CHAPTER 1.CURVES FIGURE 3.1 is that .(Note,however.that this depends on the arclength parametrization of the original curve and is not a parametrization-independent condition on the image curve IC.)We do,nevertheless. have the following geometric consequence of this condition.For any (unit)vector A.we have AT()ds (T()-Ads. and so the average value of T.A must be 0.In particular,the tangent indicatrix must cross the great circle with normal vector A.That is,if the curveis to be a tangent indicatrix,it must be"balanced"with respect to every direction A.It is natural to ask for the shortest curve(s)with this property. Ifs∈∑，lets+denote the(oriented)great circle with normal vector. Proposition 3.2(Crofton's formula).LetT be a piecewise-el curve on the sphere.Then engh0=i2rng中ds =x(the average number of intersections ofI with all great circles). (Hered represents the usual element of surface area on.) Proof.We leave this to the reader in Exercise 12. Remark.Although we don't stop to justify it here,the set of for which #(is infinite is a set of measure ero,and so the integral makes sense Applying this to the case of the tangent indicatrix of a closed space curve,we deduce the following classical result. Theorem 3.3(Fenchel).Thet otal c curvature of any closed space curve is at least.and equality holds ifand only ifthe curve is a(convex)planar curve Proof.Let I be the tangent indicatrix of our space curve.If C is a closed plane curve,then I is a great circle on the sphere.As we shall see in the next section,convexity of the curve can be interpreted as syingeverywhere,so the tangent indicarix traverses the great circle exactly once and (ef.Theorem 3.5 in the next section). To prove the converse,note that,by our earlier remarks,I must cross almost every and hence must intersect it at least twice.and so it follows from Proposition 32 that kds=length()≥
24 CHAPTER 1. CURVES T FIGURE 3.1 is that Z L 0 T.s/ds D 0. (Note, however, that this depends on the arclength parametrization of the original curve and is not a parametrization-independent condition on the image curve †.) We do, nevertheless, have the following geometric consequence of this condition. For any (unit) vector A, we have 0 D A Z L 0 T.s/ds D Z L 0 .T.s/ A/ds; and so the average value of T A must be 0. In particular, the tangent indicatrix must cross the great circle with normal vector A. That is, if the curve is to be a tangent indicatrix, it must be “balanced” with respect to every direction A. It is natural to ask for the shortest curve(s) with this property. If 2 †, let ? denote the (oriented) great circle with normal vector . Proposition 3.2 (Crofton’s formula). Let be a piecewise-C 1 curve on the sphere. Then length./ D 1 4 Z † #. \ ?/d D .the average number of intersections of with all great circles/: (Here d represents the usual element of surface area on †.) Proof. We leave this to the reader in Exercise 12. Remark. Although we don’t stop to justify it here, the set of for which #. \ ?/ is infinite is a set of measure zero, and so the integral makes sense. Applying this to the case of the tangent indicatrix of a closed space curve, we deduce the following classical result. Theorem 3.3 (Fenchel). The total curvature of any closed space curve is at least 2, and equality holds if and only if the curve is a (convex) planar curve. Proof. Let be the tangent indicatrix of our space curve. If C is a closed plane curve, then is a great circle on the sphere. As we shall see in the next section, convexity of the curve can be interpreted as saying > 0 everywhere, so the tangent indicatrix traverses the great circle exactly once and Z C ds D 2 (cf. Theorem 3.5 in the next section). To prove the converse, note that, by our earlier remarks, must cross almost every 2 † and hence must intersect it at least twice, and so it follows from Proposition 3.2 that Z ds D length./
