Annals of Mathematics On the Total Curvature of Knots Author(s): J. W. Milnor Reviewed work(s): Source: The Annals of Mathematics, Second Series, Vol. 52, No. 2 (Sep., 1950), pp. 248-257 Published by: Annals of Mathematics Stable URL: http://www.jstor.org/stable/1969467 . Accessed: 06/12/2011 20:57 Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at . http://www.jstor.org/page/info/about/policies/terms.jsp JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range of content in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new forms of scholarship. For more information about JSTOR, please contact [email protected]. Annals of Mathematics is collaborating with JSTOR to digitize, preserve and extend access to The Annals of Mathematics. http://www.jstor.org OF MATHEMATICS ANNwAzs Vol. 52, No. 2, September,1950 ON THE TOTAL CURVATURE OF KNOTS BY J. W. MILNOR (Received October5, 1949) f The totalcurvature Introduction S"(s) I ds of a closed curve C of class C", a quantity which measures the total turningof the tangent vector, was studied by W. Fenchel,who proved,in 1929,that, in threedimensionalspace, I]"(s) I ds _ 2'n,equalityholdingonlyforplane convexcurves.K. Borsuk,in 1947,extended this resultto n dimensionalspace, and, in the same paper, conjecturedthat the total curvatureof a knot in threedimensionalspace must exceed 47r.A proof ofthis conjectureis presentedbelow.' suggestedby R. H. use willbe made ofa definition, In provingthisproposition, Fox, of total curvaturewhich is applicable to any closed curve. This general definitionis validated by showingthat the generalizedtotal curvatureK(C) is equal to I \"() I ds for any closed curve C of class C". Furthermore,the theoremofFencheland Borsukis trueforany closed curve,ifthe new definition of total curvatureis used. ( ), the Closelyrelatedto the conceptof total curvatureis a new invarient/Z ina is either This positive curves. ( of closed of the isotopytype crookedness In a is or is polygon. as by the not represented (E type tegeror oc, according alternative an to is it provide of crookedness possible the of concept terms formulationof the generalizedtotal curvatureas a Lebesgue integralover an (n - 1) dimensionalsphere.The crookedness,4(Cs)of a type (S of simpleclosed curvesis connectedwith the total curvaturesof its representativecurves C by the fundamentalrelation2,ru( S) = g.l.b. K(C). Generallyspeaking this lower bound is not attained. In the courseof the paper several interestingincidentalresultsare obtained: ifthe total curvatureof a simpleclosed curveis finite,thenthereis an inscribed polygonequivalentto it by isotopy,and also if the curveis knottedtheremust be a plane whichintersectsit in at least six points. I am indebtedto R. H. Fox forsubstantialassistance in the preparationof thispaper. 1. The Total Curvatureof a Closed Polygon By a closedpolygonP in Euclidean n-space H', n > 1, will be meant a finite , ctm-i, am = ao , of which is required only that sequence of points a, , . f 1 Since the completionof this paper, therehas appeared an independantproof,by I. FAry,that the relation "(s) I ds 2 47rholds forall knots [6]. 248 ON THE TOTAL CURVATURE 249 OF KNOTS $i5 ai+, , and the linesegmentsaiai+1 fori = 0,1, ... , m - 1. For convenience let ai, wherei is any integer,signifya(X), where (i) is the least positiveresidue mod m. The terms point and vector will be used synonymously,with every to a commonorigin,so that ai+1 - ai means the vectorequal in vectorreferred lengthand parallel to the line segmentaiai+1 . Denote by ac the angle between 0 < as ir. By thetotalcurvature thevectorsa+?1- as and a, - ai1 satisfying sum P is meanttheangle K(P) ofa closedpolygon E., ai. 1.1 LEMMA.2 The adjunctionof a new vertexto a closedpolygoncannotdecrease its totalcurvature.The curvaturemay remainconstantif eitherthenew vertexa3 and thetwoadjacentvertices aj-1 and aj+l are collinearor ap-2 , 1Ia X ,aj+Xl , and increase. it mustdefinitely aj+2 are coplanar.Otherwise Let P' be the closed polygonwithverticesa1 , a2 , - , aC-1 , aj+, * , amC. aX , , ? X. *X , a Let P be the closed polygonwithverticesaL, a2 , .