethanvectorcalculusanddefinitelywithoutheavydifferential-geometric [ machinery. 1 WhenLéonFoucaultbuilthisfamouspendulumin1851inParis(firstintheParisObservatory, v movedalittlelatertothePanthéon), hehardlyhadinmindanydeepconnectionswithgeometry 6 andtopology. Hisaimwas,ofcourse,todemonstratebyadirectphysicalexperimenttherotation 6 6 of the earth about its axis. When the earth makes one full rotation relative to the stars (which 1 happensinapproximately23hoursand56minutesandiscalledsiderealday),theplaneinwhich 0 thependuluminParisisswinging,rotatesrelativetothegroundby271.1◦clockwise. Apendulum . 1 at the north pole would rotate by exactly 360◦ while at the equator there will be no rotation. In 0 7 general the angle of rotation is given by 360◦sinφ, where φ is the geographic latitude. The 1 explanation behind this formula is that the earth is curved and the normal vector to the surface : v at some point traces a cone as this point traces a circle on the sphere. At the same time the i tangent vector giving the direction of swinging of the pendulum undergoes a parallel transport X (tobedefinedlater)alongthecircleastherearenoforcestocauseanyrotationaroundthenormal r a vector. InasenseFoucault’spendulumtellsusnotonlythattheearthrotates,butthatitiscurved (incaseweknewGauss-Bonnet’stheorembutweren’tsureoftheearth’sshape). Thesamephenomenoncanbeviewedslightlydifferently. LetC denotethecirclecoinciding withoneofthegeographicparallelsonthesphere,atlatitudeφ,withcounterclockwiseorientation. WhenoneperformsaparalleltransportalongC ofatangentvector,thevectoringeneralwillbe rotatingrelativetoC,sinceC isnotageodesic(alargecircle),unlessitistheequator. Thevector will fail to return to its original orientation when coming to the initial point. The angle between the initial and the final vectors is sometimes called deficit angle (or angular deficit or angular defect). Forthecaseathandthecalculationgives2π(1−sinφ). Ifonecalculatestheareaonthe sphere,boundedbyC,onefindsoutthatitisgivenby2π(1−sinφ)R2,whereRistheradiusof ∗AmericanUniversityinBulgaria,2700Blagoevgrad,Bulgaria 1 thesphere. ThisisaveryspecialcaseoftheGauss-BonnetformulaandthebehaviorofFoucault’s pendulumdemonstratesitsvalidity. Inthenextsectionwewilldefineinasimpleandratherintuitivewaythenotionsofageodesic andparalleltransport. WewillderivetheformulaforthedeficitanglealongthecurveC asabove. Then, using appropriate technique for calculation, we will obtain the Gauss-Bonnet formula for an arbitrary closed simple curve on the sphere. Not surprisingly this derivation invokes Stokes’ theorem. Thegeneralcase-arbitraryclosedorientedtwo-dimensionalsurfaceembeddedinR3 is treatedinthelastsection. ThekeyistoconsidercarefullytheGaussmapgivenbytheunitnormal vectortothesurface,whichsendseachpointonthesurfacetoapointontheunitsphere. Itturns out that the general case is reduced to the case of the sphere by a simple change of variables formula. The Gauss curvature on the surface appears in this setting as the Jacobian of the Gauss map. 2 Geodesics, parallel transport, flat and curved surfaces Considerasmoothtwo-dimensionalsurfaceσ embeddedinR3. Intuitivelyageodesiccurveonσ is a smooth curve C ⊂ σ, such that if you travel along it with constant speed, at any given point therewillbenocomponentoftheaccelerationinthetangentplanetothesurface. Ifweimagine rolling a ball on the surface (e.g., some adhesive force causes the ball to stick to the surface but doesnotrestrictitinanyotherway)andwedothisinweightlessness,theballwilltracepreciselya geodesic. Iftherewasnorestriction,thegeodesicwouldbeastraightlineinR3. Theconditionto