$3.SOME GLOBAL RESULTS (2)(4x)=2x.Now,we claim that if T is a connected,closed curve in E of length 2,then I lies in a closed hemisphere.It will follow,then,that ifis a tangent indicatrix of be a great circle (For if lies in the hemisphere A.x≥0，Ts)Ads=0 forcesT·A=0,so is the great circle that is planarand the tangent traverses the great cir preeisely one time,which means that>0and the curve is convex.(See the next section for more details on this.) FIGURE3.2 To prove the claim,we proceed as follows.Suppose length()<2m.Choose P and inI so that the ares=Pand T=OP have the same length.Choose N bisecting the shorter great circle are from P to ,as shown in Figure 3.2.For convenience,we rotate the picture so that N is the north pole of the sphere Suppose now that the curve I were to enter the southern hemisphere;let T denote the reflection ofr across the north pole(following arcs of great circle through N).Now,IU is a closed curve containing a pair of antipodal points and therefore is longer than a great circle.(See Exercise 1.)has the same length as I',we see that length(I)>2,which is a contradiction.Therefore I indeed lies in the northem hemisphere. Wenow sketch the proof of a result that has led to many interesting questions in higher dimensions.We say a simple (non-self-intersecting)closed space curve is knotted if we cannot fill it in with a disk. Theorem 3.4(Fary-Milnor).Ifa simple closed space curve is knotted,then its total curvature is at least 4n. Sketeh of proof.Suppose the total curvature of C is less than 4x.Then the average number )<4.Since this is generically an even number2(whenever the great circle isn't tangent to )there must be an open set of &'s for which we have #()=2.Choose one such,o.This means that the tangent vector to C is only perpendicular totwice,so the function f(x)=xhas only two critical points.That is,the planes perpendicular toowill(by Rolle's Theorem)intersect C either in a line segment or in a single point (at the top and bottom):that is.by moving these planes from the bottom ofC to the top we fill in a disk.soC is unknotted. 3.2.Plane Curves.We conclude this chapter with some results on plane curves.Now we assign a sign to the curvature:Given an arclength-parametrized curve,(re)define N(s)so that (T(s).N(s)is a right- handed basis for R2(ie,one turns unterclockwise from T(s)to N(s)),and then set T(s)=(s)N(s) as before.So0 when T is twisting counterclockwise and0 when T is twisting clockwise. Although the total curvature ofa simple cosd plane curve my be quiteb arger then
÷3. SOME GLOBAL RESULTS 25 1 4 .2/.4/ D 2. Now, we claim that if is a connected, closed curve in † of length 2, then lies in a closed hemisphere. It will follow, then, that if is a tangent indicatrix of length 2, it must be a great circle. (For if lies in the hemisphere A x 0, Z L 0 T.s/ Ads D 0 forces T A D 0, so is the great circle A x D 0.) It follows that the curve is planar and the tangent indicatrix traverses the great circle precisely one time, which means that > 0 and the curve is convex. (See the next section for more details on this.) FIGURE 3.2 To prove the claim, we proceed as follows. Suppose length./ 2. Choose P and Q in so that the arcs 1 D _ PQ and 2 D _ QP have the same length. Choose N bisecting the shorter great circle arc from P to Q, as shown in Figure 3.2. For convenience, we rotate the picture so that N is the north pole of the sphere. Suppose now that the curve 1 were to enter the southern hemisphere; let 1 denote the reflection of 1 across the north pole (following arcs of great circle through N ). Now, 1 [ 1 is a closed curve containing a pair of antipodal points and therefore is longer than a great circle. (See Exercise 1.) Since 1 [ 1 has the same length as , we see that length./ > 2, which is a contradiction. Therefore indeed lies in the northern hemisphere. We now sketch the proof of a result that has led to many interesting questions in higher dimensions. We say a simple (non-self-intersecting) closed space curve is knotted if we cannot fill it in with a disk. Theorem 3.4 (F´ary-Milnor). If a simple closed space curve is knotted, then its total curvature is at least 4. Sketch of proof. Suppose the total curvature of C is less than 4. Then the average number #. \ ?