* X1 ,jl =1, 2, i a' for by Denote . vertex the P' adjoining obtained from by aj -2(X. FIG. 1 j - 1,j + 1, *, m the respectiveexteriorangles of P', and by ai for i = , m the respectiveexterioranglesof P. Denote by 1, 2, * j - 1,j, j + 1, , the angle betweena -cai1 and aj+l - j , and by 3+ the angle between aj+l- aj- and aj+l - aj. (See Fig. 1.) > aj-a , wherethe 4By the triangleinequalityforsphericaltrianglesai-, + equality can hold only if the threeangles lie in a plane, that is, only if aj.2, j , cja, and aj+l are coplanar.Similarlyaj+l + A+ > aia+, wherethe equality can hold only if aj-1, aj, a+ja , and aj42 are coplanar. From the trianglewith verticesaj1 , ai , and aj+l we have /- + /+ = a3 . Therefore K(P) - K(P') = > (aj-1 -O- ai-1) + +ex aj + (aj+l - ai4l) = O. Hence K(P) > K(P') and the equality can hold only if either aj2 , aj+l , and aj+2 are coplanar or aj-, aj , and aj+l are collinear. 2 aj-1, This proofis essentiallythe same as a proofgivenby Borsuk [1. pp. 254-256]. a , 250 J. Wv.MILNOR 1.2 COROLLARY. If thevertices ,j-1 , aj , and aj+1 of a closedpolygonP ai-2 , are notcoplanar,and thevertexaj is replacedby a vertexa' whichlies on theline segment ajaj+1 , thenK(P) is decreased. 2. The Total Curvatureof a Curve By a closedcurveC in Euclideann-spaceH' will be meant a continuousvector , Xn(t))of period 1 which is not constant in any tfunctionX(t) = (&l(t), interval.In particularany polygoncan be describedin this manner;it will be convenientto regarda polygonas a closed curve,ignoringthe distinctionbetween parameterizations.A closed curve X(t) is simple if X(t1)= X(t2) only different when (t1 - t2)/lis an integer. ' rn/ FIG. 2 , am is said to be inscribedin a closed A closed polygonP withverticesal , values tisuch that ti < ti+1 , ti+m-ti + 1, curve X(t)ifthereis a set ofparameter = i. of values and as X(ti)forall integral 2.1 LEMMA. For any closed polygon P, K(P) = l.u.b. {K(P')} where P' ranges overall polygonsinscribedin P. If P is a polygonhavingtwo or moreverticescoincident,it can be represented as the limitof a sequence of polygonswith all verticesdistinct;hence it only remainsto provethe lemmaforP withall verticesdistinct. If PO is a representativeinscribedpolygonwhose verticesinclude all but m of the verticesof P, we may adjoin the remainingverticesof P one by one to By 1.1, K(Po) < P. P, producing a sequence of polygons PO' , P' ... l.u.b. {iK(P')}. = Therefore = < but <? K(P) K(P). K(P') K(Pf); * K(Pf) TOTAL THE ON OF CURVATURE 251 KNOTS by K(C) = l.u.b. {K(P) } where For each closedcurveC definethetotalcurvature P rangesover all polygonsinscribedin C. If C is itselfa polygon,the preceding lemma shows that this definitionis consistentwith the definitionof Section 1. by arclengths, 2.2 rTHEOREM. If C is a closedcurveof class C"' pararneterized then K(C) I & (s) I ds. = If act -(S = ,, '), = X(s') are the vertices of a polygon Pm inscribed in C, suchthatlim, axmaxi { (sim - s')} = boK(Pm) 0, it willfirstbeshownthatlimm (s) ds (Compare Fig. 2). Define s2 and ,'(Sm l) = 2 (si t'(S2). + S'+1) foreveryi and m. Denote by 62 the angle between The vector X'(s) describesa curve L of length X" (s) I ds f on theunitsphereS'-'. The vectors '(s2) formthe verticesof a sphericalpolym gon of length O2= 62 which is inscribedin L. Thereforelimnb m 62 = I c X'(s) Ids. continuous,foreach E > 0 thereis a 6 > 0 suchthat Since X"(s) is uniformly < e forall u - v < 6. From the identity "(v) m + (S i+ (s2) (s2 sT),,(,m) [X"(u)- X"(0)]du dv i S i+ m~~~~~~~S ~ l "(u)- - f f - Li fn ? E""(s2) &(u)]du d - we have si+i-si m |(s)m - ) (- < X(9) +- s) 'E 4 whenever maxi { (sm+, - s7) } < 6. is the angle between (sm+) - (sm)and '(S2) then sin m2< (s2+, - s') If 2m s2) lies withina E/4,since the end point of the vector (T(s524+i)- X(s2if))(sm2+ sphereof radius (Smi+, - S'i) E/4 about the end point of the unitvector X'(V2). small E wehave '2 < 2 sin '2 < (s2i sit) E/2. The angle Hence forsufficiently between X(si+i) - ~(sT,) and X(sT) - x(s2m-)is aT , while that between X'(9T) and so m+ ,mi- and 62 < aim + sp? +'pi, ai <= O + l) is ri . Therefore X'(S, 62 < lE, where Z am -ZEr=, Em.