stayonthesurfacemakesthetrajectorycurvedingeneral,butinsuchawaythattheacceleration stays normal to the surface, since the only force is the normal adhesive force. Suppose C is a smooth curve on σ and let c(t) be a smooth parametrization of C, such that the corresponding velocity vector v(t) := dc has constant norm. (In fact if we parametrize C by arc-length, the dt normofv(t)willbeone.) Leta(t) := dv betheaccelerationanddenotebya (t)itsprojection dt T inthetangentplane(atthepointc(t)). SobydefinitionC iscalledageodesicifa (t) = 0, ∀t. T Ifweconsiderasanexampleacircleonasphereandimagineapointmovingwithconstantspeed along this circle, the acceleration is obviously a vector pointing towards the center of the circle. Unlessthecircleisalargecircle(liketheequator)itscenterdoesnotcoincidewiththecenterof the sphere and the acceleration has nonzero tangential component. In fact the only geodesics on thesphereare(partsof)largecircles. Definingparalleltransportalongageodesicissimple. Ifyouhaveafamilyoftangentvectors w(t) defined for each point c(t) of the geodesic C and (cid:107)w(t)(cid:107) = const., then we say that this family of vectors has been obtained by parallel transport along C of one vector, say w(0) if the angle between w(t) and the geodesic stays constant. More precisely we should have constant anglebetweenw(t)andv(t). Obviouslythevelocityvectorsv(t)alongC constitutethesimplest example of parallel transport. When the curve C is not a geodesic we may use the intuition coming from Foucault’s pendulum. Let w(0) be a unit tangent vector at the initial point c(0) giving the direction of swinging of the pendulum. When we move the latter along C, there will be no rotation of the pendulum around the normal vector. In this way we obtain a unit vector w(t) for t and we should say that this has been obtained from w(0) by parallel transport along C. Sotheconditionissimilartotheoneweimposedforthevelocityv(t)whenweweredefining geodesics, i.e. (dw) = 0, orinwords, therateofchangeofw(t)hasnotangentialcomponent. dt T Now, because C is no longer a geodesic, the angle between w(t) and C will be changing. The rate of change of this angle, denoted further by ω(t), has magnitude equal to the rate of rotation ofv(t)inthetangentplaneandoppositesign. Thereforewehave: (cid:107)a (t)(cid:107) T |ω(t)| = . (cid:107)v(t)(cid:107) 2 Let’s take the closed curve C to be the geographic parallel at geographic latitude φ (with coun- terclockwise direction) and calculate the total angle of rotation (relative to C) of a vector w transported parallel to itself. We can parametrize C with the path c(t) = cos2πtcosφ i + sin2πtcosφj+sinφk,t ∈ [0,1]. Straightforwardcalculationgivesa(t) = −4π2cosφ(cos2πti+ sin2πtj)anda (t) = −4π2cosφ(cos2πt sin2φi+sin2πt sin2φj−sinφcosφk). Finally T we obtain |ω(t)| = 2π|sinφ| and therefore the total angle of rotation of w when coming to the initialpoint,whichisobtainedbyintegratingω(t)between0and1(andfiguringthecorrectsign) isequalto−2πsinφ.Thisimpliesthattheanglebetweenw(0)andw(1),i.e. thedeficitangle, isgivenby Ω = 2π(1−sinφ). (1) The same result can be obtained (see, e.g. [1]) using a simple geometric approach by con- sidering the circular cone tangential to the sphere along the curve C (Fig. 1). (When C is the equatortheconedegeneratesintoa cylinder, i.e. aconewithvertex(apex) atinfinity.) Sincethe notions of geodesics and parallel transport along C depend only on the tangent planes along C and these are common for both surfaces, a curve will be a geodesic on the sphere if and only if it is a geodesic on the cone and parallel transport gives the same result for both surfaces. This is a