/ < 4. Since this is generically an even number 2 (whenever the great circle isn’t tangent to ), there must be an open set of ’s for which we have #. \ ?/ D 2. Choose one such, 0 . This means that the tangent vector to C is only perpendicular to 0 twice, so the function f .x/ D x 0 has only two critical points. That is, the planes perpendicular to 0 will (by Rolle’s Theorem) intersect C either in a line segment or in a single point (at the top and bottom); that is, by moving these planes from the bottom of C to the top, we fill in a disk, so C is unknotted. 3.2. Plane Curves. We conclude this chapter with some results on plane curves. Now we assign a sign to the curvature: Given an arclength-parametrized curve ˛, (re)define N.s/ so that fT.s/; N.s/g is a righthanded basis for R 2 (i.e., one turns counterclockwise from T.s/ to N.s/), and then set T 0 .s/ D .s/N.s/, as before. So > 0 when T is twisting counterclockwise and < 0 when T is twisting clockwise. Although the total curvature Z C j.s/jds of a simple closed plane curve may be quite a bit larger then 2, it
CHAPTER 1.CURVES K>0 FIGURE3.3 is intuitively plausible that the tangent vector must make precisely one full rotation,either counterelockwise or clockwise,and thus we have Theore .5(Hopf Umlaufsatz)IfC is a simple closed plane curve,thend The crucial ingredient is to keep track of a contimous total angle through which the tangent vector has turned.That is,we need the following an arclength-parametrized plane curve.Then there isa function 0:0,L]→R so that T(s)=(cos0(s),sin(s)）for all s∈0，.Moreover,for any two such functions,. 0 and 0*,we have (L)-0(0)=0*(L)-0*(0).The number (0(L)-0(0))/2n is called the rotation index of Proof.Consider the four open semicircles U =(xy):x U2=(x.y)1: x<0,U3={(x,y)es1 y>0,and U4=(x,y)S1:y <0.Then the functions vi.n(x.y)arctan(y/x)+2nx v2.n(x.y)=arctan(y/x)+(2n +1)x v3.n(x.y)=-arctan(x/y)+(2n+) 4n(x.y)=-arctan(x/y)+(2n-) are smooth mapsin:UR with the property that cos(in(x.y)).sin(in(x.y)))=(x.y)for every i =1.2.3.4 and n EZ. Define0(O)so that T(0)=(cosf(o),sinf()）.LctS={s∈0，)：9 is defined and el on [0.s],and let so sup S.Suppose first that so L.Choose i so that T(so)Ui,and choose n Z so that(T(so))=lim(s).Because T is continuous at so.there that T(s)Ui for all s with Is-so<8.Then setting (s)=in(T(s))for all sos<so+8 gives us a el function defined on [0.so +8/2].so we cannot have so L.(Note that 0(s)=vin(T(s))for all so-8 <s so.Why?) But the same argument shows that when so=L.the function is el on all of [0.L]. Now,since T(L)=T(0).we know that (L)-(0)must be an integral multiple of.Moreover for any other function with the same properties,we have *(s)=0(s)+2n(s)for some integer n(s). Since andare both continuous,n must be a continuous function as well:since it takes on only integer values,it must be a constant function.Therefore,(L)(0)=0(L)(0),as required
26 CHAPTER 1. CURVES T T T N N N FIGURE 3.3 is intuitively plausible that the tangent vector must make precisely one full rotation, either counterclockwise or clockwise, and thus we have Theorem 3.5 (Hopf Umlaufsatz). If C is a simple closed plane curve, then Z C ds D ˙2. The crucial ingredient is to keep track of a continuous total angle through which the tangent vector has turned. That is, we need the following Lemma 3.6. Let ˛W Œ0; L ! R 2 be an arclength-parametrized plane curve. Then there is a C 1 function W Œ0; L ! R so that T.s/ D cos .s/;sin .s/ for all s 2 Œ0; L. Moreover, for any two such functions, and , we