=1 that I 0 - 62 < ('in+, - sm,-) E/2, and 1 is the length of C. Therefore - - lim K(Pm) m-4o00 lim In orderto showthat m m E m-00 i=1 = I remainsto show that K(P) ai = lim E mn-c00 = I Jc |(s) ds. C (s) ds = l.u.b. {K(P) } forP inscribedin C, it only < I " '(s) tds. 252 J. W. MILNOR Given any polygonPk inscribedin C, we may forma sequence of polygons Pmform = k, k + 1, ... by adjoiningverticesto Pk so that limma.,maxi {(si+1 - s,)} = 0. By 1.1 K(Pk) thereforeK(P7-) ? K(Pk+1) ? fs < --* but limid.. K(Pm) (s) = ds, and "(s) ds. 3. The Crookednessof a Closed Curve For each closedcurveC and each unitvectorb, defineA(C, b) to be the number of maxima of the functionb- (t) (i.e. the numberof parametervalues tofor which b* (to) ? b- (t) for t withinsome neighborhoodof to) in a fundamental period. For each closed curve C definebt(C) = minb{Wt(C,b)}. We may call of C. ,u(C) the crookedness For everyvectorai+1 - ai inthe space H' definebi = (ai+1 - ai)/ I ai+1 - ai Accordingto the conventionintroducedearlier,bi also denotes a point on the unit sphere S"1, the sphericalimage of ai+1 - ai. Given a polygonP with verticesal , a2, ... , am, a spherical polygon Q is formed on Sn'- by joining each bi-, to bi by a great circle arc of length ai . This spherical polygon Q is called a sphericalimage of P, and is unique unless for some j the vector bj = - j+l . Note that it may happen that bj = bj+l . 3.1 THEOREM.3 For any closedcurveC in Hn, n > 2, the Lebesgueintegral u(C, b)dS, whereb ranges over the unit sphere, exists and is equal to is themeasureof S" 1. where3in_ = (27r"12)/F(n/2) (Mn_1K(C))/21r, curve is a polygonP. For every which the in case the We will firstconsider of S"'_ whichhas a pole at b. the S,-2 denote sphere great point b of S"', let if if Sn-2 and b only crosses An edge bj-lbj ofQ (aj+l - aj) and b (aj - aj1) have of b (t). Therefore,if or minimum is maximum a that b opposite sign, so aj to b) the number P is of perpendiuclar no edge of (i.e. vertex Q, Sb-2 containsno b for which Sb2 The of set points is b). 2,u(P, with of Q of intersections Sb2 of the finite collection great spheres of is union the of Q vertex containssome of Sn-1 to Ui Sbn,2the with respect the complement n2; of in each component Sb | f function2M(P, b) is constant.The integral 2p(P, b)dS, where dS is the sn-1 elementof surfaceon Sn-1,is thereforedefined.The set of points b forwhich Sb-2 meets a given segmentbi-,bi of length0 ? ai _ r is a "double lune" boundedby the greatspheresS'-27 and Ssn2. Thus the contributionof bi-,bi to 2,4(P, b) is 1 if b is an interiorpoint of this lune and 0 if b is an exteriorpoint. whereMn,1 is the measure of the entire The measureof this lune is (afiMIn,1)/r sphere. Consequently f ,-1 8 2y(P, W)dS = __E m it 1 ~~~~7r ai 7r This theoremis relatedto Crofton'sformula.Cf. [3. p. 811. K(P ON THE TOTAL CURVATURE 253 OF KNOTS If C is an arbitraryclosed curve g(t), let Pm be a set of inscribedpolygons = g(t') such that each Pm containsall a', , gm(t)with verticesa' = the vertices of Pm-1 and satisfyinglimnm.AK(Pm) = K(C) and limm.xmaxi (tm,-tm) } = 0. The values of b forwhich b (t) or any b*m(t) has an interval of constancyforma set of measure zero, and thereforehave no effecton the integral.Such values willbe ignoredforthe remainderofthe proof. We firstshow that ,t(C, b) = limm-oo A(Pm b). It is certainly true that j(C, b) > i(Pm , b) > i(Pm-i X b). If ,t(C, b) < o, it is possibleto selecta neighborhoodof each of the A(C, b) maximaof b*g(t) and of each of its minimasufficientlysmall so that a polygonwith a vertex in each of these neighborhoods musthave at least A(C, b) maxima;whichis certainlytrueofPmform sufficiently large.If A(C, b) = Oo,the set of values of t forwhichb (t) is a maximummust containa denumerablesubset {t2i}such that eitherto< t2< . .. < liming t2i < to+ 1orto> t2> .