general property for any two surfaces touching each other along a curve. Suppose now that wecuttheconealongsomeline, fromtheboundarytotheapexandlayitflatontheplane. This process is called "developing" the surface and surfaces allowing this are called developable. We willgiveadefinitionofthelattertermbutintuitivelythesearesurfaceswhichcanbeobtainedby gluing patches cut from sheets of paper. It is clear that the process of developing a surface does Ω Ω Figure1: Paralleltransportalongacircleonthesphere,usingatangentialcone not influence parallel transport - a spacial curve C transforms into a plane curve C(cid:48), the normal componentofa(t)alongC turnsintozerounderthistransformationwhilethetangentialcompo- nent of a(t) remains unchanged. The developed cone becomes a disc with a cut-out sector with angle Ω, hence the name deficit angle. Parallel transport of a vector w along the curve C(cid:48) is the usualparalleltransportintheplane. Theanglebetweenw(0)andw(1)isΩ. Arelativelysimple exercise in elementary geometry allows us to calculate this angle and obtain the result stated in Equation(1). Theideaofthedeficitanglehasfar-reachingramifications. Adevelopablesmoothsurfacelike the cone with its apex removed is very similar to the plane in that it can be laid flat on the plane andifyoutransportavectorparalleltoitselfalongacontractiblecurveC onthissurfaceitcomes backtoitself. WecallsuchsurfacesGaussian-flat,orintrinsicallyflat. Bycontrast,nopieceofthe sphere can be laid flat on the plane and if you perform a parallel transport of a vector along any (non-constant)simpleclosedcurveonthesphere,thereturningvectorwillgenerallyhavedifferent directionfromtheinitial. WesaythatsuchsurfaceshavenonzeroGaussiancurvature. Theprecise definitionofGaussiancurvaturewillbegivenlaterbutitissomelocalquantityK definedateach pointofasmoothsurfaceandmeasuringthedeviationofthesurfacefrombeingdevelopable. For the sphere it is clear that K should be the same at each point since the neighborhood of every point looks exactly the same as the neighborhood of every other. The essential part of the proof 3 oftheGauss-BonnettheoremistoshowthatifC isasimple,positivelyorientedandcontractible curve on a surface, then the deficit angle, i.e. the angle between the initial vector w(0) and the final vector w(1) is given by the integral of K over the part of the surface surrounded by C. In otherwordswehave (cid:90)(cid:90) Ω = KdA, (2) σ σ being the part of the surface surrounded by C. This formula, by the way, immediately shows that for a sphere with radius R we must have K ∝ 1 . Coming back to our argument when R2 calculating the deficit angle along the geographic parallel C we may view replacing the upper partofthespherebytheconetouchingitalongC asthelimitofaprocesswherewereplacethe upperpartofthespherebyaconicalsurfacewithitsapexcutandcappedbyasmallerandsmaller sphericalpart. SincetheconicalpartisGaussian-flatthecurvatureisconcentratedatthecapand in the limit it becomes infinite but in such a way that the integral remains equal to Ω. In other wordswemayviewourconeasacurvedsurfaceifwesaythatthecurvatureiszeroeverywhere exceptattheapexwhereithasaδ-function-likesingularity. TakeanarbitraryclosedsurfaceS andchooseatriangulationofS,i.e. coveritbycurvilinear triangles. LetV,E andF denotethenumberofvertices,edgesandtriangles,respectively. Then theEulercharacteristicχ(S) := V −E+T isindependentofthechoiceoftriangulationandisa topologicalinvariant. Nowreplacethecurvededgesbystraightonesandthecurvilineartriangles byflattriangles. WegetapolyhedronS(cid:48) havingthesameEulercharacteristic. Ifwecalculatethe