have .L/ .0/ D .L/ .0/. The number ..L/ .0//=2 is called the rotation index of ˛. Proof. Consider the four open semicircles U1 D f.x; y/ 2 S 1 W x > 0g, U2 D f.x; y/ 2 S 1 W x < 0g, U3 D f.x; y/ 2 S 1 W y > 0g, and U4 D f.x; y/ 2 S 1 W y < 0g. Then the functions 1;n.x; y/ D arctan.y=x/ C 2n 2;n.x; y/ D arctan.y=x/ C .2n C 1/ 3;n.x; y/ D arctan.x=y/ C .2n C 1 2 / 4;n.x; y/ D arctan.x=y/ C .2n 1 2 / are smooth maps i;nWUi ! R with the property that cos. i;n.x; y//;sin. i;n.x; y// D .x; y/ for every i D 1; 2; 3; 4 and n 2 Z. Define .0/ so that T.0/ D cos .0/;sin .0/ . Let S D fs 2 Œ0; L W is defined and C 1 on Œ0; sg, and let s0 D sup S. Suppose first that s0 < L. Choose i so that T.s0/ 2 Ui , and choose n 2 Z so that i;n.T.s0// D lims!s 0 .s/. Because T is continuous at s0, there is ı > 0 so that T.s/ 2 Ui for all s with js s0j < ı. Then setting .s/ D i;n.T.s// for all s0 s < s0 C ı gives us a C 1 function defined on Œ0; s0 C ı=2, so we cannot have s0 < L. (Note that .s/ D i;n.T.s// for all s0 ı < s < s0. Why?) But the same argument shows that when s0 D L, the function is C 1 on all of Œ0; L. Now, since T.L/ D T.0/, we know that .L/ .0/ must be an integral multiple of 2. Moreover, for any other function with the same properties, we have .s/ D .s/ C 2 n.s/ for some integer n.s/. Since and are both continuous, n must be a continuous function as well; since it takes on only integer values, it must be a constant function. Therefore, .L/ .0/ D .L/ .0/, as required.
$3.SOME GLOBAL RESULTS 之 Sketch of proof of Theorem 3.5.Note first that if T(s)=(cos(s).sin(s)),then T'(s)= 0'(s)(-sin 0(s).cos0(s)).soK(s)=0'(s)and t6)ds=g6'6od=6）-0is2 times the rotation index of the closed curve Let△={s,t):0≤s≤t≤L}.Consider the secant map h:△→Sdefined by T(s). 3=t h(s.t)= -T(0), (s,)=(0,L) ()-(S) a0)-a(s) otherwise Then it follows from Proposition 2.6(using Taylor's Theorem to calculate()=(s)+(t-s)a'(s)+.) that h is continuous.A more sophisticated version of the proof of Lemma 3.6 will establish(see Exercise l3)that there is a continuous function:△→R so that h(s,t）=(cosd(s,t).sin(s.t)）for all(s.t）∈△. It then follows from Lemma 3.6 that kds=9）-60=iL.)-60,0=0.-00.0+L.-0,. N N2 Without loss of generality,we assume that (is the lowest point on the curve(i.e.,the one whose y-coordinate is smallest)and,then,that a(0)is the origin and T(0)=e1,as shown in Figure 3.4.(The (0 FIGURE 3.4 last may require reversing the orientation of the curve.)Now,Ni is the angle through which the position vector of the curve tums,starting at and ending at since the curve lies in the upper half-plane,we must have N=.But N2 is likewise the angle through which the negative of the positior vector tums,so N2=NI=.With these assumptions,we see that the rotation index of the curve is 1.Allowing for the possible change in orientation,the rotation index must therefore be+1,as required. Corollary 3.7.If C is a closed curve,for any point PC there is a point C where the unit tangent vector is opposite that at p Proof.Let T(s)=(cos0(s).sin 0(s))for a el function 0:[0.L]R,as in Lemma 3.6.Say P= (s).and let )()is an integer mutiple must either 0(s1)=00+or 0(s1)=6o-x.Take Q a(s1). Recall that one of the ways of characterizing a convex function f:RR is that its graph lie on one side of each of its tangent lines.So we make the following