*. > limli- t2i > to - 1. In eithercase we may selecta series ofintermediatevalues t2i+1such that each b (t2i) > b (t2i+1) and b (t2i) > b-*(t2i-1). Given any 2j < oc we may select neighborhoodsof the g(ti), for i < 2j, so small that any polygonwithat least one vertexin each neighborhood large.Therefore has at leastj - 1 maxima;whichis trueof Pmform sufficiently A(Pm, b) increases without finitebound as m -* o. Each of the integrals - , f Pu(Pm,b) dS exists; and the nondecreasingsequence of positive functions Ai(Pm X b) approaches,4(C,b). Therefore4theintegral equals 3.2 Since 1ime-.t(Pm, K(C) COROLLARY. C Kt_1 K(C) 2ir b) dS = limm_:+ (Mn-1) fI (C, b) dS K(Pm) =(Mn1) exists and K(C)- ? 2irA(C). = f-1 1(Ca) dS? fn- lu(C)ds = Mn11i(C) By a convexcurvewill be meant a closed plane curve describedby g(t) such that any line contains g(t) eitherfor not more than two values of t withina fundamentalperiodor forall values of t withinsome interval. conditionthata closedpolygonP in 3.3 LEMMA. The necessaryand sufficient H2 be convexis thatforeveryb eitherA(P, b) = 1 or,A(P,b) = It is clear that this conditionis necessary.Suppose that P is a closed plane polygonsuch that eitherli(P, b) = 1 or 4(P, b) = oo foreach b in the plane. For any b such that A(P, b) = 1, any line perpendicularto b will intersectP in at most two points.If b is a vectorforwhichlu(P, b) = Oo, and if H' is a line perpendicularto b whichintersectsP a finitenumber,say r, of times,then it is withP in the always possibleto rotateH1 about one of its pointsof intersection properdirectionso that r is not decreased. Hence thereis a b and a H' perpendicular to b such that Az(P,b) < cc and the number of intersectionsof H' r ? 2. withP is f ? r. But by the firstcase, f < 2, and therefore o 4 [4. Theorem12.6p. 281. J. W. MILNOR 254 3.4 THEOREM. For any closedcurveC, K(C) ? 2r. The equalityholdsifand only if C is convex. Since ,g(C) ? 1 foreverycurve,or since everycurvehas an inscribedpolygon P forwhichK(P) = 2r, we have K(C) ? 2r. Since any curvewhichis not convex has an inscribedpolygonwhichis not convex,and sincea polygoninscribedin a convexcurvemustbe convex,it onlyremainsto provethe secondportionof the theoremfor polygons. It is provedin plane geometrythat the sum of the exteriorangles of a convex polygonis 2r. If therewere a non-planarpolygonP forwhich K(P) = 2r, we could selectfourconsecutivenon-coplanarvertices(neglectingverticesforwhich ai = 0). By 1.2 therewould be a new polygonP' such that K(P') < K(P) = 27, which is impossible.If there were a non-convexplane polygon P for which K(P) = 2w, then 3.3 states that there would be a directionb for which 1 < o but thereis a neighborhoodof any such 6 withinwhich1u(P,b) is so, ,u(P, b) constant.This meansthat K(P) (P,b) dS = > f dS = 2X. 4. The Curvatureand Crookednessof IsotopyTypes of Curves In H' the closed curve describedby X(t)of period1,and the closed curve describedby ;(t) of period 1 are said to be equivalentby isotopyif there is some isotopyof Hn onto itself,whichtransforms X(ul) into ~(ul) forall u. By a curvetype S in Hn is meant an equivalence class of closed curvesunder isotopy.A curve type is simpleif the representativeclosed curves are simple. if (E is that A simplecurve type S and its membersare said to be unknotted type whichcontainsall circles.If a simple curve type containsno circlesthen the type and its membersare said to be knotted. A curvetype C and its membersare said to be tame'if G containsa