deficitangleatoneoftheverticeswehave (cid:88)ni Ω = 2π− γ , i ik k=1 where n is the number of triangles meeting at the vertex v . Note that the sum of the angles at i i somevertexmayexceed2π andthedeficitanglewillbenegativeinthiscase. Thecorresponding surface near this vertex will look like an (uncomfortable edgy) saddle, possibly with multiple "ridges"and"troughs". Summingoverallverticeswegetthetotaldeficitangle: (cid:88)V (cid:88)V (cid:88)ni Ω = Ω = 2πV − γ . i ik i=1 i=1 k=1 Thedoublesumontherightisinfactthesumofallanglesofalltrianglesofourpolyhedronand thereforewecanwrite F Ω = 2π(V − ) = 2π(V −E +F) = 2πχ(S(cid:48)). (3) 2 (Weusethefactthateachtrianglecontainsthreeedgesandeachedgeiscommontotwotriangles andthusE = 3F/2.) TheresultinEquation(3)isthecontentofEuler’stheorem,namelythatthe totaldeficitangleforanarbitrarypolyhedronis2π timestheEulercharacteristic. Note: WeprovedEquation(3)fortriangleswhilethefacesofapolyhedronareusuallyarbitrary polygons and the number F in the definition of the Euler characteristic is the number of faces. This, however does not cause any difficulty as each polygon can be broken down into triangles by adding some edges. As the reader may easily check this process does not change the Euler characteristic. Thus the theorem of Euler about polygons may be viewed as a discrete analog of Gauss- Bonnet’s theorem if we think of the total deficit angle as the integral of the curvature over the surfaceofthepolyhedronandthecurvatureis"concentrated"atthevertices. Let now C be an arbitrary simple (i.e., having no self-intersections) smooth closed curve on theunitsphere. WeaimtodemonstratethattheformulainEquation(2)isstillvalidinthiscase. Without loss of generality we may assume that C does not pass through the north pole of the 4 sphere. We approximate C by a piecewise smooth curve C(cid:48) consisting of pieces of meridians (which are geodesics) and pieces of parallels (which are not geodesics). Since during parallel transport along a geodesic a vector w preserves its angle relative to it, it is clear that we must sumthecontributionstothedeficitangleΩ(cid:48) fromthemotionalongthepiecesofparallels. From our previous calculationwe have that the contribution along a piece of a parallelat a geographic latitude φ, corresponding to azimuthal change ∆θ is equal to (1 − sinφ)∆θ = (cosφ)−1(1 − sinφ)∆s,where∆sisthecorrespondingarclength. Inotherwordswehave N N Ω(cid:48) = (cid:88)(1−sinφ )∆θ = (cid:88) 1−sinφi∆s , i i i cosφ i i=1 i=1 which is a Riemann sum for the line integral of a suitable vector field F along C. We need a vectorfieldwhichpicksonlythepartsofC(cid:48) alongparallels,soFmustbealongparallels. Aunit (cid:112) vectorfieldinR3 withthispropertyisgivenby 1(−yi+xj),wherer = x2+y2. Takinginto r (cid:112) accountthatonthespherecosφ = r = x2+y2,weseethatavectorfieldthatdoesthejobis (cid:112) 1− 1−x2−y2 F = (−yi+xj). x2+y2 Thereforeweobtain,applyingStokes’theoremanddenotingbyσ thesurfacesurroundedbyC: (cid:90) (cid:90)(cid:90) (cid:90)(cid:90) Ω = F·ds = curlF·dA = curlF·NdA. (4) C σ σ Aroutinecalculationshowsthat 1 curlF = k (cid:112) 1−x2−y2 andforapointonthespherewithsphericalcoordinates(θ,φ)wecanwrite 1 curlF·N = k·N = 1. sinφ ThusEquation(4)reducesto (cid:90)(cid:90) Ω = dA, (5) σ i.e.,thedeficitangleonthesphereisgivenbytheareaofthesurfacesurroundedbyC. Remark 1 We chose to present the somewhat clumsy derivation above, since it assumes just familiaritywithclassicalvectorcalculus. Thesameresultcanbeobtainedusingdifferentialforms. Namely, if we introduce the one-form, which in spherical coordinates is given by α = (1 − sinφ)dθ,itiseasytoseethat (cid:90) Ω = α. C ApplyingthegeneralizedStokes’theoremforforms,wehave (cid:90) (cid:90)(cid:90) (cid:90)(cid:90) (cid:90)(cid:90) (cid:90)(cid:90) (cid:90)(cid:90) Ω = α = dα = −cosφdφ∧dθ = cosφdθ∧dφ = cosφdθdφ = dA. C σ σ σ σ σ (Theareaelementonthesphereinsphericalcoordinatesis,ofcourse,dA = cosφdθdφ. Strictly speaking, the one-form α may seem not to be defined at the north pole and indeed the spherical coordinates(θ,φ)don’tprovidealocalchart,butnoticethattheformbecomes0atthispointand thusαisinfactwell-defined.) Remark2ThesameconclusionremainsvalidifweallowC tobepiecewise-smoothcurve,i.e.,a curvilinearpolygon. Itisobvioushowtodoparalleltransportofwacrossavertexofthepolygon - the angle between w and C jumps to a new value, the change being equal to minus the angle betweenthepositivedirectionsofC beforeandafterthevertex. 5 3 Gaussian curvature. Gauss-Bonnet theorem for arbitrary closed surfaces LetnowS beanarbitraryorientedsmoothsurfaceembeddedinR3. TheGaussmapG : S → S2 isdefinedasfollows-foreachpointonS taketheunitnormalvectortothesurfaceatthispoint andidentifythelatterwiththecorrespondingpointontheunitsphereS2. ClearlyGisasmooth map which is not one-to-one. A contractible simple closed curve C ⊂ S will be mapped to a closed curve C(cid:48) ⊂ S2 which can have self-intersections or even degenerate to a point. (See Fig. 2for anillustration.) ThemapGdoesnot preserve ingeneralthe orientation ofacurve. Infact, as Fig. 2 suggests, a positively oriented closed curve on the interior half of the torus, where the curvatureisnegative,ismappedbyGtoanegativelyorientedcurveonS2. Gaussmap Figure2: TheeffectoftheGaussmapforacurveonthetorus The main result in this section will be a proof of Equation (2) with K being the Jacobian of theGaussmapG. Asafirststepweshowthefollowing Proposition 1. With the same notations as above, the deficit angle Ω along C is equal to the deficitangleΩ(cid:48) alongC(cid:48). Proof. Let c(t), t ∈ [0,1] be a parametrization of C. We will approximate the surface S in a neighborhood of C by a Gaussian-flat strip in the following way: Divide [0,1] into n equal subintervalsandlett betheendpointoftheithinterval. DenotebyN theunitnormalvectorat i i thepointc(t ). UsingthesphericalanglesθandφwehaveN = cosθ cosφ i+sinθ cosφ j+ i i i i i i sinφ k. (Notethatweusethelesscommondefinitionforφastheanglebetweenthehorizontal i planeandthevector. Theso-calledpolarangle,i.e. theanglebetweenthez-axisandthevector,is π−φ.) Ateachpointc(t )take(arectangularpieceof)thetangentplaneP . Considernowtwo i i N1 N1' N N ' 1 1 P1 P1' Gauss map N 2 P N 2 2 Figure3: ApproximatingthesurfaceS inaneighborhoodofC byaGaussian-flatstrip consecutiveplanesP andP . Ifφ = φ weconnectthembyaconicalsurfacedefinedinthe i i+1 i i+1 obviousway. Inthegenericcasewhenφ (cid:54)= φ wetakeanauxiliaryplaneP(cid:48),perpendicularto i i+1 i theauxiliaryvectorN(cid:48) = cosθ cosφ i+sinθ cosφ j+sinφ kandthenconnectthelatter i i+1 i i+1 i i toP usingacylindricalsurface(seeFig. 3). (TheauxiliaryplaneP(cid:48) isnotnecessarilytangent i+1 i to S.) We close the strip by connecting P to P . In this way we obtain a differentiable surface n 1 -thestripS(cid:48),whichwillbedevelopableandwillbetangenttoS atthepointsc(t ). Thesmooth i closedcurveC ⊂ S canbeapproximatedbyaclosedpiecewise-smoothcurveC ⊂ S(cid:48),e.g.,by n 6 takingaconstant-φcurvefromc(t )totheauxiliarypointc(cid:48)(t )andthenaconstant-θcurvefrom i i c(cid:48)(t )toc(t ). Whenweperformaparalleltransportofavectorw alongC ⊂ S(cid:48), thedeficit i i+1 n angleΩ ,i.e.,theanglebetweenw(0)andw(1),dependsonlyonthestripitself. Namely,ifwe n cutthestripalongalineandlayitflat,Ω