÷3. SOME GLOBAL RESULTS 27 Sketch of proof of Theorem 3.5. Note first that if T.s/ D cos .s/;sin .s/ , then T 0 .s/ D 0 .s/ sin .s/; cos .s/ , so .s/ D 0 .s/ and Z L 0 .s/ds D Z L 0 0 .s/ds D .L/ .0/ is 2 times the rotation index of the closed curve ˛. Let D f.s; t / W 0 s t Lg. Consider the secant map hW ! S 1 defined by h.s; t / D 8 ˆˆˆ< ˆˆˆ: T.s/; s D t T.0/; .s; t / D .0; L/ ˛.t / ˛.s/ k˛.t / ˛.s/k ; otherwise : Then it follows from Proposition 2.6 (using Taylor’s Theorem to calculate ˛.t / D ˛.s/C.t s/˛ 0 .s/C: : :) that h is continuous. A more sophisticated version of the proof of Lemma 3.6 will establish (see Exercise 13) that there is a continuous function QW ! R so that h.s; t / D cos Q .s; t /;sin Q .s; t / for all .s; t / 2 . It then follows from Lemma 3.6 that Z C ds D .L/ .0/ D Q .L; L/ Q .0; 0/ D Q .0; L/ Q .0; 0/ „ ƒ‚ . N1 C Q.L; L/ Q .0; L/ „ ƒ‚ . N2 : Without loss of generality, we assume that ˛.0/ is the lowest point on the curve (i.e., the one whose y-coordinate is smallest) and, then, that ˛.0/ is the origin and T.0/ D e1, as shown in Figure 3.4. (The FIGURE 3.4 last may require reversing the orientation of the curve.) Now, N1 is the angle through which the position vector of the curve turns, starting at 0 and ending at ; since the curve lies in the upper half-plane, we must have N1 D . But N2 is likewise the angle through which the negative of the position vector turns, so N2 D N1 D . With these assumptions, we see that the rotation index of the curve is 1. Allowing for the possible change in orientation, the rotation index must therefore be ˙1, as required. Corollary 3.7. If C is a closed curve, for any point P 2 C there is a point Q 2 C where the unit tangent vector is opposite that at P. Proof. Let T.s/ D cos .s/;sin .s/ for a C 1 function W Œ0; L ! R, as in Lemma 3.6. Say P D ˛.s0/, and let .s0/ D 0. Since .L/ .0/ is an integer multiple of 2, there must be s1 2 Œ0; L with either .s1/ D 0 C or .s1/ D 0 . Take Q D ˛.s1/. Recall that one of the ways of characterizing a convex function f W R ! R is that its graph lie on one side of each of its tangent lines. So we make the following
CHAPTER 1.CURVES Definition.The regular closed plane curve is comex if it lies on one side of its tangent line at each point. Propositi n 3.8.A simple closed regular plane curve C is convex if and only if we can choose the orientation of the curve so that K 0 everywhere. Remark.We leave it to the reader in Exercise 2 to give a non-simple closed curve for which this result is false. Proof.Assume,without loss of generality,that T()=(1,0)and the curve is oriented counterclock- wise.Using the functionconstructed in Lemma 6,the condition thatis equivalent to the condition that is a nondecreasing function with (L)=2 Suppose first that is nondecreasing and C is not convex.Then we can find a point P=a(so)on the curve and valuesso that and(lie on opposite sides of the tangent line to C at P.Then. by the maximum value theorem,there are valuess and s2 so that(s)is the greatest distance"above the tangent line and (s2)is the greatest distance"below."Consider the unit tangent vectors T(so),T(s), and T(s2).Since these vectors are either parallel or anti-parallel,some pair must be identical.Letting the respective values of s be *we have 0(s*)=0(s*)(since 0 is nondecreasing and (L)=2,the values cannot differ by a multiple of 2),and therefore (s)=0(s*)for all s [s*.s**]. This means that that portion of C between(s*)and (s*)is a line segment parallel to the tangent line of Cat P:this is a contradiction Conversely,suppose C is convex and (s1)=0(s2)for some s1<s2.By Corollary 3.7 there must be s3 with T(s3)=-T(s1)=-T(s2).Since C is convex,the tangent line at two of a(s1),a(s2),and a(53) ust be the same,say at a(s)=Pand (s)=Q.If PO does not lie entirely in C,choose RP RC.Since C is convex,the line through R perpendicular to P must intersect C in at least two points, say M and N.with N farther from PO than M.Since M lies in the interior of ANPO,all three vertices of the triangle can n ever lie on the same side of any line through M.In particular.N.P.and cannot lie on the same side of the tangent line to C at M.Thus,it must be that PcC.so (s)=0(s)=0(s2) for alls [s1.s2].Therefore,0 is nondecreasing,and we are done. Definition.A critical point of is called a vertex of the curve C A closed curve must have at least two vertices:the maximum and minimum points of.Every point of isa vertex.We conclude with the following Proposition 3.9(Four Vertex Theorem).A closed convex plane curve has at least four vertices. Proof.Suppose that C has fewer than four vertices.As we see from Figure 3.5.either must have two critical points (maximum and minimum)or k must have three critical points (maximum,minimum and inflection point).More precisely,suppose that k increases from P to and decreases from to P. Without loss of generality,let P be the origin and suppose the equation=0.Choose A so that '(s)0precisely when A.