polygon. Otherwisethey are said to be wild.' It is well known that in H' every simple closed curve is unknotted.In H nforn > 3, every simple tame curve is unknotted. For each curvetype A, defineK(G) = g.l.b. K(C) and A(G-)= min ,(C), where C rangesover all membersof (S. <K r, thereis an iso4.1 LEMMA. For each c and p in Jfn-1such thatI cn-1 c into P and leaves ontoitselfwhichtransforms tu 1, of H topy,f (T),I 0 2)-sphereof radius r and centerc, such fixedall pointsof H'-1 outsidethe (n thatfco(T) is a continuousfunctionof u, I, c, and p. For example: = - ui X 7 ] (P - c) for I -c forj - c! _ r. 4.2 THEOREM. For any simple closed curve C, such that 1t(C) < toC byisotopy. polygonP inscribedin C and equivalent 5 This definitionwas given by Fox and Artin [51. I r. xc, there is a ON THE TOTAL CURVATURE OF KNOTS 255 If b is a unit vectorforwhich AL(C,b) < xo,thereare a finiteset of values < t2A(Cb) < t&+ 1,forwhichb* (t) has a maximumor minimum. < t2 < . About each point X(t-)constructa cylinderZ-1(0) with generatorsparallel to b whichintersectsC in exactlytwo pointsX(Q) and i(tt) such that both lie on a base of the cylinderand such that ((t+) is the centerof this base. It will firstbe shown that there is an isotopy of the closed n-cell bounded by Zi'-(0) onto itselfwhich leaves Z '(0) fixed and transformsthe curve segment X(t) for C < t < t+ onto the polygonalline X(t), X(ti),X(tt). Each hyperplaneH'-1 perpendicularto b whichintersectsZ,-1(O) intersects it in a sphereS,-2. Performthe isotopyofeach H'-' onto itselfwhichtransforms the curve segmentX(t)forti _ t <tt into the axis of the cylinder,and which leaves all pointsoutside of the n-cellbounded by the cylinderfixed,as defined by 4.1. Select a continuoussequence of coaxial cylindersZ"'(v), 0 ? v < xc, such that any Zl'G(3) is containedwithinall Z'-1(v) withv < v, such that each cylinderintersectsC onlyin the centerof one base and in one otherpointofthat base, and such that Z-'(v) tends to the point x(ti) as v -* oo. Rotate each into Z -l(v) about its axis so that each point &(t)forC ? t < ti is transformed the plane determinedby g(t7) and the axis ofthe cylinders.Since we have transformedg(t) fortU ? t < tt onto a plane curve withinZn-1(0), it is certainly possible to transformit onto the polygonalline (t), g(ti), g(tj), still within Z-1 (0), producingan equivalentcurveC describedby f(t). This curve is divided into 4,g(C,b) = 4g(C, b) distinctsegmentsby the points i(t+) and g(t7). If g > 0 is the g.l.b. of the distancesbetweendistinctand nonconsecutivecurve Zn-1 we may segments,thenforeach point of C whichis not withinany of the Zi a in lies hyperplane constructthe sphereSn 2 whichhas its centerat the point, perpendicularto b, and has radius g/3. Since no two of these spherescan interconstructeda tube around each curvesegmentoutsidethe sect,we have in effect, It is now possibleto inscribea polygcylinders,withno two tubes intersecting. onal line,lyingcompletelywithinthe tube, in each segmentof C. Performthe each ofthese isotopyofeach H'-1 perpendicularto b ontoitselfwhichtransforms leaves fixed which and line curve segmentsonto the correspondingpolygonal and C, into therefore thus have transformed C, of the Sn2. CWte all pointsoutside that reference for future (Note isotopy. by an inscribedpolygonP, equivalent t1 (C, b) = A (P, b).) 4.3 COROLLARY. The necessary and sufficientcondition that a simple curve type (S be tameis that4(S) < oo 4.4 COROLLARY.The total curvatureof a tame knot cannot equal the curvatureof its type. Assumethat C is a tame knotoftype (EwithK(C) = K(Q,). Let P be a polygon of type S inscribedin C. Then K(P) < K(C). Since P cannotlie in any plane, we may selectfourconsecutivenon-coplanarvertices(ifwe ignoreverticesforwhich ai= 0). By 1.2 we may selecta new polygonP, stilla memberof A, and having K(C) = K(G); which is impossible. 