istheanglebetweenthefinalandthebeginningedges n ofthecut(Fig. 4). Thesituationisexactlythesameaswhenconsideringtheconetangenttothe spherealonga geographicparallelatangleφ. Wehavetosum upthecontributionsto thedeficit angleofalltheconicalpieces. Thuswehave n (cid:88) Ω = (1−sinφ )∆θ , n i i i=1 where ∆θ = θ − θ . But the same deficit angle will be obtained for the piecewise-smooth i i+1 i curve C(cid:48) ⊂ S2 which is obtained by connecting each point N on the sphere to the next point n i N byfirstmovingalongtheparallel(constantφ),thenalongthemeridian(constantθ). Taking i+1 the limit n → ∞, the deficit angle Ω will approach Ω along C ⊂ S and at the same time will n approachΩ(cid:48) alongC(cid:48) ⊂ S2. Thiscompletestheproof. Ω a a Figure4: AGaussian-flatapproximatingstrip,cutandlaidflat IftheGaussmapGwasone-to-one,Equation(2)wouldfollowimmediatelyfromProposition 1 and Equation (5) by a simple change of variables. Namely, if you take a simple closed con- tractiblecurveC ⊂ S andσ ⊂ S isthesurfacesurroundedbyit,Gwillmapσ toσ(cid:48) ⊂ S2 andC to C(cid:48) ⊂ S2, which will be the boundary of σ(cid:48). Then, by Proposition 1, the deficit angle Ω along C is equal to the deficit angle Ω(cid:48) along C(cid:48), which by Equation (5) is the surface integral over σ(cid:48) ofthefunction1andthiswouldbeequaltotheintegraloverσ oftheJacobianofG. Thusifwe setK,theGaussiancurvature,tobetheJacobianoftheGaussmap(insomelocalcoordinateson S and S2) we obtain exactly Equation (2). Note that G reverses the orientation if and only if K isnegativeandthecorrectapplicationofthechangeofvariablesrequiresthatwetakeK andnot |K|astheintegrand. The Inverse Function Theorem applied to the map G says that it is one-to-one in a (small enough)neighborhoodofanypointforwhichK (cid:54)= 0.Theinteriorofthesurfaceσwillbeaunion of, possibly countably many, open simply-connected sets where K > 0, open simply-connected sets where K < 0, and closed sets where K = 0. The boundaries of these open sets will be (piecewisesmooth)curveswhichwecanorientpositively. ThedeficitangleΩalonganoriented curveC hasthepropertiesofalineintegral–ifyousplitthecurveintopiecesthetotalangleisthe sumofthecontributionsalongthepiecesandifyoureversetheorientationofC thedeficitangle changesitssign. Thusbyastandardtechnique,wereplacethedeficitanglecalculatedalongC by asumofanglescalculatedalongboundariesofinteriorregions. Thecontributionofcurvesinthe 7 interiorofσwillcancelsinceeachsuchcurveparticipatestwicewithoppositeorientation. Wesee that it is enough to consider curves C which surround regions where K > 0 everywhere except possibly on the boundary, or K < 0 everywhere except possibly on the boundary, or regions (if there are such) where K = 0. Notice that this can be performed one step at a time (see Fig. 5 foranillustration)–wereplacetheoriginalcurveC bytwoclosedcurvesC andC , whereC 1 2 1 surrounds a single region as above and C surrounds all the rest. The deficit angle will be given 2 byapossiblyinfiniteconvergentsumofcontributionsforwhichEquation(2)holds. C 1 C 2 K>0 K<0 K=0 C Figure5: BreakingtheregionsurroundedbyC intoregionsofdefinitesignofthecurvature Before we proceed, it is helpful to derive a convenient (and quite familiar) explicit expres- sion for K as the Jacobian of G. Recall that the Jacobian of a differentiable map from a (two- dimensional)manifoldtoanothermanifoldofthesamedimensioncanbeviewedasthedetermi- nantofthematrix,correspondingtothedifferentialofthatmap. Wecanwritesymbolically K = det(dG) The differential is a linear map from the tangent space, at some point, of the first manifold, to the tangent