(s)0.Then'(s)(A.(s))ds>0.Let A be the vector obtained by rotating A through an angle of/2.Then,integrating by parts,we have -'s)(A-a(ds=()(A.T(ds=(
28 CHAPTER 1. CURVES Definition. The regular closed plane curve ˛ is convex if it lies on one side of its tangent line at each point. Proposition 3.8. A simple closed regular plane curve C is convex if and only if we can choose the orientation of the curve so that 0 everywhere. Remark. We leave it to the reader in Exercise 2 to give a non-simple closed curve for which this result is false. Proof. Assume, without loss of generality, that T.0/ D .1; 0/ and the curve is oriented counterclockwise. Using the function constructed in Lemma 3.6, the condition that 0 is equivalent to the condition that is a nondecreasing function with .L/ D 2. Suppose first that is nondecreasing and C is not convex. Then we can find a point P D ˛.s0/ on the curve and values s 0 1 , s 0 2 so that ˛.s0 1 / and ˛.s0 2 / lie on opposite sides of the tangent line to C at P. Then, by the maximum value theorem, there are values s1 and s2 so that ˛.s1/ is the greatest distance “above” the tangent line and ˛.s2/ is the greatest distance “below.” Consider the unit tangent vectors T.s0/, T.s1/, and T.s2/. Since these vectors are either parallel or anti-parallel, some pair must be identical. Letting the respective values of s be s and s with s < s, we have .s / D .s/ (since is nondecreasing and .L/ D 2, the values cannot differ by a multiple of 2), and therefore .s/ D .s / for all s 2 Œs ; s. This means that that portion of C between ˛.s / and ˛.s/ is a line segment parallel to the tangent line of C at P; this is a contradiction. Conversely, suppose C is convex and .s1/ D .s2/ for some s1 < s2. By Corollary 3.7 there must be s3 with T.s3/ D T.s1/ D T.s2/. Since C is convex, the tangent line at two of ˛.s1/, ˛.s2/, and ˛.s3/ must be the same, say at ˛.s / D P and ˛.s/ D Q. If PQ does not lie entirely in C, choose R 2 PQ, R . C. Since C is convex, the line through R perpendicular to ! PQ must intersect C in at least two points, say M and N , with N farther from ! PQ than M. Since M lies in the interior of 4NPQ, all three vertices of the triangle can never lie on the same side of any line through M. In particular, N , P, and Q cannot lie on the same side of the tangent line to C at M. Thus, it must be that PQ C, so .s/ D .s1/ D .s2/ for all s 2 Œs1; s2. Therefore, is nondecreasing, and we are done. Definition. A critical point of is called a vertex of the curve C. A closed curve must have at least two vertices: the maximum and minimum points of . Every point of a circle is a vertex. We conclude with the following Proposition 3.9 (Four Vertex Theorem). A closed convex plane curve has at least four vertices. Proof. Suppose that C has fewer than four vertices. As we see from Figure 3.5, either must have two critical points (maximum and minimum) or must have three critical points (maximum, minimum, and inflection point). More precisely, suppose that increases from P to Q and decreases from Q to P. Without loss of generality, let P be the origin and suppose the equation of ! PQ is A x D 0. Choose A so that 0 .s/ 0 precisely when A ˛.s/ 0. Then Z C 0 .s/.A ˛.s//ds > 0. Let AQ be the vector obtained by rotating A through an angle of =2. Then, integrating by parts, we have Z C 0 .s/.A ˛.s//ds D Z C .s/.A T.s//ds D Z C .s/.AQ N.s//ds