4.5 COROLLARY.The crookednessof any knotis greaterthan or equal to 2. K(P) < K(P) < If C is a curvewith,u(C) = 1, then,in the proofof 4.2 we can select the two 256 J. W. MILNOR cylinderswith a commonbase. The firstisotopywill then transformC into a plane quadrilateral,whichis certainlyunknotted. 4.6 COROLLARY. The totalcurvature ofany knotis greaterthan4%r. 4.7. THEOREM. If ( is a simplecurvetype,thenK( A) = 2iry(C). It has already been shownthat K(C) > 2iryu(C) forany C e G and therefore that K(G) > 2ry(G). If A(G) = co, this provesthe proposition.If A(G) < co, we may select a curve C of G and a directionb such that ,u(C, b) = Au(G). By 4.2 thereis a polygonP whichis a memberof G such that ,A(P,b) = ,A(C,b). For conveniencewe will select a new coordinatesystemso that b is parallel to the xl axis. We may then definethe isotopy FU(x, X2, X3,I..., Xn) = , P into an (X1 , UX2 , UX3, ... ux*,) for0 < u < 1. This evidentlytransforms \ tt\V ;F:AR:AEPZ 7': AE$C FIG. 3 equivalent polygon Pu, such that ,'(P , b) = ,u(P, b). If ai , 1 < i < m, is the set ofverticesof P. , we may divideit intofoursubsets: (a) verticesa'Usuch that b*a,_I < b*ai < b*as+X (b) vertices ai' such that b*ait 1 > b *aS' > b *a (c) verticesai' such that braid < b a~' > b-a4X (d) verticesas such that b*a!' > b*at < b*a'+. (If an equalitywereto hold,we wouldhave Au(P., b) = co.) Evidentlythenumber of verticesin (c) equals the numberin (d) equals Au( ). Howeverformembersof (c) and (d), lim~o (a') = ir; whereasfor(a) and (b), limb~o(a') = 0. Therefore limb-o K(Pu) = 2rp( C). As anotherinterestingconsequence of Theorem 4.2, we have the following. 4.8 THEOREM. Givena knotC in H3 forwhichji(C) < oo,thereis a plane whose withC consistsof at leastsix components. intersection Since every such C has an inscribedpolygonwhich is knotted,and since a plane intersectsa curve at least as many times as it intersectsan inscribed polygon,it onlyremainsto provethe theoremforknottedpolygons.If thereis a polygonwhichdoes not satisfythetheorem,theremustbe one havinga minimum numberof sides. If P is such a polygon,we have: 4r < K(P) = 2 Aj(P, b)dS. Thereforethere must be some unit vector b such that 2 < ji(P, b) and also ON THE TOTAL CURVATURE OF KNOTS 257 b a3 # ba*k foreverypair of distinctverticesaj and ak of P. If we selecta plane perpendicularto b and move it parallel to itselfin the directionof b until it intersectsP, it must firstintersectP in a minimum(i.e. an aj such that b aj < b -a.,). Afterthis it will intersectP in two points,until it intersectsanother minimum,afterwhichit will have fourintersections.If it next reachesanother minimum,the theoremis proved.If it next reachesa maximum,therewill then be only two intersections. Jointhese two pointsby a line segment,so that two new polygonsare formedby this segmentand the sides of P. (See Fig. 3.) At least one of these new polygons,P1 , must be knotted.Since /u(P) > 3, each of thenewpolygonsP, and P2 musthave at least fivesides.Since P2 has fiveor more sides,P1 musthave fewersides than P; and thereforetheremust be some plane intersectingP, in six or more components.It is clear that this plane must intersectthe originalpolygonP itselfin six or more components. BIBLIOGRAPHY [1] BORSUK, K., Sur la courburetotaledes courbesfermtes.Annales de la Soc. Polonaise 20 (1947),251-265. Math. Ann. Raumkurven. [2] FENCHEL, W., Uber Krummungund Windunggeschlossener 101 (1929),238-252. [3] BLASCHKE, W., Vorlesungen uber Integralgeometrie, zweites heft.HamburgerMathe- matische Einselschriften, 22 Heft, 1937. [4] SAKS, S., Theory of the Integral (2ndrevised edition). Warsaw 1937. space. Ann. [5] Fox, R. H. and ARTIN, E., Some WildCells and Spheresin three-dimensional of Math. Vol. 49 (1948), 979-990. Bulletinde [6] FXRY, M. I., Sur la CourbureTotaled'une CourbeGaucheFaisant un Nweud. la Soc. Math. de France LXXVII (1949), 128-138.
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