space at the image point, of the second manifold. Namely, for a tangent vector v at somepointx ∈ S wechooseasmoothpathc(t) ∈ S withc(0) = x and dc(0) = v. Thenthe 0 0 dt differentialofGatx isdefinedby 0 d dG (v) := (G◦c)(0) x0 dt andtheimageisatangentvectortoS2atthepointG(x ). TheGaussmapGisspecificinthatthe 0 two tangent spaces literally coincide - the vectors, tangent to S at some point x are orthogonal 0 (as vectors in R3) to the normal vector N(x ) and therefore they are tangent to the sphere S2 at 0 thepointN(x ). Thuswegetalinearmap(operator)fromR2 intoitself: 0 W v := −dG (v) , x0 x0 calledtheshapeoperatororWeingartenmap. (Theminussignisaquestionofconvention.) The correspondingbilinearformonthetangentspaceatx ,definedbytheformula 0 (v,w) := (W v)·w II x0 isknownasthesecondfundamentalformofthesurface. Thisformissymmetric(orequivalently, the shape operator is self-adjoint). It is enough to check symmetry for vectors forming a basis. Letr : D ⊂ R2 → S bealocalparametrizationofS andlet(u,v)bethelocalcoordinates. The two tangent vectors r := ∂r and r := ∂r are linearly independent and therefore give a (not u ∂u v ∂v 8 necessarilyorthonormal)basisforthetangentspaceatanypoint(u ,v ). Theunitnormalvector 0 0 atthispointis(supressingfurtherinthenotationsthedependenceonthepoint) r ×r u v N = (cid:107)r ×r (cid:107) u v and, using the definition of the differential and properties of the triple product of vectors, we calculate ∂N (r ,r ) = (W r )·r = −(dG(r ))·r = − ·r = r ·N = (r ,r ) , u v II u v u v ∂u v uv v u II where r := ∂ru = ∂2r = r . Analogously we have (r ,r ) = r ·N and (r ,r ) = uv ∂v ∂u∂v vu u u II uu v v II r ·N. Weobtainasymmetric2×2matrix,expressingthesecondfundamentalforminthebasis vv {r ,r }. Ifthebasisisorthonormalthematrixwillcoincidewiththematrixcorrespondingtothe u v operator W and the Gaussian curvature K will be given by its determinant.The two eigenvalues k andk arecalledtheprincipalcurvaturesatthepoint. Theygivethecurvaturesofthecurves 1 2 onthesurfacealongthetwoeigenvectorsandmeasuretherateofrotationofaunitnormalvector alongeitherofthesecurves. In general, let {e ,e } be an orthonormal basis in the tangent space and let A be the matrix 1 2 with entries a = (e ,e ) . Let r = r and r = r and B be the matrix with entries b = ij i j II 1 u 2 v (cid:80) ij (r ,r ) . Wehaveb = r ·N, b = b = r ·Nandb = r ·N. Writingr = r e , i j II 11 uu 12 21 uv 22 vv i i ij j thecomponentsr formamatrixR. Usingbilinearityofthesecondfundamentalform,weobtain ij B = RARt , detB = (detR)2detA, g := r ·r = (RRt) . ij i j ij Thesymmetricmatrixgwithentriesg istheRiemanneanmetric,inthelocalcoordinates(u,v), ij inducedbytheEuclideanmetricinR3. Weconcludethat detB detB b b −(b )2 11 22 12 K = detA = = = . (detR)2 detg g g −(g )2 11 22 12 Proposition2. TheimageundertheGaussmapGofanyclosedsetwhereK = 0isaclosedset inS2 whichhasnointerior. Proof. It is enough to consider a finite simply-connected surface σ with boundary, having everywhere Gaussian curvature K = 0. The points of σ where both principal curvatures are zero are called flat points. These are precisely the points where the so-called mean curvature K := k + k becomes zero. They form a closed subset of σ, which we denote by U. Its m 1 2 complement Uc is an open subset of σ, consisting of the non-flat points. Through each point of Uc passes a unique line, which must extend to the boundary of σ in each direction, as we shall show (see. e.g., [2]). Indeed, for each non-flat point we must have either k = 0, k (cid:54)= 0 or 1 2 k = 0, k (cid:54)= 0 and we can put on Uc a smooth vector field given by the eigenvectors in the 2 1 "flat" direction. The integral curve C of this vector field is the desired line. Notice that from the definitionoftheGaussmapthismeansthatthenormalvectorNisconstantalongC andtherefore thetangentplanesatallitspointscoincide,i.e.,thereisacommontangentplanetouchingσalong the whole curve C. For example a plane can touch the torus along two circles - one on top, the otheratthebottom. Thedifferencebetweenthisandoursituationisthatthecurvatureofthetorus is zero only along these two circles. To see that in our case C is actually a line, choose a nearby integral curve C(cid:48). The tangent planes at C and C(cid:48) are definitely different if we choose C(cid:48) to be "close enough" but different from C. These two planes intersect along a line. In the limit, when C(cid:48) approachesC,thislinewillapproachbothC andC(cid:48). ThereexistsonUc asecondvectorfield, orthogonal to the one above - at each point of Uc choose (continuously) a unit tangent vector givenbythesecondeigenvector(inthenon-flatdirection)oftheshapeoperatorW. Theintegral 9 curves of this second vector field are orthogonal to the lines constructed earlier. For an arbitrary point x ∈ Uc construct an open "trapeze" around it by first taking the path along the non-flat 0 direction l(t), t ∈ (−(cid:15),(cid:15)) with l(0) = x , (t being the arc length), then for each l(t) taking the 0 unique(parametrizedbyarclengths)linel(s,t),s ∈ (−δ,δ)withl(0,t) = l(t). Byvaryingtwe obtain a family of lines in Uc which have the property that each of them has a common tangent plane. Thisimpliesthatforanyfixedtthevelocityvectorsv(s,t) := dl(s,t)areparallelforall dt s. Further, these velocities must be a linear function of s or otherwise they would not produce a family of lines (See Figure. 6). (The coefficients of this linear function depend in general on t.) Asaconsequencethearclengthofthecurvetracedbyl(s,t),t ∈ [t ,t ],sfixed,isalinearfunc- 1 2 tionofs. Indeed,wecanwritev(s,t) = v (t)(α(t)+sβ(t))andassumethatα(t)+sβ(t) ≥ 0. 0 Then Figure6: AconicalsurfaceisagenericsurfacewithK = 0 (cid:90) t2 (cid:90) t2 (cid:90) t2 L (t ,t ) = (cid:107)v(s,t)(cid:107)dt = α(t)(cid:107)v (t)(cid:107)dt+s β(t)(cid:107)v (t)(cid:107)dt. s 1 2 0 0 t1 t1 t1 LetI beanopenintervalonwhichβ(t) (cid:54)= 0. Thenforanytwot ,t ∈ I thearclengthL (t ,t ) 1 2 s 1 2 is a non-constant linear function of s and therefore the extension of the two lines l(s,t ) and 1 l(s,t ) must intersect for some s. In fact if we take three such lines (the extensions of) any two 2 of them must intersect which is only possible if all of them intersect at a common point and we haveaconicalsurface. IfwehaveaclosedintervalJ onwhichβ(t) = 0thenforanyt ,t ∈ J 3 4 thecorrespondinglinesl(s,t )andl(s,t )willbeparallelandwegetacylindricalsurface. Now 3 4 wecanshowanimportantclassicalresult([2]). Lemma1. Thereciprocalofthenonzeroprincipalcurvatureisalinearfunctionofthearclength alonganylineinUc. Proof. LetC bealineinUc andx ∈ C anarbitrarypoint. Constructanopen"trapeze"asabove 0 parametrizedbyl(s,t),sothatl(s,0)tracesC andl(0,0) = x . Notethattheparametersisarc 0 length but the parameter t is arc length only for s = 0. Thus, according to the property shown earlieraboutlinearityofthearclengthwithrespecttos,if(s,τ)isanotherparametrizationwithτ -arclengthalongthenon-flatdirections,wehaveτ(0,t) = t,τ(s,0) = 0andthereforewemust have τ(s,t) = t(1 + sα(t)) for some coefficient function α(t). If k(s,0) denotes the nonzero principalcurvatureatthepointl(s,0)wecanwrite (cid:107)N(s,τ)−N(s,0)(cid:107) (cid:107)N(0,t)−N(0,0)(cid:107) 1 k(s,0) = lim = lim = k(0,0) . τ→0 τ t→0 t(1+sα(t)) 1+sα(0) 10